Gengsheng Wang
email: mwangcug@outlook.com

Ming Wang

Can Zhang
email: zhangcansx@163.com

Yubiao Zhang
email: yubiao¯zhang@yeah.net

Observable set, observability, interpolation inequality and spectral inequality for the heat equation in R n

Keywords: Characteristic of observable sets, observability inequality, Hölder-type interpolation inequality, spectral inequality, heat equation AMS subject classifications: 49J20, 49K20, 93D20

This paper studies connections among observable sets, the observability inequality, the Hölder-type interpolation inequality and the spectral inequality for the heat equation in R n . We present the characteristic of observable sets for the heat equation. In more detail, we show that a measurable set in R n satisfies the observability inequality if and only if it is γ-thick at scale L for some γ > 0 and L > 0. We also build up the equivalence among the abovementioned three inequalities. More precisely, we obtain that if a measurable set in R n satisfies one of these inequalities, then it satisfies others. Finally, we get some weak observability inequalities and weak interpolation inequalities where observations are made over a ball.

Introduction

In this paper, we consider the heat equation in the whole physical space R n :

∂ t u -u = 0 in (0, ∞) × R n , u(0, •) ∈ L 2 (R n ).
(1.1)

For this equation, we will characterize the observable sets and build up connections among several important inequalities which are introduced in the next subsection.

Notation Write C(• • • ) for a positive constant that depends on what are included in the brackets and may vary in different contexts. The same can be said about C (• • • ), C 1 (• • • ) and so on. Use V n to denote the volume of the unit ball in R n . Let B r (x), with x ∈ R n and r > 0, be the open ball in R n , centered at x and of radius r. (Simply write B r = B r (0).) Let S n-1 be the unit spherical surface in R n . Let N := {0, 1, 2, . . . }. Denote by Q the open unit cube in R n , centered at the origin. Let x + LQ, with x ∈ R n and L > 0, be the set {x + Ly : y ∈ Q}. For each measurable set D ⊂ R n , denote by |D| and D c its Lebesgue measure and complement set respectively. For any set G, we write χ G for the characteristic function of G. Given f ∈ L 2 (R n ), write f for its Fourier transform 1 . Given a measurable function f over R n , we denote by supp f the support of f , which is the set of all points (in R n ) where f does not vanish. Let {e t : t ≥ 0} be the semigroup generated by the Laplacian operator in R n . Given x = (x 1 , . . . , x n ) ∈ R n , let |x| :

= n i=1 x 2 i 1/2
and x := 1 + |x| 2 .

Thick sets and several inequalities

We start with introducing sets of γ-thickness at scale L.

Sets of γ-thickness at scale L A measurable set E ⊂ R n is said to be γ-thick at scale L for some γ > 0 and L > 0, if

E (x + LQ) ≥ γL n for each x ∈ R n . (1.2)
About sets of γ-thickness at scale L, several remarks are given in order.

(a 1 ) To our best knowledge, this definition arose from studies of the uncertainty principle. We quote it from [START_REF] Bonami | A survey on uncertainty principles related to quadratic forms[END_REF] (see Page 5 in [START_REF] Bonami | A survey on uncertainty principles related to quadratic forms[END_REF]). Before [START_REF] Bonami | A survey on uncertainty principles related to quadratic forms[END_REF], some very similar concepts were proposed.

For instance, the definition of relative dense sets was given in [START_REF] Kacnelśon | Equivalent norms in spaces of entire functions[END_REF] (see also Page 113 in [START_REF] Havin | The Uncertainty Principle in Harmonic Analysis[END_REF]); the definition of thick sets was introduced in [START_REF] Kovrijkine | Some results related to the Logvinenko-Sereda theorem[END_REF].

(a 2 ) Each set E of γ-thickness at scale L has the following properties: First, in each cube with the length L, |E| is bigger than or equals to γL n . Second, E is also a set of γ-thickness at scale 2L, but the reverse is not true. Third, we necessarily have that γ ≤ 1.

Next, we introduce an observability inequality for the equation (1.1).

The observability inequality A measurable set E ⊂ R n is said to satisfy the observability inequality for the equation (1.1), if for any T > 0 there exists a positive constant C obs = C obs (n, T, E) so that when u solves (1.1),

R n |u(T, x)| 2 dx ≤ C obs T 0 E |u(t, x)| 2 dx dt. (1.3)
When a measurable E ⊂ R n satisfies (1.3), it is called an observable set for (1.1).

Several notes on the observability inequality (1.3) are given in order.

(b 1 ) By treating the integral on the left hand side as a recovering term, and the integral on the right hand side as an observation term, we can understand the inequality (1.3) as follows: one can recover a solution of (1.1) at time T , through observing it on the set E and in the time interval (0, T ). From perspective of control theory, the inequality (1.3) is equivalent to the following null controllability: For any u 0 ∈ L 2 (R n ) and T > 0, there exists a control f ∈ L 2 ((0, T ) × R n ) driving the solution u to the controlled equation: ∂ t u -u = χ E f in (0, T ) × R n , from the initial state u 0 to the state 0 at time T .

(b 2 ) We can compare (1.3) with the observability inequality for the heat equation on a bounded physical domain. Let Ω be a bounded C 2 (or Lipschitz and locally star-shaped, see [START_REF] Apraiz | Observability inequalities and measurable sets[END_REF]) domain in R n . Consider the equation:

     ∂ t u -u = 0 in (0, ∞) × Ω, u = 0 on (0, ∞) × ∂Ω, u(0, •) ∈ L 2 (Ω).
(1.4)

We say that a measurable set ω ⊂ Ω satisfies the observability inequality for (1.4), if given T > 0, there is a constant C(n, T, ω, Ω) so that when u solves (1.4), When a measurable set ω ⊂ Ω satisfies (1.5), it is called an observable set for (1.4).

The inequality (1.5) has been widely studied. See [START_REF] Fursikov | Controllability of evolution equations[END_REF][START_REF] Lebeau | Contrôle exact de léquation de la chaleur[END_REF][START_REF] Lopez | Null controllability of the heat equation as singular limit of the exact controllability of dissipative wave equations[END_REF] for the case where ω is open; [START_REF] Apraiz | Null-control and measurable sets[END_REF][START_REF] Apraiz | Observability inequalities and measurable sets[END_REF][START_REF] Escauriaza | Observation from measurable sets for parabolic analytic evolutions and applications[END_REF] for the case when ω is measurable.

(b 3 ) When Ω is an unbounded domain and ω is a bounded and open subset of Ω, the inequality (1.5) may not be true. This was showed in [START_REF] Micu | On the lack of null-controllability of the heat equation on the half-line[END_REF] for the heat equation in the physical domain R + . Similar results have been obtained for higher dimension cases in [START_REF] Micu | On the lack of null-controllability of the heat equation on the half space[END_REF]. For the heat equation in an unbounded domain, [START_REF] Miller | On the null-controllability of the heat equation in unbounded domains[END_REF] imposed a condition, in terms of the Gaussian kernel, on the set ω so that the observability inequality does not hold. In particular, [START_REF] Miller | On the null-controllability of the heat equation in unbounded domains[END_REF] showed that the observability inequality fails when Ω is unbounded and |ω| < ∞. Notice that any set E ⊂ R n of finite measure does not have the characteristic on observable sets of (1.1). This characteristic is indeed the γ-thickness at scale L for some γ > 0 and L > 0. (See Theorem 1.1 of this paper.)

About works on sufficient conditions of observable sets for heat equations in unbounded domains, we would like to mention the work [START_REF] Cabanillas | Null controllability in unbounded domains for the semilinear heat equation with nonlinearities involving gradient terms[END_REF]. It showed that, for some parabolic equations in an unbounded domain Ω ⊂ R n , the observability inequality holds when observations are made over a subset E ⊂ Ω, with Ω\E bounded. For other similar results, we refer the reader to [START_REF] Miller | Unique continuation estimates for the Laplacian and the heat equation on noncompact manifolds[END_REF]. When Ω = R n , such a set E has the characteristic on observable sets of (1.1) mentioned before.

(b 4 ) An interesting phenomenon is that some potentials (growing at infinity) in heat equations may change the above-mentioned characteristic on observable sets for the heat equations with potentials. In [START_REF] Miller | Unique continuation estimates for sums of semiclassical eigenfunctions and nullcontrollability from cones[END_REF][START_REF] Duyckaerts | Resolvent conditions for the control of parabolic equations[END_REF], the authors realized the following fact: Let A = + V , where V (x) := -|x| 2k , x ∈ R n , with 2 ≤ k ∈ N. Write e tA t≥0 for the semigroup generated by the operator A. Let r 0 ≥ 0 and let Θ 0 be an open subset of S

n-1 . Let Γ = {x ∈ R n : |x| ≥ r 0 , x/|x| ∈ Θ 0 }. Then there is C(n, T, Θ 0 , r 0 , k) so that R n e T A u 0 2 dx ≤ C(n, T, Θ 0 , r 0 , k) T 0 Γ e tA u 0 2 dx dt for all u 0 ∈ L 2 (R n ). (1.6)
The cone Γ does not have the characteristic on observable sets mentioned before, but still holds the observability inequality (1.6). The main reason is as follows: The unbounded potential V changes the behaviour of the solution of the pure heat equation (1.1). This plays an important role in the proof of (1.6) (see [START_REF] Miller | Unique continuation estimates for sums of semiclassical eigenfunctions and nullcontrollability from cones[END_REF][START_REF] Duyckaerts | Resolvent conditions for the control of parabolic equations[END_REF]). It should be pointed out that when

V (x) = -|x| 2 , x ∈ R n
(which means that the potential grows more slowly at infinity), (1.6) does not hold for the above cone. We refer the readers to [START_REF] Miller | Unique continuation estimates for sums of semiclassical eigenfunctions and nullcontrollability from cones[END_REF][START_REF] Duyckaerts | Resolvent conditions for the control of parabolic equations[END_REF] for more details on this issue. Besides, we also would like to mention [START_REF] Barbu | Exact null internal controllability for the heat equation on unbounded convex domains[END_REF] for this subject.

An interesting question now arises: How do potentials influence characteristics of observable sets?

We wish to answer this question in our future studies.

We then introduce an interpolation inequality for the equation (1.1).

The Hölder-type interpolation inequality A measurable set E ⊂ R n is said to satisfy the Hölder-type interpolation inequality for the heat equation (1.1), if for any θ ∈ (0, 1), there is C Hold = C Hold (n, E, θ) so that for each T > 0 and each solution u to the equation (1.1),

R n |u(T, x)| 2 dx ≤ e C Hold (1+ 1 T ) E |u(T, x)| 2 dx θ R n |u(0, x)| 2 dx 1-θ . (1.7)
Several remarks on the Hölder-type interpolation inequality (1.7) are given in order.

(c 1 ) The above Hölder-type interpolation inequality is equivalent to what follows: There is θ = θ(n, E) ∈ (0, 1) and C Hold = C Hold (n, E) so that (1.7) holds for all T > 0 and solutions u to (1.1). This can be verified by the similar way used in the proof of [START_REF] Phung | Impulse output rapid stabilization for heat equations[END_REF]Theorem 2.1].

(c 2 ) The inequality (1.7) is a kind of quantitative unique continuation for the heat equation (1.1). It provides a Hölder-type propagation of smallness for solutions of the heat equation (1.1).

In fact, if

E |u(T, x)| 2 dx = δ, then we derive from (1.7) that R n |u(T, x)| 2 dx is bounded by Cδ θ for some constant C > 0. Consequently, u(T, •) = 0 over R n provided that it is zero over E.
(c 3 ) From perspective of control theory, the inequality (1.7) implies the approximate null controllability with cost for impulse controlled heat equations, i.e., given T > τ > 0, ε > 0, there is

C = C(n, E, T, τ, ε) such that for each u 0 ∈ L 2 (R n ), there is f ∈ L 2 (R n ) so that f L 2 (R n ) ≤ C u 0 L 2 (R n ) and u(T, •) L 2 (R n ) ≤ ε u 0 L 2 (R n ) ,
where u is the solution to the impulse controlled equation:

∂ t u -∆u = δ {t=τ } χ E f in (0, T ) × R n , with the initial condition u(0, x) = u 0 (x), x ∈ R n . (See [47, Theorem 3.1].) (c 4 )
The Hölder-type interpolation inequality (1.7) can imply the observability inequality (1.3). Moreover, it leads to the following stronger version of (1.3):

R n |u(T, x)| 2 dx ≤ C obs F E |u(t, x)| 2 dx dt, with C obs = C obs (n, T, E, F ), (1.8)
where F ⊂ (0, T ) is a subset of positive measure. This will be presented in Lemma 2.4. We derive (1.8) from (1.7), through using the telescoping series method developed in [START_REF] Phung | An observability estimate for parabolic equations from a measurable set in time and its applications[END_REF] (see also [START_REF] Phung | Bang-bang property for time optimal control of semilinear heat equation[END_REF][START_REF] Apraiz | Observability inequalities and measurable sets[END_REF]) for heat equations in bounded domains.

(c 5 ) We can compare (1.7) with an interpolation inequality for the heat equation (1.4). A measurable set ω ⊂ Ω is said to satisfy the Hölder-type interpolation inequality for the equation (1.4), if for any θ ∈ (0, 1), there is C = C(n, Ω, ω, θ) so that for any T > 0 and any solution u to (1.4),

Ω |u(T, x)| 2 dx ≤ e C(1+ 1
T ) ω |u(T, x)| 2 dx θ Ω |u(0, x)| 2 dx 1-θ
.

(1.9)

In [START_REF] Phung | Quantitative unique continuation for the semilinear heat equation in a convex domain[END_REF], the authors proved that any open and nonempty subset ω ⊂ Ω satisfies the Höldertype interpolation inequality (1.9) for heat equations with potentials in bounded and convex domains. The frequency function method used in [START_REF] Phung | Quantitative unique continuation for the semilinear heat equation in a convex domain[END_REF] was partially borrowed from [START_REF] Escauriaza | Doubling properties of caloric functions[END_REF].

In [START_REF] Apraiz | Observability inequalities and measurable sets[END_REF], the authors proved that any subset ω of positive measure satisfies the Hölder-type interpolation inequality (1.9) for the heat equation (1.4) where Ω is a bounded Lipschitz and locally star-shaped domain in R n . More about this inequality for heat equations in bounded domains, we referee the readers to [START_REF] Phung | An observability estimate for parabolic equations from a measurable set in time and its applications[END_REF][START_REF] Phung | Impulse output rapid stabilization for heat equations[END_REF][START_REF] Phung | Bang-bang property for time optimal control of semilinear heat equation[END_REF].

Finally, we will introduce a spectral inequality for some functions in L 2 (R n ).

The spectral inequality A measurable set E ⊂ R n is said to satisfy the spectral inequality, if there is a positive constant C spec = C spec (n, E) so that for each N > 0,

R n |f (x)| 2 dx ≤ e Cspec(1+N ) E |f (x)| 2 dx for all f ∈ L 2 (R n ) with supp f ⊂ B N . (1.10)
Several notes on the spectral inequality (1.10) are given in order.

(d 1 ) Recall the Lebeau-Robbiano spectral inequality (see [START_REF] Lebeau | Contrôle exact de léquation de la chaleur[END_REF][START_REF] Lebeau | Null-controllability of a system of linear thermoelasticity[END_REF]): Let Ω be a bounded smooth domain in R n and let ω be a nonempty open subset of Ω. Write Ω for the Laplacian operator on L 2 (Ω) with Domain(

Ω ) = H 1 0 (Ω) H 2 (Ω). Let {λ j } j≥1 (with 0 < λ 1 < λ 2 ≤ • • •
) be the eigenvalues of -Ω and let {φ j } j≥1 be the corresponding eigenfunctions. Then there is a positive constant C(n, Ω, ω) so that for each λ > 0, .11) This inequality was extended to the case where Ω is a bounded C 2 domain via a simpler way in [START_REF] Lü | A lower bound on local energy of partial sum of eigenfunctions for Laplace-Beltrami operators[END_REF]. Then it was extended to the case that Ω is a bounded Lipschitz and locally star-shaped domain; ω is a subset of positive measure so that ω ⊂ B R (x 0 ) ⊂ B 4R (x 0 ) ⊂ Ω for some R > 0 and x 0 ∈ Ω; and

Ω |f (x)| 2 dx ≤ e C(n,Ω,ω)(1+ √ λ) ω |f (x)| 2 dx for all f ∈ span{φ j : λ j < λ}. ( 1
C(n, Ω, ω) = C(n, Ω, |ω|/|B R |) (see [2, Theorem 5 and Theorem 3]).
By our understanding, the inequality (1.10) is comparable to (1.11) from two perspectives as follows: First, the inequality (1.10) is satisfied by functions in the subspace:

E N f ∈ L 2 (R n ) : supp f ⊂ B N with N > 0,
while the inequality (1.11) is satisfied by functions in the subspace:

F λ λj <λ f = a j φ j ∈ L 2 (Ω) : {a j } j≥1 ⊂ R with λ > 0.
From the definition of the spectral projection in the abstract setting given in [START_REF] Simon | Methods of Modern Mathematical Physics I: Functional Analysis[END_REF] (see Pages 262-263 in [START_REF] Simon | Methods of Modern Mathematical Physics I: Functional Analysis[END_REF]), we can define two spectral projections:

χ [0,N 2 ) (-∆) and χ [0,λ) (-∆ Ω ) on L 2 (R n ) and L 2 (Ω)
, respectively. Then after some computations, we find that E N and F λ are the ranges of χ [0,N 2 ) (-∆) and χ [0,λ) (-∆ Ω ), respectively. Second, the square root of the integral of χ [0,N 2 ) over R is N which corresponds to the N in (1.10), while the square root of the integral of

χ [0,λ) over R is √ λ which corresponds to the √ λ in (1.11).
(d 2 ) Though the inequality (1.10) was first named as the spectral inequality in [START_REF] Rousseau | Null-controllability of the Kolmogorov equation in the whole phase space[END_REF] (to our best knowledge), it has been extensively studied for long time. (See, for instance, [START_REF] Bonami | A survey on uncertainty principles related to quadratic forms[END_REF][START_REF] Havin | The Uncertainty Principle in Harmonic Analysis[END_REF][START_REF] Kacnelśon | Equivalent norms in spaces of entire functions[END_REF][START_REF] Kovrijkine | Some results related to the Logvinenko-Sereda theorem[END_REF][START_REF] Nazarov | Local estimates for exponential polynomials and their applications to inequalities of the uncertainty principle type[END_REF][START_REF] Logvinenko | Equivalent norms in spaces of entire functions of exponential type[END_REF][START_REF] Paneyah | Some estimates of functions of exponential type and apriori estimates for gerneral dikerential operators[END_REF][START_REF] Paneyah | On some theorems of Paley-Wiener type[END_REF][START_REF] Reznikov | Sharp constants in the PaneyahCLogvinenkoCSereda theorem[END_REF].) In [START_REF] Kovrijkine | Some results related to the Logvinenko-Sereda theorem[END_REF], the author announced that if E is γ-thick at scale L for some γ > 0 and L > 0, then E satisfies the spectral inequality (1.10), and further proved this announcement for the case when n = 1. Earlier, the authors of [START_REF] Logvinenko | Equivalent norms in spaces of entire functions of exponential type[END_REF] (see also [START_REF] Havin | The Uncertainty Principle in Harmonic Analysis[END_REF]) proved that E is γ-thick at scale L for some γ > 0 and L > 0 if and only if E satisfies the following inequality: For each N > 0, there is a positive constant C(n, E, N ) so that

R n |f (x)| 2 dx ≤ C(n, E, N ) E |f (x)| 2 dx for each f ∈ L 2 (R n ) with supp f ⊂ B N . (1.12)
This result is often referred as the Logvinenko-Sereda theorem. Before [START_REF] Logvinenko | Equivalent norms in spaces of entire functions of exponential type[END_REF], the above equivalence was proved by B. P. Paneyah for the case that n = 1 (see [START_REF] Paneyah | On some theorems of Paley-Wiener type[END_REF][START_REF] Paneyah | Some estimates of functions of exponential type and apriori estimates for gerneral dikerential operators[END_REF][START_REF] Havin | The Uncertainty Principle in Harmonic Analysis[END_REF]). In [START_REF] Kacnelśon | Equivalent norms in spaces of entire functions[END_REF], the author claimed (1.12), with C(n, E, N ) = e Cspec(1+N ) , and proved this claim for the case when n = 1.

In the proof of our main theorem of this paper, the expression C(n, E, N ) = e Cspec(1+N ) will play an important role. From this point of view, (1.12) is weaker than the spectral inequality (1.10).

(d 3 ) The inequality (1.12) is also important. It is closely related to the uncertainty principle (which is an extensive research topic in the theory of harmonic analysis and says roughly that a nonzero function and its Fourier transform cannot be both sharply localized, see [START_REF] Folland | The uncertainty principle: a mathematical survey[END_REF]).

In fact, a measurable set E satisfies the inequality (1.12) if and only if it satisfies the following uncertainty principle:

R n |f (x)| 2 dx ≤ C (n, E, N ) E |f (x)| 2 dx + B c N | f (ξ)| 2 dξ for all f ∈ L 2 (R n ).
We refer the interested readers to [START_REF] Bonami | A survey on uncertainty principles related to quadratic forms[END_REF][START_REF] Havin | The Uncertainty Principle in Harmonic Analysis[END_REF][START_REF] Jaming | Nazarov's uncertainty principles in higher dimension[END_REF][START_REF] Nazarov | Local estimates for exponential polynomials and their applications to inequalities of the uncertainty principle type[END_REF] for the proof of the above result, as well as more general uncertainty principle, where E and B c N are replaced by more general sets. It deserves mentioning what follows: The uncertainty principle can help us to get the exact controllability for the Schrödinger equation with controls located outside of two balls and at two time points. This was realized in [START_REF] Wang | Observability and unique continuation inequalities for the Schrödinger equation[END_REF]. (See [START_REF] Huang | Uncertainty principle, minimal escape velocities and observability inequalities for schrö dinger equations[END_REF] for more general cases.) (d 4 ) By using a global Carleman estimate, the authors in [START_REF] Rousseau | Null-controllability of the Kolmogorov equation in the whole phase space[END_REF] proved the spectral inequality (1.10) for such an open subset E that satisfies the property: there exists δ > 0 and r > 0 so that

∀y ∈ R n , ∃ y ∈ E such that B r (y ) ⊂ E and |y -y | ≤ δ. (1.13)
It is clear that a set with the above property (1.13) is a set of γ-thick at scale L for some γ > 0 and L > 02 .

(d 5 ) With the aid of the spectral inequality (1.10), one can use the same strategy given in [START_REF] Lebeau | Contrôle exact de léquation de la chaleur[END_REF] to derive the null controllability described in the note (b 1 ).

Aim, motivation and main result

Aim According to the note (d 2 ) in the previous subsection, the characteristic of a measurable set holding the spectral inequality (1.10) is the γ-thickness at scale L for some γ > 0 and L > 0.

Natural and interesting questions are as follows: What is the characteristic of observable sets for (1.1)? How to characterize a measurable set E satisfying the Hölder-type interpolation inequality (1.7)? What are the connections among inequalities (1.3), (1.7) and (1.10)? The aim of this paper is to answer the above questions.

Motivation The motivations of our studies are given in order.

(i) The first motivation arises from two papers [START_REF] Bardos | Sharp sufficient conditions for the observation, control, and stabilization of waves from the boundary[END_REF] and [START_REF] Apraiz | Null-control and measurable sets[END_REF]. In [START_REF] Bardos | Sharp sufficient conditions for the observation, control, and stabilization of waves from the boundary[END_REF], the authors gave, for the wave equation in a bounded physical domain Ω ⊂ R n , a sufficient and almost necessary condition to ensure an open subset Γ ⊂ ∂Ω to be observable, (i.e., Γ satisfies the observability inequality for the wave equation with observations on Γ). This condition is exactly the well known Geometric Control Condition (GCC for short) 3 . Thus, we can say that the GCC condition is a characteristic of observable open sets on ∂Ω, though this condition is not strictly necessary (see [START_REF] Rousseau | Geometric control condition for the wave equation with a time-dependent observation domain[END_REF]). The authors in [START_REF] Apraiz | Null-control and measurable sets[END_REF] presented a sufficient and necessary condition to ensure a measurable subset ω ⊂ Ω satisfying (1.5). This condition is as: |ω| > 0. Hence, the characteristic of observable sets for the equation (1.4) is as:

|ω| > 0.
Analogically, it should be very important to characterize observable sets for the heat equation (1.1). However, it seems for us that there is no any such result in the past publications. These motivate us to find the characteristic of observable sets for the equation (1.1).

(ii) For the heat equation (1.4), the observability inequality (1.5), the Hölder-type interpolation inequality (1.9) and the spectral inequality (1.11) are equivalent. More precisely, we have that if ω ⊂ Ω is a measurable set, then |ω| > 0 ⇐⇒ ω satisfies (1.11) ⇐⇒ ω satisfies (1.9) ⇐⇒ ω satisfies (1.5).

(1.14)

The proof of (1.14) was hidden in the paper [START_REF] Apraiz | Observability inequalities and measurable sets[END_REF]. (See Theorem 5, Theorem 6, as well as its proof, Theorem 1, as well as its proof, in [START_REF] Apraiz | Observability inequalities and measurable sets[END_REF].) However, for the heat equation (1.1), the equivalence among these three inequalities has not been touched upon. These motivate us to build up the equivalence among inequalities (1.3), (1.7) and (1.10).

It deserves mentioning that for heat equations with lower terms in bounded physical domains, we do not know if (1.14) is true.

Main Result

The main result of the paper is the next Theorem 1.1.

Theorem 1.1. Let E ⊂ R n be a measurable subset. Then the following statements are equivalent:

(i)
The set E is γ-thick at scale L for some γ > 0 and L > 0.

(ii) The set E satisfies the spectral inequality (1.10).

(iii) The set E satisfies the Hölder-type interpolation inequality (1.7).

(iv) The set E satisfies the observability inequality (1.3).

Several remarks about Theorem 1.1 are given in order.

(e 1 ) The equivalence of statements (i) and (iv) in Theorem 1.1 tells us: the characteristic of observable sets for the heat equation (1.1) is the γ-thickness at scale L for some γ > 0 and L > 0. This seems to be new for us.

(e 2 ) The equivalence among statements (ii), (iii) and (iv) in Theorem 1.1 presents closed connections of the three inequalities. This seems also to be new for us.

(e 3 ) We find the following way to prove Theorem 1.1: (i) ⇒ (ii) ⇒ (iii) ⇒ (iv) ⇒ (i). We prove (i) ⇒ (ii) by some ideas from [START_REF] Kovrijkine | Some results related to the Logvinenko-Sereda theorem[END_REF]. Indeed, this result was announced in [START_REF] Kovrijkine | Some results related to the Logvinenko-Sereda theorem[END_REF] and then proved for the case that n = 1 in the same reference. We prove (ii) ⇒ (iii) ⇒ (iv), though using some ideas and techniques from [START_REF] Apraiz | Observability inequalities and measurable sets[END_REF][START_REF] Phung | An observability estimate for parabolic equations from a measurable set in time and its applications[END_REF]. Finally, we show (iv) ⇒ (i) via the structure of a special solution to the equation (1.1).

(e 4 ) We noticed that four days after we put our current work in arXiv, the paper [START_REF] Egidi | Sharp geometric condition for null-controllability of the heat equation on R d and consistent estimates on the control cost[END_REF] appeared in arXiv. In [START_REF] Egidi | Sharp geometric condition for null-controllability of the heat equation on R d and consistent estimates on the control cost[END_REF], the authors independently got the equivalence (i) and (iv) in Theorem 1.1.

Extensions to bounded observable sets

From Theorem 1.1, we see that in order to have (1.3) or (1.7), the set E has to be γ-thick at scale L for some γ > 0 and L > 0. Then a natural and interesting question arises: What are possible substitutions of (1.3) or (1.7), when E is replaced by a ball in R n ? (It deserves to mention that any ball in R n does not satisfy the thick condition (1.2).) We try to find the substitutes from two perspectives as follows:

(i) We try to add weights on the left hand side and ask ourself if the following inequalities hold for all solutions of (1.1):

R n χ B r (x)|u(T, x)| 2 dx ≤ C(T, r , r, n) T 0 Br |u(t, x)| 2 dx dt (1.15) and R n ρ(x)|u(T, x)| 2 dx ≤ C(T, ρ, r, n) T 0 Br |u(t, x)| 2 dx dt, (1.16)
where ρ(x) = x -ν or e -|x| . On one hand, we proved that (1.15) is true when r < r, while (1.15) is not true when r > r (see Theorem 3.2 in Subsection 3.2). Unfortunately, we do not know if (1.15) holds when r = r. On the other hand, we showed that (1.16) fails for all r > 0 (see Corollary 3.2 in Subsection 3.2).

(ii) We try to find a class of initial data so that (1.3) (where E is replaced by B r ) holds for all solutions of (1.1) with initial data in this class. We have obtained some results on this issue (see Theorem 3.3 in Subsection 3.2). More interesting question is as: what is the biggest class of initial data so that (1.3) (where E is replaced by B r ) holds for all solutions of the heat equation (1.4) with initial data in this class? Unfortunately, we are not able to answer it.

We now turn to possible substitutions of (1.7) where E is replaced by B 1 . We expect to find b(ε) > 0 for each ε ∈ (0, 1) so that for any T > 0, there is C(n, T ) > 0 such that when u solves (1.1),

R n |u(T, x)| 2 dx ≤ C(n, T ) ε R n |u(0, x)| 2 dx + b(ε) B1 |u(T, x)| 2 dx . (1.17)
Let us explain why (1.17) deserves to be expected. Reason One. Let θ ∈ (0, 1) and T > 0.

Then the next two inequalities are equivalent. The first inequality is as: there is C(n, T, θ) so that when u solves (1.1),

R n |u(T, x)| 2 dx ≤ C(n, T, θ) B1 |u(T, x)| 2 dx θ R n |u(0, x)| 2 dx 1-θ , (1.18)
while the second inequality is as: there is C(n, T, θ) > 0 so that for any ε ∈ (0, 1) and any solution u to (1.1),

R n |u(T, x)| 2 dx ≤ C(n, T, θ) ε R n |u(0, x)| 2 dx + ε -1-θ θ B1 |u(T, x)| 2 dx . (1.19)
However, (1.19) is not true, for otherwise, we can use the same method developed in [START_REF] Phung | An observability estimate for parabolic equations from a measurable set in time and its applications[END_REF] (see also [START_REF] Apraiz | Observability inequalities and measurable sets[END_REF]) to derive (1.3) (where E is replaced by B 1 ) which contradicts the equivalence of (i) and (iv) in Theorem 1.1. Thus, b(ε) in (1.17) cannot grow like a polynomial of ε. But it seems not to be hopeless for us to find some kind of b(ε) so that (1.17) holds. Reason Two.

The space-like strong unique continuation of the heat equation (1.1) (see [START_REF] Escauriaza | Doubling properties of caloric functions[END_REF]) yields that if u(T, •) = 0 on the ball B 1 , then u(T, •) = 0 over R n . The inequality (1.17) is a quantitative version of the aforementioned unique continuation.

Though we have not found any b(ε) so that (1.17) is true, we obtained some b(ε) so that (1.17) holds for all solutions to (1.1) with initial data having some slight decay (see Theorem 3.1 in Subsection 3.1).

Finally, We would like to mention what follows: With the aid of an abstract lemma (i.e., Lemma 5.1 in [START_REF] Wang | Observability and unique continuation inequalities for the Schrödinger equation[END_REF]), each of extended inequalities mentioned above corresponds to a kind of controllability for the heat equation (1.1). We are not going to repeat the details on this issue in the current paper.

Plan of the paper

The paper is organized as follows: In Section 2, we prove Theorem 1.1. In Section 3, we present several weak observability inequalities and weak interpolation inequalities, where observations are made in a ball of R n .

2 Proof of Theorem 1.1

We are going to prove Theorem 1.1 in the following way:

(i) ⇒ (ii) ⇒ (iii) ⇒ (iv) ⇒ (i).
The above steps are based on several lemmas: Lemmas 2.1, 2.3, 2.4 and 2.5. We begin with Lemma 2.1 connecting the spectral inequality with sets of γ-thickness at scale L.

Lemma 2.1. Suppose that a measurable set E ⊂ R n is γ-thick at scale L for some γ > 0 and L > 0. Then E satisfies the spectral inequality (1.10), with

C spec (n, E) = C(1 + L) 1 + ln 1 γ for some C = C(n).
Remark 2.1. The manner that the constant e Cspec(n,E)(1+N ) (in (1.10)) depends on N is comparable with the manner that the constant e C √ λ in (1.11) depends on λ. (This has been explained in the remark (d 1 ) in Subsection 1.1.) The latter one played an important role in the proof of the Hölder-type interpolation inequality (1.9) for the heat equation (1.4) (see [START_REF] Apraiz | Observability inequalities and measurable sets[END_REF]). Analogically, the previous one will play an important role in the proof of Theorem 1.1.

Remark 2.2. In [START_REF] Kovrijkine | Some results related to the Logvinenko-Sereda theorem[END_REF], the author announced the result in Lemma 2.1 and proved it for the case when n = 1. For the completeness of the paper, we give a detailed proof for Lemma 2.1, based on some ideas and techniques in [START_REF] Kovrijkine | Some results related to the Logvinenko-Sereda theorem[END_REF].

To show Lemma 2.1, we need the following result on analytic functions: We are now in the position to prove Lemma 2.1.

Proof of Lemma 2.1. We only need to prove this lemma for the case when L = 1. In fact, suppose that this is done. Let E be γ-thick at scale L > 0. Define a new set:

L -1 E := L -1 x : x ∈ E . One can easily check that L -1 E is γ-thick at scale 1. Given N > 0 and f ∈ L 2 (R n ) with supp f ⊂ B N , let g(x) := f (Lx), x ∈ R n .
One can directly check that

g(ξ) = L -n f (L -1 ξ), ξ ∈ R n ; supp g ⊂ B LN .
From these, we can apply Lemma 2.1 (with L = 1) to the set L -1 E and the function g to find

C = C(n) so that R n |g(x)| 2 dx ≤ e 2C(1+ln 1 γ )(1+LN ) L -1 E |g(x)| 2 dx ≤ e 2C(1+L)(1+ln 1 γ )(1+N ) L -1 E |g(x)| 2 dx. (2.1)
Meanwhile, by changing variable x → Lx, we deduce that

R n |f (x)| 2 dx = L -n R n |g(x)| 2 dx; E |f (x)| 2 dx = L -n L -1 E |g(x)| 2 dx. (2.2)
Hence, from (2.1) and (2.2), we find that

R n |f (x)| 2 dx ≤ e Cspec(n,E)(1+N ) E |f (x)| 2 dx, with C spec (n, E) = 2C(1 + L) 1 + ln 1 γ .
This proves the lemma for the general case that L > 0.

We now show Lemma 2.1 for the case when L = 1 by several steps. First of all, we arbitrarily fix N > 0 and f ∈ L 2 (R n ) with supp f ⊂ B N . Without loss of generality, we can assume that f = 0.

Step 1. Bad and good cubes.

For each multi-index j = (j 1 , j 2 , • • • , j n ) ∈ Z n , let Q(j) := {x = (x 1 , . . . , x n ) ∈ R n : |x i -j i | < 1/2 for all i = 1, 2, • • • , n} .
It is clear that

Q(j) Q(k) = Ø for all j = k ∈ Z n ; R n = j∈Z n Q(j),
where Q(j) denotes the closure of Q(j). From these, we have that

R n |f (x)| 2 dx = j∈Z n Q(j) |f (x)| 2 dx. (2.3)
We will divide {Q(j) : j ∈ Z n } into two disjoint parts whose elements are respectively called "good cubes" and "bad cubes". And then we compare

R n |f | 2 with Q(j) is bad |f | 2 and Q(j) is good |f | 2
, respectively. First, we define the function:

h(s) := s n (s -1) -n -1, s ∈ [2, +∞).
It is a continuous and strictly decreasing function satisfying that

h(2) ≥ 1, lim s→+∞ h(s) = 0.
Thus we can take A 0 as the unique point in [2, +∞) so that h(A 0 ) = 1/2. Clearly, A 0 depends only on n, i.e.,

A 0 = A 0 (n). Given j ∈ Z n , Q(j) is said to be a good cube, if for each β ∈ N n , Q(j) |∂ β x f (x)| 2 dx ≤ A |β| 0 N 2|β| Q(j) |f (x)| 2 dx. (2.4)
When Q(j) is not a good cube, it is called as a bad cube. Thus, when Q(j) is a bad cube, there is β ∈ N n , with |β| > 0, so that

Q(j) |∂ β x f (x)| 2 dx > A |β| 0 N 2|β| Q(j) |f (x)| 2 dx. (2.5)
Using the Plancherel theorem and the assumption that supp f ⊂ B N (0), we obtain that for each

β ∈ N n , R n |∂ β x f (x)| 2 dx = R n ∂ β x f (ξ) 2 dξ = R n (iξ) β f (ξ) 2 dξ = |ξ|≤N ξ β f (ξ) 2 dξ ≤ N 2|β| |ξ|≤N | f (ξ)| 2 dξ = N 2|β| R n |f (x)| 2 . (2.6)
Meanwhile, it follows by (2.5) that when Q(j) is a bad cube,

Q(j) |f (x)| 2 dx ≤ β∈N n ,|β|>0 A -|β| 0 N -2|β| Q(j) |∂ β x f (x)| 2 dx. (2.7)
Since Q(j), j ∈ Z n , are disjoint, by taking the sum in (2.7) for all bad cubes, we find that

Q(j) is bad |f (x)| 2 dx ≤ β∈N n ,|β|>0 A -|β| 0 N -2|β| Q(j) is bad |∂ β x f (x)| 2 dx ≤ β∈N n ,|β|>0 A -|β| 0 N -2|β| R n |∂ β x f (x)| 2 dx. (2.8)
From (2.6) and (2.8), we have that

Q(j) is bad |f (x)| 2 dx ≤ β∈N n ,|β|>0 A -|β| 0 R n |f (x)| 2 dx = A n 0 (A 0 -1) -n -1 R n |f (x)| 2 dx. (2.9) Since h(A 0 ) = 1 2 , it follows from (2.9) that Q(j) is bad |f (x)| 2 dx ≤ 1 2 R n |f (x)| 2 dx. (2.10)
By (2.3) and (2.10), we obtain that

Q(j) is good |f (x)| 2 dx ≥ 1 2 R n |f (x)| 2 dx. (2.11)
Step 2. Properties on good cubes. Arbitrarily fix a good cube Q(j). We will prove some properties related to Q(j). First of all, we claim that there is C 0 (n) > 0 so that

∂ β x f L ∞ (Q(j)) ≤ C 0 (n)(1 + N ) n A 0 N |β| f L 2 (Q(j)) for all β ∈ N n .
(2.12)

In fact, according to (2.4), there is C 1 (n) > 0 so that

∂ β x f W n,2 (Q(j)) = µ∈N n ,|µ|≤n Q(j) |∂ β+µ x f (x)| 2 dx 1/2 ≤ µ∈N n ,|µ|≤n A |β+µ| 2 0 N |β+µ| f L 2 (Q(j)) ≤ C 1 (n)(1 + N ) n A 0 N |β| f L 2 (Q(j)) for all β ∈ N n . (2.13)
Meanwhile, because Q(j) satisfies the cone condition, we can apply the Sobolev embedding theorem

W n,2 (Q(j)) → L ∞ (Q(j)) to find C 2 (n) > 0 so that ϕ L ∞ (Q(j)) ≤ C 2 (n) ϕ W n,2 (Q(j)) for all ϕ ∈ W n,2 (Q(j)).
This, along with (2.13), leads to (2.12). Next, we let y ∈ Q(j) satisfy that

f L ∞ (Q(j)) = |f (y)|. ( 2
|E Q(j)| = ∞ 0 dr |x-y|=r χ E Q(j) (x)dσ = √ n 0 dr |x-y|=r χ E Q(j) (x)dσ = √ n 1 0 dr |x-y|= √ nr χ E Q(j) (x)dσ. (2.15)
In (2.15), we change the variable:

x = y + √ nrw with w ∈ S n-1 ,
and then obtain that

|E Q(j)| = √ n 1 0 ( √ nr) n-1 dr S n-1 χ E Q(j) (y + √ nrw)dσ ≤ n n/2 1 0 dr S n-1 χ E Q(j) (y + √ nrw)dσ. (2.16)
For each w ∈ S n-1 , let

I w r ∈ [0, 1] : y + √ nrw ∈ E Q(j) . (2.17)
Since

|S n-1 | = 2π n 2 Γ( n 2 )
, where Γ(•) is the Gamma function, it follows from (1.2) and (2.16) that

|I ω0 | ≥ |E Q(j)| n n/2 |S n-1 | ≥ Γ( n 2 ) 2(nπ) n/2 γ for some w 0 ∈ S n-1 .
(2.18)

Then we define a function φ(•) over [0, 1] by

φ(t) = f (y + √ ntw 0 ) f L 2 (Q(j)) , t ∈ [0, 1]. (2.19) (Since f ∈ L 2 (R n ) satisfies that supp f ⊂ B N , we have that f is analytic over R n . Consequently, f L 2 (Q(j)) = 0 because we assumed that f = 0 over R n .)
We claim that φ(t) can be extended to an entire function in the complex plane. In fact, by (2.19), one can directly check that [START_REF] Havin | The Uncertainty Principle in Harmonic Analysis[END_REF]) and (2.12), we see that

φ (k) (0) ≤ n By (2.
φ (k) (0) ≤ C 0 (n)(1 + N ) n n 3 2 A 0 N k for all k ≥ 0.
(2.21) From (2.21), we find that

φ(t) = φ(0) + φ (0)t + • • • + φ (k) (0) k! t k + • • • , t ∈ [0, 1], (2.22)
and that the series in (2.22), with t being replaced by any z ∈ C, is convergent. Thus, the above claim is true. From now on, we will use φ(z) to denote the extension of φ(t) over C.

Step 3. Recovery of the L 2 (R n ) norm. We will finish our proof in this step. Applying Lemma 2.2, where I = [0, 1], Ê = I w0 (defined by (2.17) and (2.18)) and Φ = φ, and then using (2.18), we can find

C 3 = C 3 (n) so that sup t∈[0,1] |φ(t)| ≤ (C/|I w0 |) ln M ln 2 sup t∈Iw 0 |φ(t)| ≤ 2C(nπ) n/2 γΓ( n 2 ) ln M ln 2 sup t∈Iw 0 |φ(t)| ≤ M C3(1+ln 1 γ ) sup t∈Iw 0 |φ(t)|, (2.23)
where

M = max |z|≤4 |φ(z)|. (2.24)
Two facts are given in order. First, it follows from (2.14) and (2.19) that

f L ∞ (Q(j)) f L 2 (Q(j)) = |f (y)| f L 2 (Q(j)) = |φ(0)|.
Second, it follows by the definition of I w0 (see (2.17) and (2.18)) that sup

t∈Iw 0 |φ(t)| ≤ f L ∞ (E Q(j)) f L 2 (Q(j)) .
The above two facts, along with (2.23), yield that

f L ∞ (Q(j)) ≤ M C3(1+ln 1 γ ) f L ∞ (E Q(j)) . (2.25)
We next define

E x ∈ E Q(j) : |f (x)| ≤ 2 |E Q(j)| E Q(j) |f (x)| dx .
By the Chebyshev inequality, we have that

|E | ≥ |E Q(j)| 2 ≥ γ 2 .
By the same argument as that used in the proof of (2.25), one can obtain that

f L ∞ (Q(j)) ≤ M C4(1+ln 1 γ ) f L ∞ (E Q(j)) for some C 4 = C 4 (n). (2.26)
Meanwhile, it follows by the definition of E that

f L ∞ (E Q(j)) ≤ 2 |E Q(j)| E Q(j) |f (x)| dx. (2.27)
From (2.26), (2.27) and the Hölder inequality, we find that

Q(j) |f | 2 dx ≤ 4 γ M 2C4(1+ln 1 γ ) E Q(j) |f | 2 dx. (2.28)
The term M (given by (2.24)) can be estimated by (2.21) as follows:

M = max |z|≤4 |φ(z)| ≤ max |z|≤4 ∞ k=0 φ (k) (0) k! |z| k ≤ C 2 (n)(1 + N ) n ∞ k=0 4n 3 2 √ A 0 N k k! ≤ e C5(1+N ) for some C 5 = C 5 (n). (2.29)
Finally, combining (2.28) and (2.29) leads to that

Q(j) |f | 2 dx ≤ e C6(1+N )(1+ln 1 γ ) E Q(j) |f | 2 dx for some C 6 = C 6 (n). (2.30)
Taking the sum in (2.30) for all good cubes, using (2.11), we see that

R n |f | 2 dx ≤ 2 Q(j) is good Q(j) |f | 2 dx ≤ 2 Q(j) is good e C6(1+N )(1+ln 1 γ ) E Q(j) |f | 2 dx ≤ 2e C6(1+N )(1+ln 1 γ ) E |f | 2 dx,
which leads to (1.10), where

C spec (n, E) = (C 6 + 1)(1 + L) 1 + ln 1 γ , with L = 1.
This ends the proof of Lemma 2.1.

Lemma 2.3 and Lemma 2.4 deal with connections among the spectral inequality (1.10), the Hölder-type interpolation inequality (1.7) and the observability inequality (1.8). In their proofs, we borrowed some ideas and techniques from [START_REF] Apraiz | Observability inequalities and measurable sets[END_REF][START_REF] Phung | An observability estimate for parabolic equations from a measurable set in time and its applications[END_REF]. Lemma 2.3. Suppose that a measurable set E ⊂ R n satisfies the spectral inequality (1.10). Then E satisfies the Hölder-type interpolation (1.7), with

C Hold = 1 1 -θ (C spec + 1) 2 + ln 12.
Proof. Let E ⊂ R n satisfy the spectral inequality (1.10). Arbitrarily fix T > 0, θ ∈ (0, 1) and a solution u to (1.1). Write u 0 (x) = u(0, x), x ∈ R n .

Then we have that u(T, x) = e T u 0 (x) for all x ∈ R n .

Given N > 0, write respectively χ ≤N (D) and χ >N (D) for the multiplier operators with the symbols χ {|ξ|≤N } and χ {|ξ|>N } . Namely, for each

g ∈ L 2 (R n ), χ ≤N (D)g(ξ) = χ {|ξ|≤N } g(ξ) and χ >N (D)g(ξ) = χ {|ξ|>N } g(ξ) for a.e. ξ ∈ R n .
Then we can express u 0 as:

u 0 = χ ≤N (D)u 0 + χ >N (D)u 0 .
From this and (1.10), we can easily check that

R n |u(T, x)| 2 dx ≤ 2 R n e T χ ≤N (D)u 0 (x) 2 dx + 2 R n e T χ >N (D)u 0 (x) 2 dx ≤ 2e Cspec(1+N ) E e T χ ≤N (D)u 0 (x) 2 dx + 2 R n e T χ >N (D)u 0 (x) 2 dx ≤ 4e Cspec(1+N ) E e T u 0 (x) 2 dx + 2 + 4e Cspec(1+N ) R n e T χ >N (D)u 0 (x) 2 dx. (2.31) Since R n e T χ >N (D)u 0 (x) 2 dx = R n e -T |ξ| 2 χ >N (ξ) u 0 (ξ) 2 dξ ≤ e -T N 2 R n u 0 (ξ) 2 dξ = e -T N 2 R n |u 0 (x)| 2 dx, it follows from (2.31) that R n |u(T, x)| 2 dx ≤ 4e Cspec(1+N ) E |u(T, x)| 2 dx + 2 + 4e Cspec(1+N ) e -T N 2 R n |u 0 (x)| 2 dx ≤ 6e Cspec e CspecN E |u(T, x)| 2 dx + e CspecN -T N 2 R n |u 0 (x)| 2 dx . (2.32) Given ε ∈ (0, 1), choose N = N (ε) so that exp C spec N -T N 2 = ε.
(This can be done since the set: {C spec s -T s 2 : s > 0} contains (-∞, 0].) With the above choice of N , we have that

N = C spec + C 2 spec + 4T ln 1 ε 2T ≤ 1 T C spec + T ln 1 ε .
Thus, with θ ∈ (0, 1) fixed before, we see that

exp [C spec N ] ≤ exp C 2 spec T exp C spec √ T ln 1 ε ≤ exp C 2 spec T exp 1 -θ θ ln 1 ε + θ 1 -θ C 2 spec T = exp C 2 spec (1 -θ)T ε -1-θ θ .
From this and (2.32), we find that for every ε ∈ (0, 1),

R n |u(T, x)| 2 dx ≤ 6e Cspec e C 2 spec (1-θ)T ε -1-θ θ E |u(T, x)| 2 dx + ε R n |u 0 (x)| 2 dx .
Choosing in the above

ε = E |u(T, x)| 2 dx R n |u 0 (x)| 2 dx θ , we obtain that R n |u(T, x)| 2 dx ≤ 12e Cspec e C 2 spec (1-θ)T E |u(T, x)| 2 dx θ R n |u 0 (x)| 2 dx 1-θ ≤ e ( 1 1-θ (Cspec+1) 2 +ln 12)(1+ 1 T ) E |u(T, x)| 2 dx θ R n |u 0 (x)| 2 dx 1-θ
, which leads to (1.7) with

C Hold = 1 1 -θ (C spec + 1) 2 + ln 12.
This ends the proof of Lemma 2.3.

Lemma 2.4. Suppose that a measurable set E ⊂ R n has the property: there is a positive constant C Hold = C Hold (n, E) so that for any T > 0,

R n |u(T, x)| 2 dx ≤ e C Hold 1+ 1 T E |u(T, x)| 2 dx 1/2 R n |u(0, x)| 2 dx 1/2 , (2.33)
when u solves the equation (1.1). Then for each T > 0 and each subset F ⊂ (0, T ) of positive measure, there is a positive constant C obs = C obs (n, T, F, C Hold ) so that when u solves (1.1),

R n |u(T, x)| 2 dx ≤ C obs F E |u(s, x)| 2 dxds. ( 2

.34)

In particular, if F = (0, T ) then the constant C obs in (2.34) can be expressed as:

C obs = exp [36(1 + 3C Hold )(1 + 1/T )] .
Proof. Suppose that E ⊂ R n satisfies (2.33). Arbitrarily fix T > 0 and F ⊂ (0, T ) of positive measure. Applying Cauchy's inequality to (2.33), we find that for all t > 0 and ε > 0,

R n |u(t, x)| 2 dx ≤ 1 ε e 2C Hold (1+ 1 t ) E |u(t, x)| 2 dx + ε R n |u 0 (x)| 2 dx. (2.35)
By a translation in time, we find from (2.35) that for all 0 < t 1 < t 2 and ε > 0,

R n |u(t 2 , x)| 2 dx ≤ 1 ε e 2C Hold 1+ 1 t 2 -t 1 E |u(t 2 , x)| 2 dx + ε R n |u(t 1 , x)| 2 dx. (2.36)
Let l be a Lebesgue density point of F . Then according to [46, Proposition 2.1], for each λ ∈ (0, 1), there is a sequence {l m } ∞ l=1 ⊂ (l, T ) so that for each m ∈ N + ,

l m+1 -l = λ m (l 1 -l) (2.37)
and

F (l m+1 , l m ) ≥ 1 3 (l m -l m+1 ). (2.38)
Arbitrarily fix m ∈ N + . Take s so that 0 < l m+2 < l m+1 ≤ s < l m < T.

Using (2.36) (with t 1 = l m+2 and t 2 = s) and noting that

R n |u (l m , x)| 2 dx ≤ R n |u(s, x)| 2 dx and l m+1 -l m+2 ≤ s -l m+2 ,
we see that

R n |u (l m , x)| 2 dx ≤ 1 ε e 2C Hold 1+ 1 l m+1 -l m+2 E |u(s, x)| 2 dx + ε R n |u (l m+2 , x)| 2 dx. (2.39)
Integrating with s over F (l m+1 , l m ) in (2.39) implies that

R n |u (l m , x)| 2 dx ≤ ε R n |u (l m+2 , x)| 2 dx + 1 ε 1 |F (l m+1 , l m )| e 2C Hold 1+ 1 l m+1 -l m+2 F (lm+1,lm) E |u(s, x)| 2 dxds. (2.40)
Since it follows by (2.38) that

F (l m+1 , l m ) ≥ 1 3 (l m -l m+1 ) ≥ 1 3 e - 1 lm-l m+1 ,
we obtain from (2.40) that

R n |u (l m , x)| 2 dx ≤ ε R n |u (l m+2 , x)| 2 dx (2.41) + 3 ε e 1 lm-l m+1 +2C Hold 1+ 1 l m+1 -l m+2 F (lm+1,lm) E |u(s, x)| 2 dxds.
Meanwhile, it follows by (2.37) that

l m -l m+1 = 1 1 + λ (l m -l m+2 ) (2.42)
and

l m+1 -l m+2 = λ 1 + λ (l m -l m+2 ) . (2.43)
Inserting (2.42) and (2.43) into (2.41), we find that

R n |u (l m , x)| 2 dx ≤ ε R n |u (l m+2 , x)| 2 dx + 3e 2C Hold 1 ε e C lm-l m+2 F (lm+1,lm) E |u(s, x)| 2 dxds
(2.44)

with

C = 1 + λ + 2C Hold (1 + λ) λ . (2.45) Rewrite (2.44) as εe - C lm -l m+2 R n |u (l m , x)| 2 dx -ε 2 e - C lm -l m+2 R n |u (l m+2 , x)| 2 dx ≤ 3e 2C Hold F (lm+1,lm) E |u(s, x)| 2 dxds. (2.46) Next, we fix λ ∈ (1/ √ 2, 1). Let µ := 1 2-λ -2 . Then µ > 1. Setting, in (2.46), ε = exp - (µ -1)C l m -l m+2 ,
we have that

e - µC lm-l m+2 R n |u (l m , x)| 2 dx -e - (2µ-1)C lm-l m+2 R n |u (l m+2 , x)| 2 dx ≤ 3e 2C Hold F (lm+1,lm) E |u(s, x)| 2 dxds. (2.47)
Meanwhile, one can easily check that exp -

(2µ -1)C l m -l m+2 = exp - µC λ 2 (l m -l m+2 ) . (2.48) Because l m+2 -l m+4 = λ 2 (l m -l m+2 ) ,
we deduce from (2.47) and (2.48) that

e - µC lm-l m+2 R n |u (l m , x)| 2 dx -e - µC l m+2 -l m+4 R n |u (l m+2 , x)| 2 dx ≤ 3e 2C Hold F (lm+1,lm) E |u(s, x)| 2 dxds.
Summing the above inequality for all odd m derives that

e -µC l 1 -l 3 R n |u (l 1 , x)| 2 dx ≤ 3e 2C Hold ∞ m=1 F (lm+1,lm) E |u(s, x)| 2 dxds ≤ 3e 2C Hold F (l,l1) E |u(s, x)| 2 dxds ≤ 3e 2C Hold F E |u(s, x)| 2 dxds.
Thus, we have that

R n |u(T, x)| 2 dx ≤ R n |u(l 1 , x)| 2 dx ≤ 3e 2C Hold e µC l 1 -l 3 F E |u(s, x)| 2 dxds, (2.49)
which leads to (2.34) with

C obs = 3 exp 2C Hold + µC l 1 -l 3 .
Finally, in the case when F = (0, T ), we set

l 1 = 2T 3 , l = T 3 and λ = 2 3 .
Then we have that (see (2.45)) This completes the proof of Lemma 2.4.

l 1 -l 3 = T 9 , µ =
The next lemma seems to be new to our best knowledge. The key of its proof is the structure of a special solution to the equation (1.1). This structure is based on the heat kernel. Lemma 2.5. Suppose that a measurable set E ⊂ R n satisfies the observability inequality (1.3). Then the set E is γ-thick at scale L for some γ > 0 and L > 0.

Proof. Let E ⊂ R n be a measurable set satisfying the observability inequality (1.3). Recall that the heat kernel is as:

K(t, x) = (4πt) -n/2 e -|x| 2 /4t , t > 0, x ∈ R n . Given u 0 ∈ L 2 (R n ), the function defined by (t, x) -→ R n K(t, x -y)u 0 (y) dy, (t, x) ∈ (0, ∞) × R n , (2.50)
is a solution to the equation (1.1) with the initial condition u(0, x) = u 0 (x), x ∈ R n .

Arbitrarily fix x 0 ∈ R n . By taking

u 0 (x) = (4π) -n/2 e -|x-x0| 2 /4 , x ∈ R n ,
in (2.50), we get the following solution to the equation (1.1):

v(t, x) = (4π(t + 1)) -n 2 e -|x-x 0 | 2 4(t+1) , t ≥ 0, x ∈ R n . (2.51)
From (2.51), we obtain by direct computations that

R n |v(1, x)| 2 dx = 4 -n π -n 2 .
(2.52)

From (2.51), we also find that for an arbitrarily fixed L > 0, Choose L > 0 in such a way that

v(t, x) ≤ (4π) -n 2 e -L
C(2π) -n 2 e -L 2 8 ≤ 1 2 4 -n π -n 2 .
Then by (2.56) and (2.51), we obtain that

1 2 4 -n π -n 2 ≤ C 1 0 E B L (x0) |v(t, x)| 2 dx dt = C 1 0 E B L (x0) (4π(t + 1)) -n e -|x-x 0 | 2 2(t+1) dx dt ≤ C 1 0 E B L (x0) (4π) -n dx dt ≤ C(4π) -n |E B L (x 0 )|, from which, it follows that |E B L (x 0 )| ≥ (2C) -1 π n 2 .
(2.57)

Since B L (x 0 ) ⊂ (x 0 + 2LQ), we see from (2.57) that

E (x 0 + 2LQ) ≥ (2C) -1 π n 2 .
From this, as well as the choice of L, we can find L > 0 and γ > 0, which are independent of the choice of x 0 , so that

E (x 0 + L Q) ≥ γ(L ) n .
(2.58)

Notice that x 0 in (2.58) was arbitrarily taken from R n . Hence, the set E is γ-thick at scale L . This ends the proof of Lemma 2.5.

We now on the position to prove Theorem 1.1.

Proof of Theorem 1.1. We can prove it in the following way:

(i) ⇒ (ii) ⇒ (iii) ⇒ (iv) ⇒ (i).
Indeed, the conclusions (i) ⇒ (ii), (ii) ⇒ (iii), (iii) ⇒ (iv) and (iv) ⇒ (i) follow respectively from Lemma 2.1, Lemma 2.3, Lemma 2.4 and Lemma 2.5. This ends the proof of Theorem 1.1.

Tracking the constants in Lemma 2.1, Lemma 2.3 and Lemma 2.4, we can easily get the following consequences of Theorem 1.1: Corollary 2.1. Let E ⊂ R n be a set of γ-thick at scale L for some γ > 0 and L > 0. Then the following conclusions are true for a constant C = C(n) > 0:

(a) The set E satisfies the Hölder-type interpolation (1.7) with

C Hold (n, E, θ) = C 1 -θ (1 + L) 2 1 + ln 1 γ 2 . (b)
The set E satisfies the observability inequality (1.3) with C obs (n, E, T ) = e 300(1+C)(1+L) 2 (1+ln 1 γ ) 2 (1+ 1 T ) .

Weak interpolation and observability inequalities

In this section, we introduce several weak observability inequalities and interpolation inequalities, where observations are made over a ball in R n . One one hand, these inequalities can be viewed as extensions of (1.3) and (1.7) in some senses, while on the other hand, they are independently interesting.

Weak interpolation inequalities with observation on the unit ball

We begin with introducing two spaces. Given a > 0 and ν > 0, we set L 2 (e a|x| ν dx) := {f : R n → R : f is measurable and f L 2 (e a|x| ν dx) < +∞}, equipped with the norm:

f L 2 (e a|x| ν dx) := R n |f (x)| 2 e a|x| ν dx 1/2 , f ∈ L 2 (e a|x| ν dx).
Given ν > 0, we set

L 2 ( x ν dx) := {f : R n → R : f is measurable and f L 2 ( x ν dx) < +∞},
equipped with the norm:

f L 2 ( x ν dx) := R n |f (x)| 2 x ν dx 1/2 , f ∈ L 2 ( x ν dx).
Notice that any function in one of the above spaces decays along the radical direction.

In this subsection, we will build up some interpolation inequalities for solutions to (1.1), with initial data in L 2 (e a|x| ν dx) (or L 2 ( x ν dx)). In these inequalities, observations are made over the unit ball in R n and at one time point. The purpose to study such observability has been explained in Subsection 1.3. Our main results about this subject are included in the following theorem: Theorem 3.1. (i) There is θ = θ(n) ∈ (0, 1) and C = C (n) such that for any ε > 0, T > 0 and a > 0,

R n |u(T, x)| 2 dx ≤ C 1 (a, T ) ε R n |u 0 (x)| 2 e a|x| dx + ε -1 e ε -4| ln θ| a B1 |u(T, x)| 2 dx ,
when u solves (1.1) with the initial condition u(0, •) = u 0 (•) ∈ L 2 (e a|x| dx). Here,

C 1 (a, T ) = e C (1+ 1 T +a+a 2 T ) 1 + a -n Γ a 2| ln θ| .
(ii) There is θ = θ(n) ∈ (0, 1) and C = C (n) so that for any ε ∈ (0, 1), T > 0 and ν ∈ (0, 1],

R n |u(T, x)| 2 dx ≤ C 2 (ν, T ) ε R n |u 0 | 2 x ν dx + e e (3| ln θ|+1) ( 1 ε ) 1 ν B1 |u(T, x)| 2 dx ,
when u solves (1.1) with the initial condition u(0, •) = u 0 (•) ∈ L 2 ( x ν dx). Here,

C 2 (ν, T ) = (1 + T ν 2 )e C (1+ 1 T ) . Remark 3.1. (a)
The condition that ν ≤ 1 in (ii) of Theorem 3.1 is not necessary. We make this assumption only for the brevity of the statement of the theorem. Indeed, from the definition of L 2 ( x ν dx), we see that L 2 ( x ν dx) → L 2 ( x dx) for any ν ≥ 1. From this and (ii) of Theorem 3.1, one can easily check that when ν > 1, any solution of (1.1) satisfies that

R n |u(T, x)| 2 dx ≤ 1 + T 1 2 e C (n)(1+ 1 T ) ε R n |u 0 | 2 x ν dx + e e (3| ln θ|+1) 1 ε B1 |u(T, x)| 2 dx .
(b) [16, Theorem 1] contains the following result: There is a universal constant C > 0 so that for each T > 0 and R > 0,

sup 0≤t≤T e t|x| 2 4(t 2 +R 2 ) u(t) L 2 (R n ) ≤ C u(0) L 2 (R n ) + e T |x| 2 4(T 2 +R 2 ) u(T ) L 2 (R n ) , (3.1)
when u solves (1.1). The first inequality in Theorem 3.1 is comparable to the above inequality (3.1). By our understanding, these two inequalities can be viewed as different versions of Hardy uncertainty principle. On one hand, the inequality (3.1) can be understood as follows: From some information on a solution to (1.1) at infinity in R n at two time points 0 and T , one can know the behaviour of this solution at infinity in R n at each time t ∈ [0, T ]. On the other hand, the first inequality in Theorem 3.1 can be explained in the following way: From some information on a solution to (1.1) at infinity in R n at time 0, and in the ball B 1 in R n at time T , one can know the behaviour of this solution at infinity in R n at time T . Similarly, we can compare the second inequality in Theorem 3.1 with (3.1). It deserves to mention that we can only prove inequalities in Theorem 3.1 for the pure heat equation (1.1), while [START_REF] Escauriaza | Hardy Uncertainty Principle, Convexity and Parabolic Evolutions[END_REF]Theorem 1] gave the inequality (3.1) for heat equations with general potentials.

(c) The first inequality in Theorem 3.1 can also be understood as follows: If we know in advance that the initial datum of a solution to (1.1) is in the unit ball of L 2 (e a|x| dx), then by observing this solution in the unit ball of R n at time T , we can approximately recover this solution over R n at the same time T , with the error C 1 (a, T )ε. The second inequality in Theorem 3.1 can be explained in a very similar way.

To show Theorem 3.1, we need some preliminaries. We begin with some auxiliary lemmas on the persistence of the heat semigroup in the spaces L 2 (e a|x| ν dx) and L 2 ( x ν dx). Lemma 3.1. Let a > 0 and 0 < ν ≤ 1. Then when u 0 ∈ L 2 (e a|x| ν dx),

e t u 0 L 2 (e a|x| ν dx) ≤ 2 n 2 e a 2 2-ν t ν 2-ν u 0 L 2 (e a|x| ν dx) for all t > 0.
Proof. Arbitrarily fix a > 0, 0 < ν ≤ 1 and u 0 ∈ L 2 (e a|x| ν dx). Using the fundamental solution of (1.1) and the definition of L 2 (e a|x| ν dx), we have that

e t u 0 L 2 (e a|x| ν dx) = R n e a|x| ν 2 (4πt) -n/2 R n e -|x-y| 2 4t u 0 (y) dy 2 dx 1 2 . (3.2) Since |x| ν ≤ (|x -y| + |y|) ν ≤ |x -y| ν + |y| ν for all x, y ∈ R n ,
(Here, we used the elementary inequality: (τ + s) ν ≤ τ ν + s ν , τ, s > 0.) it follows from (3.2) that

e t u 0 L 2 (e a|x| ν dx) ≤ R n (4πt) -n/2 R n e -|x-y| 2 4t + a|x-y| ν 2 e a|y| ν 2 |u 0 (y)| dy 2 dx 1 2 ≤ R n (4πt) -n/2 e -|x| 2 4t + a|x| ν 2 dx • R n e a|y| ν 2 u 0 (y) 2 dy 1 2 = R n (4πt) -n/2 e 1 4t (-|x| 2 +2ta|x| ν ) dx • u 0 L 2 (e a|x| ν dx) . (3.3)
Meanwhile, by the Young inequality:

2ta|x| ν ≤ |x| 2 2 ν + (2ta) 2 2-ν 2 2-ν ≤ 1 2 |x| 2 + (2ta) 2 2-ν , we obtain that R n (4πt) -n/2 e 1 4t (-|x| 2 +2ta|x| ν ) dx ≤ R n e (2ta) 2 2-ν 4t (4πt) -n/2 e -|x| 2 8t dx = 2 n 2 e (2ta) 2 2-ν 4t ≤ 2 n 2 e a 2 2-ν t ν 2-ν . (3.4)
Now, the desired inequality follows from (3.3) and (3.4). This ends the proof of Lemma 3.1.

Remark 3.2. The inequality in Lemma 3.1 does not hold for the case when ν > 1. Indeed, given ν > 1, let u 0 (x) = e -1 2 |x| ν x -n , x ∈ R n . It is clear that u 0 ∈ L 2 (e |x| ν dx). However, we have that for any t > 0, e t u 0 / ∈ L 2 (e |x| ν dx). This can be proved as follows: Arbitrarily fix t > 0. By some direct calculations, we find that when |x| ≥ 2,

e t u 0 (x) ≥ C(4πt) -n/2 e -1 4t e -1 2 (|x|-1 2 ) ν |x| -1/2 -n for some C = C(n).
This leads to that

e t u 0 L 2 (e |x| ν dx) ≥ C(4πt) -n/2 e -1 4t |x|≥2 e |x| ν -(|x|-1 2 ) ν |x| -1/2 -2n dx 1/2 . (3.5)
Meanwhile, one can easily find a constant M > 2 so that

|x| ν -(|x| -1/2) ν ≥ ν 4 |x| ν-1 , when |x| ≥ M.
From this and (3.5), we obtain that

e t u 0 L 2 (e |x| ν dx) ≥ C(4πt) -n/2 e -1 4t |x|≥M e ν 4 |x| ν-1 |x| -1/2 -2n dx 1/2 = ∞. Lemma 3.2. Let ν ≥ 0. Then for any u 0 ∈ L 2 ( x ν dx), e t u 0 L 2 ( x ν dx) ≤ 4 ν+2 Γ(ν/2 + n) 1 + t ν 4 u 0 L 2 ( x ν dx) for all t > 0.
Proof. The proof is similar to that of Lemma 3.1. (See also [START_REF] Simon | Schrödinger semigroups[END_REF]Lemma B.6.1].)

Lemma 3.3. Given s > 0, there is C = C(n, s) so that when f ∈ L 2 (R n ) satisfies that f ∈ L 2 (e a|ξ| s dξ) for some a > 0, D α f L ∞ (R n ) ≤ C |α|+1 a -2|α|+3n 2s (α!) 1 s f (ξ)
L 2 (e a|ξ| s dξ)

for each α ∈ N n .

(Here, we adopt the convention that

α! = α 1 !α 2 ! • • • α n !.)
Remark 3.3. From Lemma 3.3, we see that if f ∈ L 2 (R n ) satisfies that f ∈ L 2 (e a|ξ| s dξ), with s > 0 and a > 0, then f is analytic, when s = 1, while f is ultra-analytic, when s > 1.

Proof of Lemma 3.3. Arbitrarily fix s > 0, a > 0 and f ∈ L 2 (R n ), with f ∈ L 2 (e a|ξ| s dξ). Then arbitrarily fix α = (α 1 , . . . , α n ) ∈ N n , β = (β 1 , . . . , β n ) ∈ N n and γ = (γ 1 , . . . , γ n ) ∈ N n , with |γ| ≤ n. Several facts are given in order. Fact One: By direct computations, we see that

R n ξ 2β e -a|ξ| s dξ ≤ R n ξ 2β e -a(Σ n i=1 |ξi| s /n) dξ = n i=1 R ξ i |ξ i | 2βi e -a|ξi| s /n dξ i = n i=1 2 ∞ 0 r 2βi e -ar s /n dr = n i=1 2 n a 2β i +1 s ∞ 0 t 2β i +1 s -1 e -t dt = 2 n n a 2|β|+n s n i=1 Γ 2β i + 1 s .
From this, we obtain that

R n |ξ 2(α+γ) |e -a|ξ| s dξ 1/2 ≤ 2 n/2 n a 2|α|+3n 2s n i=1 Γ 2α i + 2γ i + 1 s . (3.6)
Fact Two: By the Sobolev embedding

H n (R n ) → L ∞ (R n ), we can find C 1 (n) > 0 so that D α f L ∞ (R n ) ≤ C 1 (n) γ∈N n ,|γ|≤n D α+γ f L 2 (R n ) . (3.7)
Fact Three: By the Plancheral theorem and the Hölder inequality, we obtain that

D α+γ f L 2 (R n ) = ξ α+γ f (ξ) L 2 (R n ) ≤ f (ξ)e a|ξ| s /2 L 2 (R n ) R n |ξ 2(α+γ) |e -a|ξ| s dξ 1/2 . (3.

8)

Fact Four: There exists

C 2 = C 2 (n, s) so that Γ 2α i + 2γ i + 1 s ≤ C αi+1 2 Γ 1/s (α i ) = C αi+1 2 (α i !) 1 s . (3.9)
The proof of (3.9) is as follows: From the Stirling formula, we have that

lim x→+∞ Γ(x) √ 2πe -x x x+ 1 2 = 1. (3.10)
From (3.10), we can find constants M 1 = M 1 (s) and

C 3 = C 3 (n, s) so that for all α i > M 1 , Γ 2α i + 2γ i + 1 s ≤ 2 √ 2πe -2α i +2γ i +1 s 2α i + 2γ i + 1 s 2α i +2γ i +1 s + 1 2 = 2 √ 2π • e -2α i +2γ i +1 s 2α i + 2γ i + 1 s 2γ i +1 s + 1 2 • 2α i + 2γ i + 1 s 2α i s ≤ 2 √ 2π sup x>0 e -x x 2γ i +1 s + 1 2 • 2α i + 2γ i + 1 s 2α i s = 2 √ 2π • e -( 2γ i +1 s + 1
2

) 2γ i + 1 s + 1 2 2γ i +1 s + 1 2 • 2α i + 2n + 1 s 2α i s ≤ C αi 3 α 2α i s i . (3.11)
From (3.10), we can also find an absolute constant M 2 ≥ 1 so that for all

α i > M 2 , Γ(α i ) ≥ 2 -1 √ 2πe -αi α i αi+ 1 2 . (3.12)
According to (3.11) and (3.12), there is a constant C 4 (n, s) so that

Γ 2αi+2γi+1 s Γ 1/s (α i ) ≤ [C 4 (n, s)] αi for all α i > M := max{M 1 , M 2 }. (3.13)
Meanwhile, it is clear that there is a constant C 5 (n, s) so that

Γ 2αi+2γi+1 s Γ 1/s (α i ) ≤ C 5 (n, s) for all α i ≤ M. (3.14)
(Here we agree that Γ(0) = ∞.) Combining (3.13) and (3.14) leads to (3.9). Inserting (3.9) into (3.6), noticing that |γ| ≤ n, we find that for some

C 6 = C 6 (n, s), R n ξ 2(α+γ) e -a|ξ| s dξ 1/2 ≤ 2 n/2 n a 2|α|+3n 2s n i=1 C αi+1 2 (α i !) 1 s ≤ C |α|+1 6 a -2|α|+3n 2s (α!) 1 s . (3.15)
Finally, it follows from (3.7), (3.8) and (3.15) that for some C = C(n, s),

D α f L ∞ (R n ) ≤ C 1 (n) γ∈N n ,|γ|≤n f (ξ)e a|ξ| s /2 L 2 (R n ) C |α|+1 6 (α!) 1 s ≤ C |α|+1 a -2|α|+3n 2s (α!) 1 s f (ξ) L 2 (e a|ξ| s dξ)
.

This ends the proof of Lemma 3.3.

The next corollary is a consequence of Lemma 3.3.

Corollary 3.1. There is C = C(n) > 0 so that when f ∈ L 2 (R n ) satisfies that f ∈ L 2 (e a|ξ| 2 dξ)
for some a > 0,

D α f L ∞ (R n ) ≤ e C(1+b 2 )(1+ 1 a ) |α|! b |α| f (ξ) L 2 (e a|ξ| 2 dξ)
for all b > 0 and α ∈ N n .

Remark 3.4. Let u 0 ∈ L 2 (R n ) be arbitrarily given. Set u(t, x) = e t u 0 (x), (t, x) ∈ (0, ∞)×R n .
Then u is the solution of (1.1) with u(0, •) = u 0 (•). Arbitrarily fix t > 0. By applying Corollary 3.1 (where f (•) = u(t, •) and a = 2t), we see that the radius of analyticity of u(t, •) (which is treated as a function of x) is independent of t. It is an analogy result for solutions of the heat equation in a bounded domain with an analytic boundary (see [START_REF] Apraiz | Observability inequalities and measurable sets[END_REF][START_REF] Escauriaza | Observation from measurable sets for parabolic analytic evolutions and applications[END_REF]). This property plays a very important role in the proof of the observability estimates from measurable sets when using the telescope series method developed in [START_REF] Apraiz | Observability inequalities and measurable sets[END_REF][START_REF] Escauriaza | Observation from measurable sets for parabolic analytic evolutions and applications[END_REF].

Proof of Corollary 3.1. Arbitrarily fix a > 0 and f ∈ L 2 (R n ) with f ∈ L 2 (e a|ξ| 2 dξ). Then arbitrarily fix b > 0 and α ∈ N n . According to Lemma 3.3 (with s = 2), there is

C = C (n) so that D α f L ∞ (R n ) ≤ C |α|+1 a -2|α|+3n 4 (α!) 1 2 f (ξ) L 2 (e a|ξ| 2 dξ) ≤ g(|α|) |α|! b |α| f (ξ) L 2 (e a|ξ| 2 dξ) , (3.16)
where

g(r) = C a -3n/4 bC a -1/2 r (r!) -1/2 , r ∈ N.
To estimate g(r) pointwisely, we use (3.12) to find that when r > M 2 ≥ 1 (where M 2 is given by (3.12)),

g(r) ≤ C a -3n/4 bC a -1/2 r 2 -1 √ 2πe -r r r+ 1 2 -1/2 ≤ 2 1/4 π -1/4 C a -3n/4 be 1/2 C a -1/2 r r -r/2 ≤ 2 1/4 π -1/4 C a -3n/4 sup r>0 (be 1/2 C a -1/2 ) r r -r/2 = 2 1/4 π -1/4 C a -3n/4 e b 2 C 2 2a . (3.17) Meanwhile, it is clear that when r ≤ M 2 , g(r) ≤ C a -3n/4 bC a -1/2 + 1 M2 . (3.18)
From (3.17) and (3.18), it follows that g(r) ≤ e C(1+b 2 )(1+ 1 a ) for all r = 0, 1, 2, . . . , which, together with (3.16), yields the desired inequality. This ends the proof of Corollary 3.1.

To prove Theorem 3.1, we also need the decomposition:

R n = j≥1 Ω j , with Ω j := {x ∈ R n : j -1 ≤ |x| < j}. (3.19)
The next lemma concerns with the propagation of smallness for some real-analytic functions with respect to the above decomposition of R n .

Lemma 3.4. There are constants C = C(n) > 0 and θ = θ(n) ∈ (0, 1) so that for any a > 0,

j ≥ 1 and f ∈ L 2 (R n ) with f ∈ L 2 (e a|ξ| 2 dξ), Ωj+1 |f | 2 dx ≤ j (n-1)(1-θ) e C(1+ 1 a ) Ωj |f | 2 dx θ R n | f | 2 e a|ξ| 2 dξ 1-θ .
The proof of Lemma 3.4 needs Corollary 3.1 and the next lemma which is quoted from [START_REF] Apraiz | Null-control and measurable sets[END_REF] (see also [START_REF] Apraiz | Observability inequalities and measurable sets[END_REF]Theorem 4]), but is originally from [START_REF] Vessella | A continuous dependence result in the analytic continuation problem[END_REF].

Lemma 3.5 ([1, Theorem 1.3] ). Let R > 0 and let f : B 2R ⊂ R n → R be real analytic in B 2R verifying |D α f (x)| ≤ M (ρR) -|α| |α|!, when x ∈ B 2R and α ∈ N n
for some positive numbers M and ρ ∈ (0, 1]. Let ω ⊂ B R be a subset of positive measure. Then there are two constants

C = C(ρ, |ω|/|B R |) > 0 and θ = θ(ρ, |ω|/|B R |) ∈ (0, 1) so that f L ∞ (B R ) ≤ CM 1-θ 1 |ω| ω |f (x)| dx θ .
We are now in the position to show Lemma 3.4.

Proof of Lemma 3.4. Let a > 0 and j ≥ 1 be arbitrarily given. Arbitrarily fix f ∈ L 2 (R n ) with f ∈ L 2 (e a|ξ| 2 dξ). The rest proof is divided into the following several steps.

Step 1. The decompositions of Ω j and Ω j+1 in the polar coordinates. In the polar coordinate system, we have that

Ω j = (r, ϑ 1 , • • • , ϑ n-1 ) ∈ [j -1, j) × [0, 2π) n-1 , Ω j+1 = (r, ϑ 1 , • • • , ϑ n-1 ) ∈ [j, j + 1) × [0, 2π) n-1 .
(3.20)

When j is large, the distance between two points in Ω j can be very large. This makes our studies on the propagation from Ω j to Ω j+1 harder. To pass this barrier, we need to build up a suitable refinement for each Ω j . We set

[0, 2π] = 1≤l≤j ∆ l , with ∆ l := l -1 j 2π, l j 2π . (3.21) Given (k 1 , . . . , k n-1 ) ∈ N n-1 with 1 ≤ k i ≤ j (i = 1, • • • , n -1), we set Ω j;k1,••• ,kn-1 = (r, ϑ 1 , • • • , ϑ n-1 ) ∈ Ω j : ϑ 1 ∈ ∆ k1 , • • • , ϑ n-1 ∈ ∆ kn-1 , Ω j+1;k1,••• ,kn-1 = (r, ϑ 1 , • • • , ϑ n-1 ) ∈ Ω j+1 : ϑ 1 ∈ ∆ k1 , • • • , ϑ n-1 ∈ ∆ kn-1 . (3.22)
Then one can easily check that for each ĵ ∈ {j, j + 1},

Ω ĵ;k1,••• ,kn-1 Ω ĵ;k 1 ,••• ,k n-1 = ∅, if (k 1 , • • • , k n-1 ) = (k 1 , • • • , k n-1 ), (3.23)
and

Ω ĵ = Ω ĵ;k1,••• ,kn-1 , (3.24)
where the union is taken over all different (k 1 , . . . , k n-1 ) ∈ N n-1 , with 1 ≤ k i ≤ j for all i = 1, . . . , n -

In what follows, we write d

Ω j;k1,••• ,kn-1 for the diameter of Ω j;k1,••• ,kn-1 .
Step 2. To prove the following three properties:

(O1) There are constants

c 1 = c 1 (n) and c 2 = c 2 (n) so that for any (k 1 , . . . , k n-1 ) ∈ N n-1 , with 1 ≤ k i ≤ j (i = 1, . . . , n -1)
,

c 1 V n ≤ Ω j;k1,••• ,kn-1 ≤ c 2 V n . (3.25)
(O2) We have that for any (k 1 , . . . , k n-1 ) ∈ N n-1 with 1 ≤ k i ≤ j (i = 1, . . . , n -1),

d Ω j;k1,••• ,kn-1 = d (Ω j;1,••• ,1 ) := sup x,x ∈Ωj;1,••• ,1 |x -x |. (3.26)
(O3) We have that for any fix

(k 1 , . . . , k n-1 ) ∈ N n-1 with 1 ≤ k i ≤ j (i = 1, . . . , n -1), d Ω j;k1,••• ,kn-1 ≤ 2π 1≤i≤n i 2 ≤ 2πn 3 2 ; (3.27)
d Ω j+1;k1,••• ,kn-1 ≤ 2π j + 1 j 1≤i≤n i 2 ≤ 4πn 3 2 . (3.28)
To see (O1), we use the definitions of Ω j;k1,••• ,kn-1 and Ω j to find that

Ω j;k1,••• ,kn-1 = 1 j n-1 |Ω j | = V n j n -(j -1) n j n-1 ,
which leads to (3.25).

The conclusion (O2) follows immediately from the definitions of Ω j;k1,••• ,kn-1 and d Ω j;k1,••• ,kn-1 .

To show (3.27) in (O3), we let (r, ϑ 1 , • • • , ϑ n-1 ) and (r, ϑ 1 , • • • , ϑ n-1 ) be the polar coordinates of x = (x 1 , . . . , x n ) and x = (x 1 , . . . , x n ), respectively. Then we have that

x, x ∈ Ω j;1,••• ,1 ⇐⇒ j -1 ≤ r < j, 0 ≤ ϑ l , ϑ l < 2π j for all l = 1, • • • , n -1. (3.29)
Notice that the connection between (x 1 , . . . , x n ) and (r, ϑ 1 , . . . , ϑ n-1 ) is as:

               x 1 = r cos ϑ 1 , x 2 = r sin ϑ 1 cos ϑ 2 , • • • x n-1 = r sin ϑ 1 sin ϑ 2 • • • sin ϑ n-2 cos ϑ n-1 , x n = r sin ϑ 1 sin ϑ 2 • • • sin ϑ n-2 sin ϑ n-1 .
Then, by the mean value theorem, we have that for some ζ ∈ (0, 2π/j),

|x 1 -x 1 | = r| cos ϑ 1 -cos ϑ 1 | = r| sin ζ • (ϑ 1 -ϑ 1 )| ≤ j| sin ζ| 2π j ≤ 2π.
By inserting suitable terms and using the mean value theorem, we have that

|x 2 -x 2 | ≤ r | sin ϑ 1 cos ϑ 2 -sin ϑ 1 cos ϑ 2 | + | sin ϑ 1 cos ϑ 2 -sin ϑ 1 cos ϑ 2 | ≤ 2π • 2.
Similarly, we can verify that

|x i -x i | ≤ 2π • i for all i = 3, . . . , n.
These, along with (3.26), lead to (3.27).

The inequality (3.28) in (O3) can be proved in the same way. The reason that the factor j+1 j appears in (3.28) is as follows: Since 1 ≤ k i ≤ j (i = 1, . . . , n -1), we see from the definition of

Ω j+1;1,••• ,1 that x, x ∈ Ω j+1;1,••• ,1 ⇐⇒ j ≤ r < j + 1, 0 ≤ ϑ l , ϑ l < 2π j for all l = 1, • • • , n -1.
(The above is comparable with (3.29).)

Step 3. To prove that there are constants C = C(n) > 0 and θ = θ(n) ∈ (0, 1) (both independent of a, j and f ) so that

Ω j+1;k 1 ,••• ,k n-1 |f | 2 dx ≤ e C(1+ 1 a ) Ω j;k 1 ,••• ,k n-1 |f | 2 dx θ R n | f | 2 e a|ξ| 2 dξ 1-θ . Since Ω j;k1,••• ,kn-1 Ω j+1;k1,••• ,kn-1 is connected (see (3. 22
)), it follows from (3.27) and (3.28) that

d Ω j;k1,••• ,kn-1 Ω j+1;k1,••• ,kn-1 ≤ d Ω j;k1,••• ,kn-1 + d Ω j+1;k1,••• ,kn-1 ≤ 6πn 3 2 .
Thus, there exists x ∈ R n such that

Ω j;k1,••• ,kn-1 Ω j+1;k1,••• ,kn-1 ⊂ B R0 ( x), with R 0 = 7πn
= R 0 , there is C = C(n) > 0 such that for all α ∈ N n , D α f L ∞ (R n ) ≤ e C(1+ 1 a ) |α|! R |α| 0 f (ξ)
L 2 (e a|ξ| 2 dξ) .

(3.31)

By (3.31), as well as (3.30), we can apply Lemma 3.5, where

ρ = 1, R = R 0 , B R = B R0 ( x), B 2R = B 2R0 ( x), ω = Ω j;k1,••• ,kn-1 , M = e C(1+ 1 a ) f (ξ) L 2 (e a|ξ| 2 dξ)
, to find constants C 0 = C 0 (n) > 0 and θ = θ(n) ∈ (0, 1) so that

f L 2 (B R 0 ( x)) ≤ C 0 f θ L 2 (Ωj;k 1 ,••• ,k n-1 ) e C(1+ 1 a ) f (ξ) L 2 (e a|ξ| 2 dξ) 1-θ ≤ C 0 e C(1+ 1 a ) f θ L 2 (Ωj;k 1 ,••• ,k n-1 ) f (ξ) 1-θ L 2 (e a|ξ| 2 dξ) . ( 3.32)
(Here, we used (3.25) and a coordinate translation.) Finally, the desired inequality of this step follows from (3.32) and (3.30).

Step 4. To prove the inequality of this lemma From (3.23) and (3.24), we see that Ω j is the disjoint union of all Ω j;k1,••• ,kn-1 with different (k 1 , . . . , k n-1 ) ∈ N n-1 satisfying 1 ≤ k i ≤ j for all i = 1, . . . , n -1. Meanwhile, by (3.20), (3.21) and (3.22), one can also check that Ω j+1 is the disjoint union of all Ω j+1;k1,••• ,kn-1 with different (k 1 , . . . , k n-1 ) ∈ N n-1 satisfying 1 ≤ k i ≤ j for all i = 1, . . . , n -1. These, along with Lemma 3.4, yield that for some C = C(n) > 0 and θ = θ(n),

Ωj+1 |f | 2 dx = Ω j+1;k 1 ,••• ,k n-1 |f | 2 dx ≤ e C(1+ 1 a ) Ω j;k 1 ,••• ,k n-1 |f | 2 dx θ R n | f | 2 e a|ξ| 2 dξ 1-θ ≤ e C(1+ 1 a ) Ω j;k 1 ,••• ,k n-1 |f | 2 dx θ R n | f | 2 e a|ξ| 2 dξ 1-θ = j (n-1)(1-θ) e C(1+ 1 a ) Ωj |f | 2 dx θ R n | f | 2 e a|ξ| 2 dξ 1-θ , (3.33)
where the sums are taken over all different (k 1 , . . . , k n-1 ) ∈ N n-1 with 1 ≤ k i ≤ j for all i = 1, . . . , n -1. (Notice that there are j n-1 such (k 1 , . . . , k n-1 )). Hence, the desired conclusion follows from (3.33). This ends the proof of Lemma 3.4.

Based on Lemma 3.4, we can have the next propagation result which will be used later.

Lemma 3.6. There exist constants C = C(n) > 0 and θ = θ(n) ∈ (0, 1) so that for any a > 0 and j ≥ 1,

Ωj+1 |f | 2 dx ≤ j n-1 e C(1+ 1 a ) B1 |f | 2 dx θ j R n | f | 2 e a|ξ| 2 dξ 1-θ j , when f ∈ L 2 (R n ) satisfies that f ∈ L 2 (e a|ξ| 2 dξ).
Proof. Arbitrarily fix a > 0 and j ≥ 1. And then arbitrarily fix f ∈ L 2 (R n ) with f ∈ L 2 (e a|ξ| 2 dξ). From Lemma 3.4, we can use the induction method to verify that

Ωj+1 |f | 2 dx ≤ j(j -1) θ (j -2) θ 2 • • • 2 θ j-2 1 θ j-1 (n-1)(1-θ) × e C(1+ 1 a )(1+θ+•••+θ j-1 ) Ω1 |f | 2 dx θ j R n | f | 2 e a|ξ| 2 dξ 1-θ j ≤ j n-1 e C 1-θ (1+ 1 a ) Ω1 |f | 2 dx θ j R n | f | 2 e a|ξ| 2 dξ 1-θ j . ( 3.34)
Since Ω 1 = B 1 and θ = θ(n) ∈ (0, 1), the desired conclusion in the lemma follows from (3.34). This ends the proof of Lemma 3.6.

The next proposition plays a very important role in the proof of Theorem 3.1.

Proposition 3.1. There exist constants C = C(n) > 0 and θ = θ(n) ∈ (0, 1) so that for any a > 0, t > 0 and ε > 0,

R n e -a|x| |f | 2 dx ≤ e C(1+ 1 t +a) 1 + a -n Γ a 2| ln θ| ε R n | f | 2 e t|ξ| 2 dξ + e ε -2| ln θ| a B1 |f | 2 dx , when f ∈ L 2 (R n ) satisfies f ∈ L 2 (e t|ξ| 2 dξ).
To prove Proposition 3.1, we need the following result quoted from [START_REF] Wang | Observability and unique continuation inequalities for the Schrödinger equation[END_REF]: For otherwise, when A = 0, we have that f = 0 over R n , thus the desired inequality is trivial; while when B = 0, we can use the analyticity of f (which follows from Corollary 3.1) to see that f = 0 over R n and then the desired inequality is trivial again. By (3.19), we have that We now estimate the last term of (3.36). According to Lemma 3.6, there is

C 1 = C 1 (n) > 0 and θ = θ(n) ∈ (0, 1) so that j≥1 Ωj+1 e -aj |f | 2 dx ≤ e C1(1+ 1 t ) j≥1 j n-1 e -aj B1 |f | 2 dx θ j R n | f | 2 e t|ξ| 2 dξ 1-θ j ≤ e C1(1+ 1 t ) n! (2/a) n j≥1 e -a 2 j A (B/A) θ j , (3.37)
where A and B are given by (3.35). In the proof of (3.37), we used the inequality: This, along with the fact that A/B > e, yields that This proves the desired inequality for the first case that A/B > e.

j n e -a 2 j ≤ n! (2/
A| ln(B/A)| -a 2| ln θ| ≤ A ≤ B • A/B ≤ e ε -2|
In the second case where A/B ≤ e, we derive directly that

R n e -a|x| |f | 2 dx ≤ R n |f | 2 dx ≤ R n | f | 2 e t|ξ| 2 dξ ≤ e B1 |f | 2 dx ≤ e ε R n | f | 2 e t|ξ| 2 dξ + e ε -2| ln θ| a B1 |f | 2 dx for any ε > 0.
This proves the desired inequality for the second case that A/B ≤ e. Hence, we end the proof of Proposition 3.1.

We now are on the position to show Theorem 3.1.

Proof of Theorem 3.1. (i). Arbitrarily fix u 0 ∈ L 2 (e a|x| dx). Let u(T, x) = (e T u 0 )(x), x ∈ R n . By the Hölder inequality, we have that

R n |u(T, x)| 2 dx ≤ R n |u(T, x)| 2 e a|x| dx 1/2 R n |u(T, x)| 2 e -a|x| dx 1/2 . (3.41)
We will estimate the two terms on right side of (3.41) one by one. For the first term, we apply Lemma 3.1 (with ν = 1) to obtain that

R n |u(T, x)| 2 e a|x| dx ≤ 2 n e 2a 2 T R n |u 0 (x)| 2 e a|x| dx. (3.42)
To estimate the second term (on right side of (3.41)), we first notice that e T ∆ u 0 ∈ L 2 (e T |ξ| 2 dξ), since u 0 ∈ L 2 (e a|x| dx) ⊂ L 2 (R n ). Thus, we can apply Proposition 3.1 (with f = e T u 0 and t = 2T ) to find C = C(n) > 0 and θ = θ(n) ∈ (0, 1) so that for each ε > 0,

R n e -a|x| |u(T, x)| 2 dx ≤ C(T, a, n) ε R n |u 0 (x)| 2 dx + e ε -2| ln θ| a B1 |u(T, x)| 2 dx , (3.43)
where C(T, a, n) = e C(1+

> 0 R n |u(T, x)| 2 dx ≤ C R n |u 0 (x)| 2 e a|x| dx 1/2 ε R n |u 0 (x)| 2 dx + e ε -2| ln θ| a B1 |u(T, x)| 2 dx 1/2 ≤ C ε 1/2 R n |u 0 (x)| 2 e a|x| dx 1/2 ε 1/2 R n |u 0 (x)| 2 dx + ε -1/2 e ε -2| ln θ| a B1 |u(T, x)| 2 dx 1/2 ≤ 2 -1 C ε 1/2 R n |u 0 (x)| 2 e a|x| dx + ε 1/2 R n |u 0 (x)| 2 dx + ε -1/2 e ε -2| ln θ| a B1 |u(T, x)| 2 dx ≤ C ε 1/2 R n |u 0 (x)| 2 e a|x| dx + ε -1/2 e ε -2| ln θ| a B1 |u(T, x)| 2 dx , (3.44)
where C = C(T, a, n) = 2 n/2 e a 2 T C(T, a, n).

Since ε > 0 can be arbitrary taken, we replace ε by ε 2 in (3.44) to get the desired conclusion in (i) of Theorem 3.1.

(ii). Arbitrarily fix u 0 ∈ L 2 ( x ν dx). Let u(T, x) = e T u 0 (x), x ∈ R n . Three facts are given in order. Fact One. Using the inequality:

1 ≤ ε|x| ν + e (1/ε) 1 ν e -|x| for all ε > 0 and x ∈ R n , we find that R n |u(T, x)| 2 dx ≤ ε R n |u(T, x)| 2 |x| ν dx + e (1/ε) 1 ν R n |u(T, x)| 2 e -|x| dx. ( 3.45)
Fact Two. Since 0 < ν ≤ 1, we can apply Lemma 3.2 to find

C 1 = C 1 (n) so that R n |u(T, x)| 2 |x| ν dx ≤ 4 ν+2 Γ(ν/2 + n) 1 + T ν 4 u 0 L 2 ( x ν dx) 2 ≤ C 1 1 + T ν 2 R n |u 0 (x)| 2 x ν dx. (3.46)
Fact Three. We can use (3.43) (with a = 1 and ε = µ) to find C 2 = C 2 (n) and θ = θ(n) ∈ (0, 1) so that for all µ > 0,

R n e -|x| |u(T, x)| 2 dx ≤ e C2(1+ 1 T ) µ R n |u 0 (x)| 2 dx + e µ -2| ln θ| B1 |u(T, x)| 2 dx . (3.47)
To continue the proof, we arbitrarily fix ε ∈ (0, 1). We will first use Fact Three, and then use Fact One and Fact Two. By taking µ = εe -( 1 ε )

1 ν in (3.47), we obtain that e (1/ε)

1 ν R n e -|x| |u(T, x)| 2 dx ≤ e (1/ε) 1 ν e C2(1+ 1 T ) µ R n |u 0 (x)| 2 dx + e µ -2| ln θ| B1 |u(T, x)| 2 dx = e C2(1+ 1 T ) ε R n |u 0 (x)| 2 dx + b ε B1 |u(T, x)| 2 dx , (3.48)
where b ε = exp (1/ε)

1 ν + (ε -2 e 2(1/ε) 1 ν ) | ln θ| .
Meanwhile, one can directly check the following two inequalities:

s 2 ≤ e s for all s > 0;

(3.49)

s + e 3| ln θ|s ≤ e (3| ln θ|+1)s for all s > 0.

Choosing s = ε -1 and s = (1/ε) 1 ν in (3.49) and (3.50) respectively, using 0 < ν ≤ 1, we find that

b ε ≤ exp (1/ε) 1 ν + (e 1/ε e 2(1/ε) 1 ν ) | ln θ| ≤ exp (1/ε) 1 ν + (e 3(1/ε) 1 ν ) | ln θ|
≤ exp e (3| ln θ|+1)(1/ε)

1 ν R n e -|x| |u(T, x)| 2 dx ≤ e C2(n)(1+ 1 T ) ε R n |u 0 (x)| 2 dx + e e (3| ln θ|+1)(1/ε) 1 ν B1 |u(T, x)| 2 dx . (3.52)
Finally, inserting (3.46) and (3.52) into (3.45), we obtain that for some

C 3 = C 3 (n), R n |u(T, x)| 2 dx ≤ C 1 (1 + T ν 2 ) + e C2(1+ 1
T ) ε R n |u 0 (x)| 2 x ν dx + e e (3| ln θ|+1)(1/ε) 1 ν B1 |u(T, x)| 2 dx ≤ (1 + T ν 2 )e C3(1+ 1 T ) ε R n |u 0 (x)| 2 x ν dx + e e (3| ln θ|+1)(1/ε) 1 ν B1 |u(T, x)| 2 dx ,
which leads to the desired conclusion in (ii) of Theorem 3.1. Hence, we end the proof of Theorem 3.1.

Weak observability inequalities with observations on balls

According to Theorem 1.1, it is impossible to recover a solution of (1.1) by observing it over a ball. Thus, two interesting questions arise. First, can we recover a solution of (1.1) over a ball by observing it on another ball? Second, can we have observability inequalities with observations over balls for solutions of (1.1) with some kind of initial values? The answer to the first question is almost negative, while we give partially positive answer for the second question. The first main result of this subsection is stated as follows:

Theorem 3.2. (i) There is an absolute positive constant C so that for all T > 0 and 0 < r < r, (ii) Given T > 0 and r > r > 0, there is no constant C = C(T, r , r, n) so that and so that (ii) By contradiction, we suppose that there were T > 0, r > r > 0 and C = C(T, r , r, n) > 0 so that We would use a constructive method to derive a contradiction with (3.59). For this purpose, we define, for each k ≥ 1,

ϕ(x) = 1, x ∈ B r , 0, x ∈ B c r . Set v = ϕu. Then v satisfies v t -v = -2∇ϕ • ∇u -ϕu in R + × R n , v(0, •) = ϕ(•)u 0 (•) in R n . ( 3
u k (t, x) = 1 (4π(t + 1)) n/2 e -|x 1 -k| 2 +|x | 2 4(t+1) , (t, x) = (t, x 1 , x ) ∈ [0, ∞) × R × R n-1 .
One can easily check that u k is the solution of (1.1) with initial value:

u k (0, x) = 1 (4π) n/2 e -|x 1 -k| 2 +|x | 2 4 , x = (x 1 , x ) ∈ R × R n-1 .
It is clear that {u k (0, •)} k≥1 is uniformly bounded in L 2 (R n ).

We next show that when k is large enough, u k does not satisfy (3.59) (which leads to a contradiction). To this end, we need two estimates: The second main result of this subsection is stated as follows:

Theorem 3.3. (i) There is a generic constant C so that for any T > 0, M > r > 0 and u 0 ∈ L 2 (R n ) with supp u 0 ⊂ B r ,

R n |u(T, x)| 2 dx ≤ 1 T + Cn (M -r) 2 T 0 B M |u(t, x)| 2 dx dt,
where u is the solution to (1.1) with u(0, •) = u 0 (•).

(ii) Assume that 0 ≤ u 0 ∈ L 1 (R n ) so that Br u 0 (x) dx ≥ µ R n u 0 (x) dx for some r > 0 and µ ∈ (0, 1).

Then for any T > 0, M > 0 and any solution u to (1.1) with u(0, •) = u 0 (•), Here, r ∧ M := min{r, M }.

Proof. (i) The proof is similar to that of (i) of Theorem 3.

1 9 2 9 3

 199 Thick sets and several inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.2 Aim, motivation and main result . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.3 Extensions to bounded observable sets . . . . . . . . . . . . . . . . . . . . . . . . . 8 1.4 Plan of the paper . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Proof of Theorem 1.1 Weak interpolation and observability inequalities 22 3.1 Weak interpolation inequalities with observation on the unit ball . . . . . . . . . . 22 3.2 Weak observability inequalities with observations on balls . . . . . . . . . . . . . . 35

Ω

  |u(T, x)| 2 dx ≤ C(n, T, ω, Ω)

Lemma 2 . 2 (

 22 [27, Lemma 1]). Let Φ be an analytic function in D 5 (0) (the disc in C, centered at origin and of radius 5). Let I be an interval of length 1 such that 0 ∈ I. Let Ê ⊂ I be a subset of positive measure. If |Φ(0)| ≥ 1 and M = max |z|≤4 |Φ(z)|, then there exists a generic constant C > 0 such that sup x∈I |Φ(x)| ≤ C/| Ê| ln M ln 2 sup x∈ Ê |Φ(x)|.

  2 and C ≤ 2 + 6C Hold . Now, we derive from (2.49) that R n |u(T, x)| 2 dx ≤ 3e 2C Hold e 36(1+3C Hold ) x)| 2 dxds.

Lemma 3 . 7 ([ 54 ,

 3754 Lemma 3.1]). Let a > 0, b ∈ (0, 1) and θ ∈ (0, 1).Then ∞ k=1 b θ k e -ak ≤ e a | ln θ| Γ a | ln b| | ln x| -a | ln θ| .Proof of Proposition 3.1. Arbitrarily fix a > 0 and t > 0. And then arbitrarily fix f ∈ L 2 (R n ) satisfies f ∈ L 2 (e t|ξ| 2 dξ). It suffices to show the inequality in Proposition 3.1 for the above fixed a, t, f and any ε > 0. Without loss of generality, we can assume thatA := R n | f | 2 e t|ξ|2 dξ = 0 and B := B1 |f | 2 dx = 0. (3.35)

e

  -a|x| |f | 2 dx = B1 e -a|x| |f | 2 dx + j≥1 Ωj+1 e -a|x| |f | 2 dx ≤ B1 |f | 2 dx + j≥1 Ωj+1 e -aj |f | 2 dx. (3.36)

  [START_REF] Phung | Bang-bang property for time optimal control of semilinear heat equation[END_REF]) and (3.51) leads to that e(1/ε)

2

 2 (t, x) dx dt, when u solves (1.1).

B r u 2 (

 2 T, x) dx ≤ C T 0 Br u 2 (t, x) dx dt for any solution u to (1.1).Proof. (i) Arbitrarily fix T > 0 and 0 < r < r. Arbitrarily fix a solution u to (1.1). Let u(0, x) = u 0 (x), x ∈ R n . Choose a C 2 function ϕ on R n so that for some absolute constant C > 0,0 ≤ ϕ(x) ≤ 1 over R n ; |D α ϕ(x)| ≤ C (r -r ) -|α| for all α ∈ N n , with |α| ≤ 2,(3.53)

B r u 2

 2 (T, x) dx ≤ C T 0 Br u 2 (t, x) dx dt for any solution u to (1.1). (3.59)

2 k 2 1 t + 1 + |x 1 - 2 . 3 . 2 .

 2211232 (t, x) dx dt ≤ r n T V n (4π(T + 1)) n e -(k-r)2 2(T +1) , when k > r + 2n(T + 1); (3.60)B r u 2 k (T, x) dx ≥ 1 (4π(T + 1)) n e -(k-r-σ 3) n V n , with σ = r -r, when k > r .(3.61)To show (3.60), we obtain from a direct computation that∂ t (ln u k (t, x)) = -n k| 2 + |x | 2 4(t + 1) 2 for all t ≥ 0, x = (x 1 , x ) ∈ R n ,from which, it follows that when t ≥ 0 andx = (x 1 , x ) ∈ R n , ∂ t (ln u k (t, x)) > 0 ⇐⇒ |x 1 -k| 2 + |x | 2 > 2n(t + 1itfollows that u k does not satisfy (3.59), when k is large enough. This shows the conclusion in (ii) of Theorem 3.2.Hence, we end the proof of Theorem 3.2.The next corollary is a direct consequence of Theorem 3.Corollary Given ν > 0, T > 0 and r > 0, there is no constant C = C(T, r, ν, n) > 0 so thatR n u 2 (T, x)ρ(x) dx ≤ C T 0 Br u 2 (t,x) dx dt for any u solves (1.1), (3.65)where either ρ(x) = x -ν , x ∈ R n , or ρ(x) = e -|x| , x ∈ R n .Remark 3.5. It was announced in[8, p. 384] (without proof ) that given a bounded interval E, there is no positive weight function ρ such that∞ 0 |u(T, x)| 2 ρ(x) dx ≤ C T 0 E |u(t, x)| 2 dx dtfor all solutions of the heat equation in the physical space (0, ∞). The above Corollary 3.2 presents a similar result for the heat equation in the physical space R n .

1 V

 1 n (r ∧ M ) n µ 2 e x)| 2 dx dt.

2 . 2 v 2 t 2

 2222 Arbitrarily fix T > 0, M > > 0 and u 0 ∈ L 2 (R n ) with supp u 0 ⊂ B r . Write u for the solution to (1.1) with u(0, •) = u 0 (•). Choose a C 2 function ϕ over R n so that for some absolute constant C > 0,0 ≤ ϕ(x) ≤ 1 over R n ; |D α ϕ(x)| ≤ C(M -r) -|α| for all α ∈ N n , with |α| ≤ 2 (3.66)and so thatϕ(x) = 0, x ∈ B r , 1, x ∈ B c M . Set v = ϕu. Multiplying (3.54) by v, we find that 1 -v v = -2u∇ϕ • ∇v + 2|∇ϕ| 2 -ϕ ϕ u 2 .(3.67)Integrating (3.67) over (0, T ) × R n , we obtain that 1 2 R n v 2 (T, x) dxt, x)∇ϕ(x) • ∇v(t, x) dx dt + T 0 R n 2|∇ϕ(x)| 2 -ϕ(x) ϕ(x) u 2 (t, x) dx dt.Since the support of u 0 is contained in B r , we have that v(0, •) = 0 over R n . Then by the Hölder inequality, we deduce from (3.68) thatR n v 2 (T, x) dx ≤ x)| 2 -ϕ(x) ϕ(x) u 2 (s, x) dx ds. (3.69) Note that (3.69) is still true if we replace T by any t ∈ (0, T ). This implies that x)| 2 -ϕ(x) ϕ(x) u 2 (t, x) dx dτ dt. (3.70) Since v = u on B c M , it follows from (3.66) and (3.70) that T 0 |x|≥M u 2 (t, x) dx ≤ CnT (M -r) (t, x) dx dt.

  .14) (Due to the continuity of |f | over R n , such y exists.) Because the diameter of Q(j) is √ n, we can use the spherical coordinates centered at y to obtain that

  a) n for all j ≥ 1.

	Meanwhile, by Lemma 3.7 (with b = B/A), we have that
	j≥1	e -a 2 j A(B/A) θ j	≤	e | ln θ| a 2	Γ	a 2| ln θ|	A| ln(B/A)| -a 2| ln θ| .	(3.38)
	About A/B, there are only two possibilities: either A/B > e or A/B ≤ e.
	In the first case when A/B > e, we claim that
	A| ln(B/A)| -a 2| ln θ| ≤ εA + e ε -2| ln θ| a	B for all ε > 0.	(3.39)
	In fact, when ε satisfies that	A| ln(B/A)| -a 2| ln θ| ≤ εA,
	(3.39) is trivial. One the other hand, when ε > 0 satisfies that
			A| ln(B/A)| -a 2| ln θ| > εA,
	we have that			A/B < e ε -2| ln θ|

a

.

  -tv v = -2tu∇ϕ • ∇v + t 2|∇ϕ| 2 -ϕ ϕ u 2 . -ϕ(x) ϕ(x) u 2 (t,x) dx dt. (3.58) Since v = u on B r and v = 0 on B c r , we can use (3.58) and (3.53) to get that

	Inserting (3.57) into (3.56) leads to that
	T	v 2 x) dx
		R n
		T										T
	≤ t 3|∇ϕ(x)| 2 B r 0 R n v 2 (t, x) dx dt + 2 0 R n u 2 (T, x) dx
	≤	1 T	0	T	Br	u 2 (t, x) dx dt +	2 T	0	T	Br\B r	t 3|∇ϕ(x)| 2 -ϕ(x) ϕ(x) u 2 (t, x) dx dt
	≤	1 T	+	8Cn (r -r ) 2	0	T	Br	u 2 (t, x) dx dt,
	which leads to the desired conclusion in (i) of Theorem 3.2.
														.54)
	Multiplying (3.54) by tv leads to
					1 2	tv 2	t -	1 2	v 2 (3.55)
	Integrating (3.55) over (0, T ) × R n , we have that
					1 2 R n	T v 2 (T, x) dx -	1 2	0	T	R n	v 2 (t, x) dx dt +	0	T	R n	t |∇v(t, x)|	2 dx dt	(3.56)
					T
		=					-2tu∇ϕ(x) • ∇v(t, x) dx dt
					0	R n
				+	T		t 2 |∇ϕ(x)|	2 -ϕ(x) ϕ(x) u 2 (t, x) dx dt.
					0		R n
	Meanwhile, by the Hölder inequality, we find that
						T
								-2tu(t, x)∇ϕ(x) • ∇v(t, x) dx dt
					0	R n
					≤		T		t |∇v(t, x)|	2 dx dt +	T	t |∇ϕ(x)|	2 u 2 (t, x) dx dt.	(3.57)
						0	R n						0	R n

In fact, one can choose L = 2(δ + r), γ = r n (2(δ + r)) -n Vn.

An open subset ω ⊂ Ω is said to satisfy the GCC if there exists T 0 > 0 such that any geodesic with velocity one meets ω within time T 0 (see e.g.[START_REF] Laurent | Internal control of the Schrödinger equation[END_REF]).

k max |β|=k D β f L ∞ (Q(j)) f L 2 (Q(j))for all k ≥ 0.(2.20)

Acknowledgment. This work was partially supported by the National Natural Science Foundation of China under grants 11501424 and 11701535.

This implies that k > r + 2n(T + 1) ∂ t (ln u k (t, x)) > 0 for all (t, x) × (0, T ) × B r .

From the above, we see that when k > r + 2n(T + 1), we have that for each x ∈ B r , u k (t, x) is an increasing function of t on [0, T ]. Hence, when k > r + 2n(T + 1),

which leads to (3.60). We next show (3.61). Given b > 0, use B r (b, 0 ) to denote the ball centered at (b, 0, 0, • • • , 0) and of radius r, namely

Let k > r and σ = r -r.

Meanwhile, it is clear that

which leads to the conclusion (i) of Theorem 3.3.

(ii) Let T > 0 and M > 0 arbitrarily given. Arbitrarily fix u 0 so that

u 0 (x)dx for some r > 0 and µ ∈ (0, 1).

(3.73)

Write u for the solution to (1.1) with u(0, •) = u 0 (•). We first prove that when 0 < M ≤ r,

For this purpose, we need the following two estimates:

To show (3.75), we observe that

Since M ≤ r, it follows from (3.78) that

which leads to (3.75). We now show (3.76). By (3.77) and the Young inequality, we have that

R n u 0 (x) dx, which leads to (3.76). Next, by (3.75) and (3.76), we see that