Térence Bayen
email: bayen@clipper.ens.fr

Thomas Lachand-Robert
email: thomas.lachand-robert@univ-savoie.fr

Édouard Oudet
email: edouard.oudet@univ-savoie.fr

Analytic parametrization and volume minimization of three dimensional bodies of constant width

We present a complete analytic parametrization of constant width bodies in dimension 3 based on the median surface: more precisely, we define a bijection between some space of functions and constant width bodies. We compute simple geometrical quantities like the volume and the surface area in terms of those functions. As a corollary we give a new algebraic proof of Blaschke's formula. Finally, we derive weak optimality conditions for convex bodies which minimize the volume among constant width bodies.

Introduction

A body (that is, a compact connected subset K of R n ) is said to be of constant width α if its projection on any straight line is a segment of length α ∈ R + , the same value for all lines. This can also be expressed by saying that the width map

w K : ν ∈ S n-1 -→ max x∈K ν • x -min x∈K ν • x (1)
has constant value α. This is also equivalent to the geometrical fact that two parallel support hyperplanes on K are always separated by a distance α, independent of their direction.

Obvious bodies of constant width are the balls; but they are many others. These bodies, also called orbiforms in dimension two, or spheroforms in dimension three (as in [START_REF] Bonnesen | Theory of convex bodies[END_REF]), have many interesting properties and applications. Orbiforms in particular have been studied a lot during the nineteenth century and later, particularly by Frank Reuleaux, whose name is now attached to those orbiforms you get by intesecting a finite number of disks of equal radii α, whose center are vertices of a regular polygon of diameter α.

Among the oldest problems related to these bodies of constant width are the question of which are those with maximal or minimal volume, for a given value of the width α. It is not difficult to prove that the ball (of radius α/2) has maximal volume: this follows from the isoperimetric inequality.

On the other hand, the question of which body of constant width α has minimal volume proved to be much more difficult. First notice that this problem is not correctly stated: indeed, one can remove the interior of a body to decrease its volume, without changing its constant width property. Therefore, we need to add an additional requirement for the problem to make sense (even though this is not needed for the maximization problem). The problem is well-posed if we consider only convex bodies, and this is the usual statement considered.

So let us define formally the following class:

W α := K ⊂ R n ; K compact convex and ∀ν ∈ S n-1 , w K (ν) = α . (2)
The problem of interest is now to minimize the n-dimensional volume, denoted by |K| hereafter:

Find K * ∈ W α such that |K * | = min K∈Wα |K| . (3)
Note that the existence of K * is easy to establish. Indeed W α is a compact class of sets for most reasonable topologies (for instance the Haussdorff topology), and the volume is a continuous function.

In dimension two, the problem was solved by Lebesgue and Blaschke: the solution turns out to be a Reuleaux triangle.

In dimension three, the problem is still open. Indeed the mere existence of non trivial three-dimensional bodies of constant width is not so easy to establish. In particular, no finite intersection of balls has constant width (except balls themselves), a striking difference with the two-dimensional case.

A simple construction is to consider a two dimensional body of constant width having an axis of symmetry (like the Reuleaux triangle for instance): the corresponding body of revolution obtained by rotation around this axis is a spheroform. F. Meissner proved that the rotated Reuleaux triangle has the smaller volume among bodies of revolution in W α .

Later on he was able to construct another spheroform (usually called "Meissner's tetrahedron") which does not have the symmetry of revolution. The volume of this body is smaller than any other known of constant width, so it is a good candidate as a solution to the problem [START_REF] Chakerian | Convexity and its Applications, Chap. Convex bodies of constant width[END_REF]. We describe this body in more details later on in this paper. Let us just say for the moment that it looks like an intersection of four balls centered on the vertices of a regular tetrahedron, but some of the edges are smoothed; in particular, it doesn't have all the symmetries of a regular tetrahedron.

In this paper we first present a complete analytic parametrization of constant width bodies in dimension 3 based on the median surface. More precisely, we define a bijection between the space of functions C 1,1 σ (Ω) and constant width bodies. Then, we compute simple geometrical quantities like the volume and the surface area in terms of those functions. As a corollary we give a new algebraic proof of Blaschke's formula and compute the surface and the volume of Meissner's tetrahedron. Finally, we derive weak optimality conditions for the problem (3).

The Median Surface

In this section, we introduce a geometrical tool, which we call the median surface.

Definition and basics

For a convex body K, we say that a hyperplane H is a hyperplane of support for K at x, if x ∈ K ∩ H and K is included in one of the half-spaces limited by H. If ν ∈ S n-1 is a normal vector to H, pointing outside the half space containing K, we say that ν is an outward support vector at x. Obviously if K is smooth (that is, has a differentiable boundary), then ν is just the outward unit normal at x. In this particular case, there is a map x → ν which is usually called the Gauss map.

The reverse Gauss map (which is well defined for a body of constant see for instance [START_REF] Lachand-Robert | Bodies of constant width in arbitrary dimension[END_REF]), satisfies R K (ν) -R K (-ν) = αν for all ν. We may now introduce a parallel surface to ∂K. Consider, for all ν ∈ S n-1 the point,

M K (ν) := R K (ν) - α 2 ν = R K (-ν) + α 2 ν.
Notice that M K (-ν) = M K (ν). The set of points M K (ν) is called the median surface of the body K.

Let us recall from [START_REF] Lachand-Robert | Bodies of constant width in arbitrary dimension[END_REF] one geometrical characterization of constant width bodies:

Theorem 1 Let K be closed subset of R n . Then K has constant width α if and only if it satisfies:

∀ν ∈ S n-1 , ∃x ν ∈ K, x ν + αν ∈ K and ∀y ∈ K, (y -x ν ) • ν ∈ [0, α]. (4)

Construction of constant width sets

We present in this section a construction process of constant width bodies starting from an appropriate surface, which will be their median surface. More precisely:

Theorem 2 Let α > 0 be given and M : S n-1 → R n be a continuous application satisfying

∀ν ∈ S n-1 , M(-ν) = M(ν); (5)
∀ν 0 , ν 1 ∈ S n-1 , M(ν 1 ) -M(ν 0 ) • ν 0 ≤ α 4 |ν 1 -ν 0 | 2 . ( 6
)
Define a subset K ⊂ R n as follows:

K := M(ν) + tν ; ν ∈ S n-1 , t ∈ 0, α 2 . ( 7
)
Then K is a convex body of constant width α, and M K ≡ M. Conversely, any convex body of constant width α can be described by [START_REF] Harrell | Calculations for Convex Bodies. Example: The rotated Reuleaux triangle[END_REF], where M = M K .

Notice that we could have defined K by

K := M(ν) + tν ; ν ∈ S n-1 , t ∈ - α 2 , α 2 . ( 8
)
This is equivalent to [START_REF] Harrell | Calculations for Convex Bodies. Example: The rotated Reuleaux triangle[END_REF], due to [START_REF] Dugundji | Fixed point theory[END_REF]. Similarly, taking (5) into consideration, we can rewrite [START_REF] Harrell | A direct proof ot a theorem of Blaschke and Lebesgue[END_REF] with -ν 0 , -ν 1 . We deduce that for an application M satisfying (5), ( 6) is equivalent to:

∀ν 0 , ν 1 ∈ S n-1 , M(ν 1 ) -M(ν 0 ) • ν 0 ≤ α 4 |ν 1 -ν 0 | 2 . ( 9
)
In order to prove this theorem, we make use of a lemma:

Lemma 1 Under the assumptions of Theorem 2, let K be defined by [START_REF] Harrell | Calculations for Convex Bodies. Example: The rotated Reuleaux triangle[END_REF].

Then R n = {M(ν) + tν ; ν ∈ S n-1 , t ∈ R + }, K is compact, and ∂K ⊂ M(ν) + α 2 ν ; ν ∈ S n-1 . (10)
(It will come from Theorem 2 that there is actually equality for the sets in [START_REF] Gilbarg | Elliptic Partial Differential Equations of Second Order[END_REF].) [START_REF] Dugundji | Fixed point theory[END_REF]. We consider some x ∈ R n , and assume by contradiction that x / ∈ Q(S n-1 × R). For each ν, define x ν as the projection of x onto the straight line M(ν) + Rν. Our assumption implies x = x ν . Moreover

Proof. Consider the map Q : S n-1 × R → M(ν) + tν where M satisfies (5). Since M is continuous, K = Q S n-1 × [0, α 2 ] is a compact set. Let us first prove that Q(S n-1 × R + ) = R n . Note that Q(S n-1 × R + ) = Q(S n-1 × R) from
x ν = M(ν) + t ν ν where t ν := ν • x -M(ν)
as a classical property of the projection.

Since in particular x = M(ν) for all ν, we can define a map f :

S n-1 → S n-1 by f (ν) := (x -M(ν))/ |x -M(ν)|.
Note that f is continuous, and f (-ν) = f (ν). Such a map has an even topological degree, and in particular has a fixed point [START_REF] Dugundji | Fixed point theory[END_REF]. Therefore there exists some ν such that f (ν) = ν. For such a ν, we get x ν = x, a contradiction.

We now turn on the proof of [START_REF] Gilbarg | Elliptic Partial Differential Equations of Second Order[END_REF]. Consider some x ∈ ∂K. In particular, x ∈ K, so x = M(ν 0 ) + t 0 ν 0 for some ν 0 and

t 0 ∈ [-α 2 , α 2 ]
. There exists a sequence (x n ) ⊂ R n \ K with limit x. From our previous study, we known that x n = M(ν n ) + t n ν n for some ν n ∈ S n-1 and t n ∈ R + . The assumption x n / ∈ K implies t n > α/2, but on the other hand the sequence (t n ) is bounded since (x n ) is bounded and M(S n-1 ) is compact. Therefore we may assume that the sequences (ν n ) and (t n ) are convergent. Let us denote by ν ∞ and t ∞ ≥ α 2 their limits. Since M is continuous, we have

x = M(ν ∞ ) + t ∞ ν ∞ . In particular, M(ν 0 ) = x -t 0 ν 0 = M(ν ∞ ) + t ∞ ν ∞ -t 0 ν 0 .
Let us assume with no loss of generality that ν 0 • ν ∞ ≥ 0 (otherwise we just have to change ν 0 to -ν 0 and t 0 to -t 0 ). We write [START_REF] Harrell | A direct proof ot a theorem of Blaschke and Lebesgue[END_REF] for ν ∞ , ν 0 , so

M(ν 0 ) -M(ν ∞ ) • ν ∞ ≤ α 4 |ν ∞ -ν 0 | 2 = α 2 (1 -ν 0 • ν ∞ ) ⇐⇒ t ∞ -t 0 ν 0 • ν ∞ ≤ α 2 (1 -ν 0 • ν ∞ ) ⇐⇒ t ∞ ≤ α 2 - α 2 -t 0 ν 0 • ν ∞ ≤ α 2 since t 0 ∈ -α 2 , α 2 . This proves that t ∞ = α 2 . Hence x ∈ Q(S n-1 , α 2 ).
Proof of Theorem 2. We begin with the proof of the reciprocal statement in the Theorem. Let K be a body of constant width. We already know that its median surface M = M K is continuous and satisfies [START_REF] Dugundji | Fixed point theory[END_REF]. Since

M K (ν) = R K (ν) -α 2 ν, and R K (ν 1 ) • ν 0 ≤ R K (ν 0 ) • ν 0 from the definition of R K , we have M(ν 1 ) -M(ν 0 ) • ν 0 ≤ α 2 (1 -ν 0 • ν 1 ) = α 4 |ν 1 -ν 0 | 2 .
This proves [START_REF] Harrell | A direct proof ot a theorem of Blaschke and Lebesgue[END_REF]. Since K is convex and M K (ν) + α 2 ν ∈ K, M K (ν) -α 2 ν ∈ K, we see that K contains the right hand side of [START_REF] Hilbert | Elf Eigenschaften der Kugel[END_REF]. Now let x ∈ K be given, and let y be the farthest point from x in K. Define ν := (yx)/ |y -x|. For any z ∈ K, we have

y • ν = |y -x| + x • ν ≥ |z -x| + x • ν ≥ (z -x) • ν + x • ν = z • ν so y = R K (ν) = M K (ν) + α 2 ν. Hence x = M K (ν) + tν with t = α 2 -|x -y|. Since |x -y| ≤ α, we have |t| ≤ α
2 , which concludes the proof of [START_REF] Hilbert | Elf Eigenschaften der Kugel[END_REF]. We now prove the direct statement in the Theorem. So consider a map M satisfying ( 5) and ( 6), and K be defined by [START_REF] Harrell | Calculations for Convex Bodies. Example: The rotated Reuleaux triangle[END_REF] (or (8) equivalently). In view of Theorem 1, we need to prove [START_REF] Do | Differential Geometry of Curves and Surfaces[END_REF].

Let ν ∈ S n-1 be given. Consider x ν := M(ν) -α 2 ν, so that x ν + αν = M(ν) + α 2 ν ∈ K from its definition. Consider any y ∈ K, so that y = M(ν) + tν. Changing ν and t to their opposite, if necessary, we may assume that ν • ν ≥ 0. Note that

(y -x ν ) • ν = M(ν) -M(ν) • ν + tν • ν + α 2 .
Using [START_REF] Hiriart-Urruty | Fundamentals of convex analysis[END_REF] with ν 0 = ν, ν 1 = ν, we get

- α 2 (1 -ν • ν) ≤ M(ν) -M(ν) • ν ≤ α 2 (1 -ν • ν). Hence, since t ∈ -α 2 , α 2 : 0 ≤ (t + α 2 ) ν • ν ≤ (y -x ν ) • ν ≤ α + (t - α 2 ) ν • ν ≤ α.
This concludes the proof of the theorem.

Applications M satisfying ( 5) and (6) will play an important role in the remaining of this paper. So let us give a few additionnal properties on them. We start here with simple inequalities, and will consider what happens on a differential level in the next section. Note that all these results apply in particular to the median surface of any convex body of constant width according to Theorem 2.

Lemma 2 Let M be a continuous application satisfying (5) and (6). Then M is α 2 -lipschitzian:

∀ν 0 , ν 1 ∈ S n-1 , |M(ν 1 ) -M(ν 0 )| ≤ α 2 |ν 1 -ν 0 | . ( 11
)
and satisfies:

∀ν 0 , ν 1 ∈ S n-1 , M(ν 1 ) + α 2 ν 1 -M(ν 0 ) - α 2 ν 0 ≤ α. (12)
Proof. According to Theorem 2, M is the median surface of some K ∈ W α defined by [START_REF] Harrell | Calculations for Convex Bodies. Example: The rotated Reuleaux triangle[END_REF]. Since K contains M(ν) + α 2 ν for any ν, and has diameter α, we get [START_REF] Rockafellar | Convex Analysis[END_REF].

Squaring the left hand side of ( 12) and expanding it, we get

|M(ν 1 ) -M(ν 0 )| 2 -α(M(ν 1 ) -M(ν 0 )) • (ν 1 -ν 0 ) ≤ α 2 4 |ν 1 + ν 0 | 2 (13) since |ν 1 -ν 0 | 2 + |ν 1 + ν 0 | 2 = 4.
The above relation is true for any pair of unit vectors, so we can write it for (ν 1 , -ν 0 ) and (-ν 1 , ν 0 ). We get, taking (5) into account:

|M(ν 1 ) -M(ν 0 )| 2 -α(M(ν 1 ) -M(ν 0 )) • (ν 1 + ν 0 ) ≤ α 2 4 |ν 1 -ν 0 | 2 |M(ν 1 ) -M(ν 0 )| 2 + α(M(ν 1 ) -M(ν 0 )) • (ν 1 + ν 0 ) ≤ α 2 4 |ν 1 -ν 0 | 2 .
Summing these relations yields (11).

Smooth median surface

In this section we reduce [START_REF] Harrell | A direct proof ot a theorem of Blaschke and Lebesgue[END_REF] to local differential properties. This is easy whenever M is differentiable, but requires more involved statements in the general case. Note that M will always be defined on the sphere S n-1 , and if differentiable, its derivative DM(ν) is defined on the tangent space to the sphere at ν, which is simply ν ⊥ := {w ∈ R n ; w • ν = 0}. In the following proposition, we consider C 2 maps ν : [0, 1] → S n-1 , and ˙ ν is the derivative of ν. Notice that only the end point ν(0) and the corresponding derivatives do matter.

Proposition 1 Let M : S n-1 → R n be given. Then M satisfies (9) if and only if it satisfies ∀ ν ∈ C 2 ([0, 1]; S n-1 ), lim sup t > →0 1 t 2 M( ν(t)) -M( ν(0)) • ν(0) ≤ α 4 ˙ ν(0) 2 . ( 14
)
If M is differentiable, then (14) is equivalent to ∀ν 0 ∈ S n-1 , ∀w ∈ ν ⊥ 0 , ν 0 • DM(ν 0 )w = 0 and |w • DM(ν 0 )w| ≤ α 2 |w| 2 . ( 15
)
We will shorten [START_REF] Yaglom | Convex Figures[END_REF] in the following by writing it ν 0 •DM(ν 0 ) = 0 (as vectors) and ±DM(ν 0 ) ≤ α 2 Id (as matrices). This expresses the fact that ν 0 is the normal vector to the surface ν 0 → M(ν 0 ) at M(ν 0 ), and that the absolute values of the curvature radii does not exceed α 2 . (See also the parametric equivalent in the next section.)

Proof. Assume first that M satisfies [START_REF] Hiriart-Urruty | Fundamentals of convex analysis[END_REF]. Let ν ∈ C 2 ([0, 1]; S n-1 ), and define

ν 0 := ν(0) for short. Note that | ν(t)| 2 = 1 for all t, so ∀t, ν(t) • ˙ ν(t) = 0 and ν(t) • ¨ ν(t) = -˙ ν(t) 2 . ( 16
)
In particular a Taylor expansion near t = 0 yields

ν(t) • ν 0 = ν 0 + t ˙ ν(0) + t 2 2 ¨ ν(0) + o(t 2 ) • ν 0 = 1 - t 2 2 ˙ ν(0) 2 + o(t 2 ).
Using [START_REF] Hiriart-Urruty | Fundamentals of convex analysis[END_REF] with ν(t) and ν 0 , we get:

M( ν(t)) -M(ν 0 ) •ν 0 ≤ α 2 (1 -ν(t) • ν 0 ) ≤ α 4 t 2 ˙ ν(0) 2 + o(t 2 ).
Dividing by t 2 , we get ( 14).

If M is differentiable and w ∈ ν ⊥ 0 , consider ν(t) := p S n-1 (ν 0 + tw) where p S n-1 : x → x/ |x| is the projection on the sphere. So ν(0) = ν 0 and ˙ ν(0) = w. Hence we have

M( ν(t)) • ν 0 = M(ν 0 ) • ν 0 + tν 0 • DM(ν 0 )w + o(t) so (14) clearly implies ν 0 • DM(ν 0 )w = 0.
Assume for a moment that M is twice differentiable and satisfies [START_REF] Toyama | Self-Parallel Constant Mean Curvature Surfaces[END_REF]. We already know that ν 0 • DM(ν 0 )w = 0 for any ν 0 and any w ∈ ν ⊥ 0 . Therefore we have, for any ν ∈ C 2 ([0, 1]; S n-1 ):

∀t, 0 = ν(t) • DM( ν(t)) ˙ ν(t).
Differentiating this relation with respect to t, we get

0 = ˙ ν(t) • DM( ν(t)) ˙ ν(t) + ν(t) • D 2 M( ν(t))( ˙ ν(t), ˙ ν(t)) since ν(t) • DM( ν(t)) = 0. Considering t = 0 and w := ˙ ν(0) ∈ ν ⊥ 0 yields ∀w ∈ ν ⊥ 0 , w • DM(ν 0 )w = -ν 0 • D 2 M(ν 0 )(w, w). ( 17
)
Therefore a Taylor expansion yields

M( ν(t)) • ν 0 = M(ν 0 ) • ν 0 + 1 2 t 2 ν 0 • D 2 M(ν 0 )(w, w) + o(t 2 ) = M(ν 0 ) • ν 0 - 1 2 t 2 w • DM(ν 0 )w + o(t 2 ). (18)
It is now clear that ( 14) implies ( 15). If M is not twice differentiable, we use an approximation argument as follows. For any β > α and any ε > 0, there exists an approximating map

M ε ∈ C 2 (S n-1 , R n ), such that M -M ε W 1,∞ (S n-1 ;R n ) ≤ ε (19)
and M ε satisfies ( 14) with α replaced by β. Hence M ε satisfies (15), also with α replaced by β. Letting ε go to zero and using (19), we deduce that M satisfies ( 15) with α replaced by β. Since this holds for any β > α, it holds for α as well.

Conversely, if M is differentiable and satisfies [START_REF] Yaglom | Convex Figures[END_REF], let us prove that it satisfies [START_REF] Toyama | Self-Parallel Constant Mean Curvature Surfaces[END_REF]. Using exactly the same approximation, we see that we just have to prove that for M twice differentiable. In such a case, ( 15) implies (17). Hence the Taylor expansion (18) holds true. This yields [START_REF] Toyama | Self-Parallel Constant Mean Curvature Surfaces[END_REF].

Let us now prove the reverse statement of the proposition, that is, a map M satisfying (14) also satisfies [START_REF] Hiriart-Urruty | Fundamentals of convex analysis[END_REF]. Again it is enough to prove it for a twice differentiable map, for (19) implies in particular uniform convergence of M ε to M.

So let us consider two vectors ν 0 , ν 1 in S n-1 and prove [START_REF] Hiriart-Urruty | Fundamentals of convex analysis[END_REF]. We consider a geodesic path ν ∈ C 2 ([0, 1]; S n-1 ) such that ν(0) = ν 0 and ν(1) = ν 1 . Such a path satisfies ν(t) ∈ (Rν 0 + Rν 1 ), and

ν 0 • ˙ ν(t) ≤ 0 for all t. The function f : t → ν 0 •M( ν(t)) has derivative f ′ (t) = ν 0 •DM( ν(t)) ˙ ν(t). Since ν(t) ∈ (Rν 0 + Rν 1 ), we have ν 0 = (ν 0 • ν(t)) ν(t) + (ν 0 • ˙ ν(t)) ˙ ν(t) 2 ˙ ν(t).
Taking [START_REF] Yaglom | Convex Figures[END_REF] and

ν 0 • ˙ ν(t) ≤ 0 into account, we get |f ′ (t)| = ν 0 • ˙ ν(t) ˙ ν(t) 2 ˙ ν(t) • DM( ν(t)) ˙ ν(t) ≤ - α 2 (ν 0 • ˙ ν(t)). Therefore M(ν 1 ) -M(ν 0 ) • ν 0 = |f (1) -f (0)| ≤ - α 2 1 0 (ν 0 • ˙ ν(t)) dt = α 2 (1 -ν 0 • ν 1 ).
This completes the proof of the proposition.

Remark 2.A. Observe that ( 14) is equivalent to

∀ ν ∈ C 2 ([0, 1]; S n-1 ), lim sup t > →0 1 t 2 M( ν(t)) -M( ν(0)) • ν(t) ≤ α 4 ˙ ν(0) 2 . ( 20
)
Indeed we just have to prove that for a smooth M again. Then we may rewrite (18) with ν 0 and ν(t) reversed:

M(ν 0 ) • ν(t) = M( ν(t)) • ν(t) - 1 2 t 2 ˙ ν(t) • DM( ν(t)) ˙ ν(t) + o(t 2 ). Since w = ˙ ν(0) = ˙ ν(t) + O(t)
and DM is continuous, we get by substracting (18):

M( ν(t)) -M( ν(0)) • ν(t) -ν(0) = o(t 2 )
as t → 0. This proves that the limits on the right hand sides in ( 14) and (20) are equal.

Let us recall a classical geometrical definition: two smooth oriented surfaces S and S ′ are said to be parallel at distance δ if S ′ is the image of S through the map x → x + δ n S (x), where n S is the normal vector field on S. It is classical that the normal vector on S ′ at x + δ n S (x) is actually n S (x). (We will give a proof of this result in the next section.) In particular, S is also a surface parallel to S ′ , at distance -δ. Moreover, if S have well defined radii of curvature ρ i (x) (i = 1, 2), then S ′ also have radii of curvature at x + δ n S (x), equal to ρ i (x) + δ.

So we see that for a body K of constant width α with median surface M K , the median surface and the boundary ∂K are parallel at distance ±α, whenever they are smooth. In general, these surfaces are not smooth, but only have Lipschitz regularity, though.

Parametrizations

In this section, we give a parametrization of the median surface of a body K of constant width. This provides a simple parametrization of the boundary of K, and gives a simple formula to compute the volume and surface area of K.

From now on we focus on the three-dimensional setting. A similar work can easily be done in dimension two, but the properties of orbiforms are already quite well known.

Isothermal parametrization of the sphere

Let us start with a parametrization of the unit sphere S 2 in the form (u, v) ∈ Ω → ν(u, v), where Ω is some subset of R 2 . We assume that this parametrization is isothermal, that is, satisfies for all (u, v) ∈ Ω:

∂ u ν(u, v) • ∂ v ν(u, v) = 0 and |∂ u ν(u, v)| = |∂ v ν(u, v)| =: 1 λ(u, v) .
(21) We also assume that the map ν : Ω → S 2 is injective and almost surjective, that is, its image set is equal to S 2 except possibly a finite number of points.

An example of such a parametrization is

(u, v) ∈ (R/2πZ) × R -→ cos u cosh v , sin u cosh v , tanh v (22)
and in such a case λ(u, v) = cosh v, and ν(Ω) = S 2 \ {(0, 0, ±1)}. However we do not rely on this particular form in the following.

For technical reasons, we will also assume that λ satisfies, for all values of (u, v), the identity

λ 2 ∇ • λ -1 ∇λ = λ∆λ -|∇λ| 2 = 1. (23)
(Gradient and Laplacian taken relative to (u, v).) This is clearly true for the particular parametrization given above.

Let us shorten the notations by not writing the dependencies on the parameters (u, v). We introduce the unit vectors ν u := λ∂ u ν, ν v := λ∂ v ν. Since ν is also a unit vector, we have ν • ∂ u ν = 0, so ν • ν u = 0; and similarly ν • ν v = 0. Hence the family (ν, ν u , ν v ) is an orthonormal basis of R 3 , taking (21) into account.

Lemma 3 For such an isothermal parametrization of the unit sphere, we have

∂ u ν u = -λ -1 ν + λ -1 ∂ v λ ν v (24) ∂ v ν u = -λ -1 ∂ u λ ν v (25) ∂ u ν v = -λ -1 ∂ v λ ν u (26) ∂ v ν v = -λ -1 ν + λ -1 ∂ u λ ν u (27) Proof. Since ν • ∂ u ν = 0, we get by differentiating ν • ∂ 2 uv ν = -∂ u ν • ∂ v ν = 0, so ∂ 2
uv ν has the form αν u + βν v . On the other hand

∂ 2 uv ν = ∂ u (∂ v ν) = ∂ u (λ -1 ν v ) = λ -1 ∂ u ν v -λ -2 ∂ u λ ν v .
Since

|ν v | = 1 implies ν v •∂ u ν v = 0, we get β = ∂ 2 uv ν •ν v = -λ -2 ∂ u λ. Similarly α = -λ -2 ∂ v λ. Putting this relation in the value of ∂ 2
uv ν above, we deduce (26). We get (25) using ∂ 2 uv ν = ∂ v (∂ u ν) in the same way. Differentiating the three relations

|ν u | 2 = 1, ν • ν u = 0 and ν u • ν v with respect to u, we get ν u • ∂ u ν u = 0, ν • ∂ u ν u = -ν u • ∂ u ν = -λ -1 |ν u | 2 = -λ -1 and ν v • ∂ u ν u = -ν u • ∂ u ν v = λ -1 ∂ v λ.
This gives (24). The proof of (27) is similar.

Let us finish this section with a note about the antipodal symmetry on S 2 that we will use in the following sections. There must be some involutive map σ : Ω → Ω such that ν • σ(u, v) = -ν(u, v) for all (u, v) ∈ Ω. For instance with the parametrization (22) we have

ν(u + π, -v) = -ν(u, v) (28)
so σ : (u, v) → (u + π, -v). We will call this map the antipodal symmetry of the parametrization. In the following, we will always assume that σ is C 1 and is consistent with the isothermal parametrization, that is satisfies:

λ • σ = λ. ( 29
) Since ν = -ν • σ, we have ∂ u ν = -∂ u σ∂ u ν • σ.
Considering the norm of both sides, we deduce with (29) that |∂ u σ| = 1. Similarly we have |∂ v σ| = 1. Since σ ∈ C 1 , we see that

∂ u σ = const. = ±1 and ∂ u σ = const. = ±1. (30)
These relations, together with the definition of ν u , ν v and (29) imply

ν u • σ = -∂ u σν u , ν v • σ = -∂ v σν v . (31)

Parametrization of the median surface

Since the three vectors (ν, ν u , ν v ) are independent, any point P ∈ R 3 can be written in the form

P = hν + h 1 ν u + h 2 ν v . If h, h 1 , h 2
are actually some smooth functions of (u, v), P depends on (u, v) and describe a surface. In this section we investigate the conditions on h, h 1 , h 2 ensuring that such a surface is the median surface of a spheroform, with support vector ν(u, v) at P (u, v).

Proposition 2 Given an isothermal parametrization ν : Ω → S 2 of the sphere, let K be a strictly convex body. There exists a

C 1 map h : Ω → R such that R K (ν) = M(h)(u, v) for all ν = ν(u, v), where
M(h) : Ω -→ R 3 (u, v) -→ h ν + λ∂ u h ν u + λ∂ v h ν v . ( 32
)
Proof. For any given ν = ν(u, v), consider P (u, v) := R K (ν(u, v)). Since the three vectors (ν, ν u , ν v ) are independent, P (u, v) can be written in the form P = hν + h 1 ν u + h 2 ν v , for some functions h, h 1 , h 2 of (u, v). These functions are continuous since R K is continuous.

Note that h K (ν(u, v)) = ν(u, v) • P (u, v) = h(u, v
). So h is just the support function of K, and in particular is of class C 1 . Moreover we have from the definition of R K :

∀(u 1 , v 1 ) ∈ Ω, P (u 1 , v 1 ) • ν ≤ P • ν.
(All values of the functions are at (u, v), unless otherwise specified.) Let us write this relation with

u 1 = u + t, v 1 = v.
For small values of t, we have from (24-27):

ν(u + t, v) = ν + tλ -1 ν u + o(t), ν u (u + t, v) = ν u -tλ -1 ν -∂ v λν v + o(t), ν v (u + t, v) = ν v -tλ -1 ∂ v λν u + o(t). Also since h is of class C 1 , we have h(u + t, v) = h + t∂ u h + o(t). Hence 0 = P • ν -h ≥ P (u 1 , v 1 ) • ν -h ≥ t ∂ u h -λ -1 h 1 (u + t, v) + o(t).
Passing to the limit t = 0 with either t > 0 or t < 0, we deduce that

h 1 (u, v) = λ∂ u h. Similarly h 2 = λ∂ v h.
Remark 3.B. Notice that M(h) is obviously linear with respect to h. Since in the previous proposition, h = M(h) • ν = R K • ν is the support function of K, then the mapping from K to h is additive (with respect to the Minkowski addition). However, not any h yields an interesting body K. In particular, if h(u, v) = w • ν(u, v) for some fixed vector w ∈ R 3 , then

∂ u h = w • ∂ u ν = λ -1 w • ν u , so M(h) =
w is constant, and the corresponding body K reduces to a point. Due to the additivity property, we see that adding w • ν to some given h is equivalent to a translation of the corresponding body K by the vector w.

We prove in the next theorem that for a constant width body, the corresponding function h is actually C 1,1 (the derivatives are lipschitzian). Here and in the following, differential operators like ∇ (gradient) or ∆ (laplacian) are taken relative to the variables (u, v). We denote by ∇ ⊥ the operator (-∂ v , ∂ u ). Whenever h is twice differentiable, we denote by D 2 h the 2 × 2 matrix of its second-order derivatives (hessian matrix).

An inequality like D 2 h(u, v) ≤ A, where A is also a 2 × 2 symmetrical matrix, means that the difference A -D 2 h(u, v) is nonnegative definite. For h ∈ C 1,1 only, the second-order derivatives do not necessarily exists, but the Taylor expansion

T [h](u, v; ξ, η) := h(u + ξ, v + η) -h(u, v) -ξ∂ u h(u, v) -η∂ v h(u, v) is of order O(ξ 2 + η 2 ) for (ξ, η) small.
Definition 1 We shall say that D 2 h(u, v) ≤ A = (a i,j ) in a generalized sense, if the following occurs:

lim sup (ξ,η)→(0,0) T [h](u, v; ξ, η) -1 2 a 11 ξ 2 + 2a 12 ξη + a 22 η 2 ξ 2 + η 2 ≤ 0. ( 33
)
Similarly we say that D 2 h(u, v) ≥ A in a generalized sense, if a similar property holds with a limit-inf ≥ 0 instead.

Clearly this is the same as the usual meaning for a twice-differentiable function h, since

T [h](u, v; ξ, η) = 1 2 ξ 2 ∂ 2 uu h(u, v) + ξη∂ 2 uv h(u, v) + 1 2 η 2 ∂ 2 vv h(u, v) + o(ξ 2 + η 2 )
in that case.

Definition 2 Given an isothermal parametrization ν : Ω → S 2 of the sphere, let σ be its antipodal symmetry. Let C 1,1 σ (Ω) be the set of all C 1,1 maps h

: Ω → R such that h • σ = -h. ( 34
)
Let C 1,1 σ,α (Ω) be the subset of functions h ∈ C 1,1 σ (Ω) satisfying everywhere on Ω in a generalized sense (see Definition 1 above):

- α 2λ 2 Id ≤ U[h] ≤ α 2λ 2 Id ( 35
)
where

U[h] := D 2 h + λ -2 h Id + λ -1 ∇λ ⊗ ∇h -λ -1 ∇ ⊥ λ ⊗ ∇ ⊥ h. (36)
Theorem 3 Given an isothermal parametrization of the sphere, let C 1,1 σ,α (Ω) be given by the Definition 2 above.

Then an application M : S 2 → R 3 is the median surface of a spheroform if and only if there exists h ∈ C 1,1 σ,α (Ω) such that M(ν) = M(h)(u, v) for all ν = ν(u, v), where the map M(h) : Ω → R 3 is defined by (32). In this case, the map M(h + α 2 ) : Ω → R 3 describes all but a finite number of the points on ∂K.

The restriction about exceptional points on ∂K comes from the fact that ν(Ω) equals S 2 , excepts some exceptional points. (The points (0, 0, ±1) with the parametrization (22).)

Proof. Given h ∈ C 1,1 σ,α (Ω), define M : S 2 → R 3 by M(ν) = M(h)(u, v) for all ν = ν(u, v).
Let us prove that M is the median surface of some spheroform. In view of Theorem 2, Proposition 1 and Remark 2.A, we just have to prove ( 5) and (20). From (29-31) we get Let us now prove (20). Any ν ∈ C 2 ([0, 1]; S 2 ) can be written in the parametrization as

M(-ν) = M(h) • σ(u, v) = -h • σν + λ∂ u h • σ∂ u σν u + λ∂ v h • σ∂ v σν v .

But (34) implies in particular ∂

u h • σ = ∂ u h/∂ u σ
ν(t) = ν(u(t), v(t)) where u(t), v(t) ∈ C 2 ([0, 1]). If ν(0) = ν(u 0 , v 0 ) =: ν 0 , we also have u(0) = u 0 , v(0) = v 0 . Let us consider ξ := u(t) -u 0 , η := v(t) -v 0 . Since ∂ u ν = λ -1 ν u , and
∂ 2 uu ν = ∂ u λ -1 ν u = λ -2 -ν + ∂ v λν v -∂ u
λν u with the help of (24). With similar relations for the other derivatives, we get the Taylor expansion of ν near t = 0:

ν(t) = ν(u 0 + ξ, v 0 + η) = ν 0 + ξλ -1 ν u + ηλ -1 ν v + 1 2λ 2 ξ 2 (-ν +∂ v λν v -∂ u λν u )-2ξη(∂ u λν v +∂ v λν u )+η 2 (-ν +∂ u λν u -∂ v λν v ) + o(ξ 2 + η 2 ), ( 37
)
where all functions on the right hand side are computed at (u 0 , v 0 ). In particular we get using ξ = u(t) -

u 0 = t u(0) + o(t) and η = v(t) -v 0 = t v(0) + o(t), ˙ ν(0) = lim t→0 1 t ( ν(t) -ν 0 ) = λ -1 ( u(0)ν u + v(0)ν v ).
This implies 1

λ 2 (ξ 2 + η 2 ) = t 2 ˙ ν(0) 2 + o(t 2 ). ( 38
) Similarly since M(ν(u, v)) • ν(u, v) = h(u, v
) from the definition of h, we have:

M( ν(t)) -M(ν 0 ) • ν(t) = h(u(t), v(t)) -M(ν 0 ) • ν(u(t), v(t)) = h(u 0 + ξ, v 0 + η) -(hν 0 + λ∂ u hν u + λ∂ v hν v ) • ν(u 0 + ξ, v 0 + η) = h(u 0 + ξ, v 0 + η) -h -ξ∂ u h -η∂ v h + h 2λ 2 (ξ 2 + η 2 ) + 1 2λ ξ 2 (-∂ v λ∂ v h + ∂ u λ∂ u h) + 2ξη(∂ u λ∂ v h + ∂ v λ∂ u h) + η 2 (-∂ u λ∂ u h + ∂ v λ∂ v h) + o(ξ 2 + η 2 ) = T [h](u 0 , v 0 ; ξ, η) + 1 2 ξ η A ξ η + o(t 2 )
where

A := λ -2 h Id + λ -1 ∇λ ⊗ ∇h -λ -1 ∇ ⊥ λ ⊗ ∇ ⊥ h.
Using the right inequality in (35), and the definition of the corresponding generalized sense, we deduce with (38):

lim sup t→0 1 t 2 M( ν(t)) -M(ν 0 ) • ν(t) ≤ lim sup t→0 α 4λ 2 t 2 (ξ 2 + η 2 ) = α 4 ˙ ν(0) .
Similarly the left inequality in (35) yields the reverse inequality, which achieves the proof of (20).

Conversely, let K be a spheroform. We know from Proposition 2 that there exists some function h ∈ C 1 (Ω) such that R K (ν) = M( h)(u, v) for all ν = ν(u, v).

Consider now the function h

:= h -α 2 . From the definition of M, it is clear that M(h)(u, v) = M( h) -α 2 ν(u, v)
, so for any ν = ν(u, v) we have

M K (ν(u, v)) = R K (ν(u, v)) - α 2 ν(u, v) = M(h)(u, v). Moreover the map ν → M K (ν) is lipschitzian from Lemma 2. Hence ∂ u h(u, v) = M K (ν) • ν u (u, v) is lipschitzian, too. And similarly for ∂ v h. So h ∈ C 1,1 . Additionally h(u, v) = M K (ν(u, v)) • ν(u, v) implies h • σ = M K (-ν) • (-ν) = -h, so h satisfies (34). Hence h ∈ C 1,1
σ (Ω). We know that M K satisfies (20). If we consider the special path ν : t → ν(u 0 + tξ, v 0 + tη), we can expand ν(t) near t = 0 as before, obtaining something similar to (37). This implies with a similar computation:

M( ν(t) -M(ν 0 ) • ν(t) = t 2 T [h](u 0 , v 0 ; ξ, η) + t 2 2 ξ η A ξ η + o(t 2 ).
Therefore (20) implies (35) in the generalized sense. This completes the proof that h ∈ C 1,1 σ,α (Ω).

Regularity of the parametrization

In this section, we investigate the consequences of (35) on h, whenever h is regular enough.

|R(h)| ≤ min α λ , α 2λ + 2λ α J(h) (39)
on ω, where R(h) and J(h) are the trace and determinant of the matrix

λ -1 hI + ∇λ ⊗ ∇h -∇ ⊥ λ ⊗ ∇ ⊥ h + λD 2 h, that is R(h) := 2h λ + λ∆h (40) J(h) := λ -2 h 2 + h∆h + λ 2 det D 2 h + λ∇ ⊥ λ • D 2 h • ∇ ⊥ h (41) -λ∇λ • D 2 h • ∇h -|∇λ| 2 |∇h| 2 .
Proof. For a C 2 function, the generalized sense for (35) is just the common pointwise sense. We can multiply by λ and get (39) since a 2 × 2 matrix is nonnegative definite, if, and only if, its trace and determinant are nonnegative.

Let us note for further references that R(h) and J(h) are the trace and determinant of a symmetric matrix. Therefore it has real eigenvalues, and in particular the discriminant of its characteristic polynomial is nonnegative:

R(h) 2 ≥ 4J(h). (42)
This holds for any C 2 function h.

Notice that for any δ ∈ R,

R(h + δ) = R(h) + 2δ λ , (43)
J(h + δ) = J(h) + δλ -1 R(h) + λ -2 δ 2 . (44)
Therefore (39) may be equivalently written

R(h + α 2 ) ≥ 0, R(h - α 2 ) ≤ 0, J(h + α 2 ) ≥ 0 and J(h - α 2 ) ≥ 0. ( 45
)
Remark 3.C. The appearance of the matrix in the previous proposition seems quite odd at first. Here is another way to obtain it, which is easier to understand, but requires again h ∈ C 2 , so we can compute the derivatives of M := M(h). We get using (24-27):

∂ u M = aν u + bν v and ∂ v M = cν u + dν v (46)
where

a := λ -1 h + ∂ u λ∂ u h -∂ v λ∂ v h + λ∂ 2 uu h b = c := ∂ v λ∂ u h + ∂ u λ∂ v h + λ∂ 2 uv h d := λ -1 h + ∂ v λ∂ v h -∂ u λ∂ u h + λ∂ 2
vv h. So we find DM ν = 0 in agreement to Proposition 1. We also see from their definition that R(h) = a + d and J(h) = adbc.

Since M = M(h) ∈ C 1 , ( 14) is equivalent to [START_REF] Yaglom | Convex Figures[END_REF] according to Proposition 1. Since for ν 0 = ν(u 0 , v 0 ), we have ν ⊥ 0 = Span(ν u (u 0 , v 0 ), ν v (u 0 , v 0 )), we just have to check [START_REF] Yaglom | Convex Figures[END_REF] for w = ξν u + ην v , with arbitrary (ξ, η). This inequality becomes then, using (24-27):

∀(ξ, η) ∈ R 2 , aξ 2 + 2bξη + dη 2 ≤ α 2λ (ξ 2 + η 2 ).
This means

- α 2λ Id ≤ a b c d ≤ α 2λ Id
in the sense of matrices, which is (35).

Remark 3.D. The previous proposition has also a geometrical meaning and can be proved using corresponding considerations. Indeed, in matrix notations, we have ∇M = a b c d ν u ν v , using again the notations of the previous remark. Consequently we get:

∇ν = ∂ u ν ∂ v ν = λ -1 ν u ν v = λ -1 a b c d -1
∇M.

By definition, the curvatures of the surface (u, v) → M(u, v) are the eigenvalues of the matrix A such that ∇ν = A∇M, since ν is normal to the surface. And the curvature radii, their inverse, are the eigenvalues of A -1 .

We see that in our case

A = λ -1 a b c d -1
. So the curvature radii are the solutions ρ i (i = 1, 2) of the equation

ρ 2 -λ 2 R(h)ρ + λ 2 J(h) = 0. (47)
Therefore, if we change h to h := h + δ in order to consider a parallel surface, we see that the curvature radii ρi on this new surface are solutions of the equation

0 = ρ2 -(2δ + 2h + λ 2 ∆h)ρ + λ 2 J(h) + δ(2h + λ 2 ∆h) + δ 2 = (ρ + δ) 2 -(2h + λ 2 ∆h)(ρ + δ) + λ 2 J(h).
Hence ρi = ρ i + δ as claimed before.

For a body of constant width α, the parallel to the median surface at distance ± α 2 are part of the boundary of K. So they are convex, with opposite directions (the outward normal vector on

M(h -α 2 )(u, v) is -ν(u, v)). Hence we must have ρ i ∈ [-α 2 , + α 2 ]
. This is equivalent to saying that the left hand side of (47) is nonnegative whenever ρ = ± α 2 , and that the sum of the roots belongs to [-α, α]. This in turn is equivalent to (39). In other words, (39) expresses the fact that the radii of curvature on the median surface are in [-α 2 , + α 2 ], whenever they are defined. The equivalent formula (45) expresses the fact that the Gaussian curvatures J(h ± α 2 ) are nonnegative, while the mean curvatures R(h ± α 2 ) have opposite signs, since the convex surfaces are opposite.

Surface area and volume

According to Theorem 3, there is a one to one correspondence between C 1,1 σ,α (Ω) and W α . We investigate now the way to compute the volume and surface area of some K ∈ W α through the corresponding function h. Proposition 4 Let ω ⊂ Ω be a symmetrical subset of the parametrization space, that is σ(ω) = ω. Let h ∈ C 1,1 σ,α (Ω) be C 2 on ∂ω, and let K be the corresponding spheroform.

The set R K (ω) ⊂ ∂K has surface area:

|R K (ω)| = α 2 4 |ν(ω)| + ω λ -2 h 2 - 1 2 |∇h| 2 + ∂ω h + 1 2 λ 2 ∆h ∇h • n - 1 4 ∂ω ∇ λ 2 |∇h| 2 • n. (48)
(Here |ν(ω)| stands for the surface area of the subset ν(ω) of S 2 .)

Proof. It is enough to prove the proposition for h ∈ C 2 (ω). Indeed an approximation argument allows to generalize to others h, since the right hand side of (48) involves second-order derivatives only on the boundary of ω.

We make use of the notations of Proposition 3 and Remark 3.C. We have

∂ u M × ∂ v M = (aν u + bν v ) × (cν u + dν v ) = J(h)ν.
Hence the area of the surface

M(h)(ω) is ω |J(h)|. Since M(h) is the median surface of K ∈ W α , M(h + α 2 ) and M(h -α 2 )
both describe the boundary of K. If we restrict the parameters to the subset ω, they both describe R K (ω) since we assumed ω = σ(ω). So the surface area of R K (ω) is equal to ω J(h ± α

2 ) . This implies, using (44) and (45):

|R K (ω)| = 1 2 ω J(h + α 2 )(u, v) du dv + 1 2 ω J(h - α 2 )(u, v) du dv = ω α 2 4λ 2 (u, v) + J(h)(u, v) du dv. = α 2 4 |ν(ω)| + ω J(h) du dv.
(The latter equality follows from ∂ u ν × ∂ v ν = λ -2 ν.) To complete the proof of the proposition, we have now to prove:

ω J(h)(u, v) du dv = ω λ -2 h 2 - 1 2 |∇h| 2 + ∂ω h + 1 2 λ 2 ∆h ∇h • n - 1 4 ∂ω ∇ λ 2 |∇h| 2 • n. ( 49
)
By expanding products in (41), we get

J(h) = λ -2 h 2 + h∆h + J 1 (h) -J 2 (h), (50)
where

J 1 (h) := λ 2 ∂ 2 uu h∂ 2 vv h -(∂ 2 uv h) 2 + λ ∂ u λ∂ u h∂ 2 vv h + ∂ v λ∂ v h∂ 2 uu h -∂ u λ∂ v h∂ 2 uv h -∂ v λ∂ u h∂ 2 uv h and J 2 (h) := (∂ u λ) 2 + (∂ v λ) 2 (∂ u h) 2 + (∂ v h) 2 + λ ∂ u λ∂ u h∂ 2 uu h + ∂ v λ∂ v h∂ 2 vv h + ∂ v λ∂ u h∂ 2 uv h + ∂ u λ∂ v h∂ 2 uv h . Let us define w 1 := ∂ u h∂ 2 uv h -∂ v h∂ 2 uu h, w 2 := ∂ u h∂ 2 vv h -∂ v h∂ 2 uv h and w := (w 1 , w 2 ). We have ∂ u w 2 -∂ v w 1 = 2(∂ 2 uu h∂ 2 vv h -(∂ 2 uv h) 2 ). Therefore 2J 1 (h) = ∂ u λ 2 w 2 -∂ v λ 2 w 1 .
This implies using Green's formula

ω J 1 (h) = 1 2 ∂ω λ 2 w • d ℓ.
Now let us denote by H the scalar function

|∇h| 2 . Since w = ∆h∇ ⊥ h-1 2 ∇ ⊥ H (where ∇ ⊥ = (-∂ v , ∂ u )), we also have ω J 1 (h) = 1 2 ∂ω λ 2 ∆h∇h - 1 2 ∇H • n ds.
Considering now J 2 , we can check easily that J 2 (h) = |∇λ| 2 H+ 1 2 λ∇λ • ∇H. Integrating by parts we get

ω J 2 (h) = ω H |∇λ| 2 - 1 2 ∇ • (λ∇λ) + 1 2 ∂ω Hλ∇λ • n ds = 1 2 ω H |∇λ| 2 -λ∆λ + 1 2 ∂ω Hλ∇λ • n ds = - 1 2 ω H + 1 2 ∂ω Hλ∇λ • n ds using (23).
Finally we have

ω h∆h = ∂ω h |∇h| • n ds - ω |∇h| 2 = ∂ω h∇h • n ds - ω H.
So integrating (50) yields (49).

We are now in position to compute the volume and surface area of any spheroform, expressed as integrals of the corresponding function h:

Theorem 4 Let h ∈ C 1,1
σ,α (Ω) be given, and K ∈ W α the corresponding spheroform. The surface area |∂K| and the volume |K| are given by

|∂K| = Ω λ -2 h 2 - 1 2 |∇h| 2 + πα 2 , ( 51
) |K| = α 2 Ω λ -2 h 2 - 1 2 |∇h| 2 + πα 3 6 . ( 52
)
Corollary 5 (Blaschke) Let K be any convex body of constant width α in dimension 3. Then the volume and surface area of K satisfy:

|K| = α 2 |∂K| - πα 3 3 . ( 53
)
We refer the reader to [1] for the original proof of this property. Here it follows directly from Theorem 4. --→ OM • n dσ, using Stokes' formula. We can choose M = M(h + α 2 )(u, v) as a parametrization, and then n = ν(u, v) and dσ = J(h + α 2 ) du dv. But we may also choose M = M(h -α

2 ), and in such a case n = -ν(u, v) since n is the outward normal in Stokes' formula, and dσ = J(h -α

2 ) du dv. So we have

|K| = 1 3 Ω (h + α 2 )J(h + α 2 ) = - 1 3 Ω (h - α 2 )J(h - α 2
).

In particular this implies, using (44) and an integration by parts:

|K| = 1 6 Ω (h + α 2 )J(h + α 2 ) -(h - α 2 )J(h - α 2 ) . = α 6 Ω J(h) + α 3 24 Ω λ -2 + α 6 Ω h(2λ -2 h + ∆h) = α 6 Ω J(h) + πα 3 6 + α 6 Ω 2λ -2 h 2 -|∇h| 2 .
This proves (52) using (49) with ω = Ω.

Figure 1: Two views of one Meissner's tetrahedron

Description of Meissner's tetrahedron

A description of this volume can be found in [START_REF] Chakerian | Convexity and its Applications, Chap. Convex bodies of constant width[END_REF], [START_REF] Yaglom | Convex Figures[END_REF] and [START_REF] Hilbert | Elf Eigenschaften der Kugel[END_REF]. We shall give a brief definition of this volume and describe its parametrization. Meissner's tetrahedron is geometrically defined in the following way: consider a body K t obtained as the intersection of four balls of radius α which centers are the vertices of a regular tetrahedron (of edge lengths α). Thus, the boundary of K t is composed of four pieces of balls connected by six arc of circles. Surprisingly, this set K t is not of constant width: geometrical considerations show that opposite circular edges are too far away. Meissner proposed to smooth three edges of K t in order to get a constant width body. Consider E the union of three circular edges which share a common vertex S. Then, the body K defined as

K = x∈E B(x, α) ∩ K t
is a body of constant width called Meissner's tetrahedron (see figure 3.4). Notice that it is possible to build an other constant width body based on the regular etrahedron by smoothing a different set of edges.

We give below an analytical representation in terms of its h function based on the parametrization of the sphere described by (22). In order to take benefit of the invariance of the previous body K by rotations of angles ±2π/3, we consider a body K built on a regular tetrahedron which has its vertex S on the z-axes and the others on the plane z = 0. Moreover, we assume that the equilateral triangle formed by other vertices on z = 0 is symmetric with respect to the y-axes. It is straightforward to check that

         2/3 tanh v -1/2, if sinh v > 2 √ 2 cos u, -1/2 -(1/2 √ 3)(cos u/ cosh v) + . . . ( √ 3/2)(cosh v 2 -sin u 2 )/ cosh v, if sinh v ≤ 2 √ 2 cos u, cosh v ≥ 2 sin u, 1/2 + (1/ √ 3) cos(u + 2π/3)/ cosh v if sinh v ≤ 2 √ 2 cos u, cosh v > 2 sin u.
Notice that it is possible to compute the volume and the surface area of Meissner's tetrahedron thanks to equations (51) and (53). After some symbolic computations, we get the formulas presented in [START_REF] Harrell | Calculations for Convex Bodies. Example: The rotated Reuleaux triangle[END_REF]:

|K| = 2π 3 - π √ 3 4 arccos 1 3 |∂K| = 2π - π √ 3 2 arccos 1 3

Local optimality

We now come back to the volume functional K → |K| in order to investigate the properties of its minimizers. A striking consequence of Theorem 4 is that minimizing the volume in W α is equivalent to minimizing the surface area. More precisely, the volume minimization problem is equivalent to min h∈C 1,1 σ,α (Ω)

L(h)

where

L(h) = Ω λ -2 h 2 - 1 2 |∇h| 2 . ( 54
)
Let us first observe that the maximum value of L is zero:

Lemma 4 For any h ∈ C 1,1 σ (Ω), we have L(h) ≤ 0.

In particular, a maximizer of L in C 1,1 σ,α (Ω) is always h = 0, which corresponds to a ball of radius α/2. Hence such a ball has maximal volume among all spheroforms, a well-known result.

Proof. Let W σ be the space of all functions h ∈ W 1,2 (Ω) satisfying (34). This is a closed subspace of the Sobolev space W 1,2 (Ω), so it is a Banach space. Let us define s ∈ R as follows:

s = inf h∈Wσ Ω |∇h| 2 Ω λ -2 h 2
. This is a "weighted Sobolev constant", and it is classical in PDE theory that the infimum is actually attained by a smooth function φ ∈ W σ satisfying the corresponding Euler equation

Ω ∇h • ∇φ = s Ω λ -2 hφ, ∀h ∈ W σ .
In other words, φ is an eigenfunction of the operator -λ 2 ∆, with the symmetry condition φ • σ = -φ. Additionally, if we choose two open sets Ω + ⊂ Ω and Ω -= σ(Ω + ) such that Ω + ∩ Ω -= ∅ and Ω = Ω + ∪ Ω -, then it follows from Krein-Rutman's theorem that there exists an eigenfunction φ satisfying φ > 0 on Ω + . One way to choose such a set Ω + is to consider some fixed vector w ∈ R 3 , and to set Ω + := (u, v) ∈ Ω ; w • ν(u, v) > 0 .

Given such a w, define g := (u, v) → w • ν(u, v). As explained in Remark 3.B, M(g) = w for all (u, v), and the body corresponding to h + g, for any h ∈ C 1,1 σ,α (Ω), is just a translation of the body corresponding to h. In particular, they have the same volume, so L(h + g) = L(h). Since L is quadratic, this means that 0 = L(g) + Ω λ -2 hg -∇h • ∇g for all h ∈ C 1,1 σ,α (Ω). In particular, L(g) = 0 since we can take h = 0, and ∆g + 2λ -2 g is orthogonal (for the L 2 scalar product) to all h ∈ C 1,1 σ,α (Ω). The latter implies it is orthogonal to W σ , since α>0 C 1,1 σ,α (Ω) contains C 2 σ (Ω). Hence it is orthogonal to φ, so we get Remark 3.E. Since balls are the unique maximizers of the volume among spheroforms of given width, it follows that for h ∈ C 1,1 σ (Ω):

L(h) = 0 ⇐⇒ ∃ w ∈ R 3 , h(u, v) = w • ν(u, v),
for all (u, v) ∈ Ω.

An interesting consequence of the previous lemma is that the functional L is actually strictly concave with respect to h (when considered on the quotient of C 1,1 σ (Ω) by the smallest subspace containing all the functions w • ν(u, v) for w ∈ R 3 ).

Indeed L is quadratic, so for any h, g ∈ C 1,1 σ (Ω) and for all t ∈ [0, 1]:

L(th + (1t)g) -tL(h) -(1t)L(g) = -t(1t)L(hg) ≥ 0.

From the remark 3.E, the equality holds if and only if it exists w ∈ R 3 such that h = g + w • ν(u, v).

The following weak optimality result applies not only to global minimizers, but also to local ones. Notice that this condition is very close from the one established in (??) for a relaxed problem of (3).

Theorem 6 Let K be a body of constant width, and a local minimizer of the volume functional. Then K is everywhere irregular in the following sense: for any A ⊂ S n-1 , one of the two subsets R K (A) or R K (-A) of ∂K is not a smooth surface.

In this context, a "smooth surface" means that the set of points can be described as the graph of a regular function. Observe that this result is obvious in dimension two for global minimizers, since these are Reuleaux triangles.

  and a similar relation for v. So M(h) • σ = M(h) and M satisfies (5).

Proof.

  The parametrization domain Ω = S 1 × R has no boundary. So if we apply (48) with ω = Ω, we get (51) since |ν(Ω)| = |S 2 | = 4π. The volume of K can be expressed as |K| = 1 3 ∂K

Figure 2 :

 2 Figure 2: Construction of Meissner's h function

2 Ωλ

 2 -2 gφ = Ω ∇g • ∇φ = s Ω λ -2 gφ.Now both functions g and φ are positive on Ω + and odd with respect to σ, soΩ λ -2 gφ = 2 Ω + λ -2 gφ > 0and therefore s = 2. This implies L(h) ≤ 0 for any h ∈ W σ , and in particular in C 1,1 σ (Ω).

convex curve. In this situation we would conclude that one of the two pieces of the boundary R K (A) or R K (-A) has measure 0. Notice that Meissner's tetrahedron satisfies the previous assumption.

Proof. Let K be a local minimizer of the volume and A ⊂ S 2 with R K (A) a smooth surface. Let h be the function of C 1,1 σ (Ω) associated to K by the proposition 2. Since every constant width bodies are strictly convex, we can assume without loss of generality that R K (A) is the graph of a strictly convex function. In this context, it is standard that the reverse Gauss map is a smooth diffeomorphism. Moreover, the function h is also locally smooth on the points of ω ⊂ Ω corresponding to A since:

Let us first establish that h saturates the pointwise constraint (35) on a subset of ω. By reducing ω to a smaller set if necessary, we suppose that ω ∩ σ(ω) = ∅. Assume by contradiction that the four inequalities are strict. Let g ∈ C 2 (ω) with compact support. We extend it to σ(ω) by symmetry, defining g(σ(u, v)) = -g(u, v), so that the new function, still denoted g, belongs to C 1,1 σ (Ω). Due to the non-saturation property, the functions f + := h + tg and f -:= htg belong to C 1,1 σ (Ω) for |t| small enough. Now L is strictly concave so we have:

for all g. Since an equality in (55) is not possible because of the remark 3.E (none of the function w • ν(u, v) has a compact support), we have that L(h) > min(L(f + ), L(f -)). This contradicts the local minimality of h. We established in subsection 3.3, that the saturation of the constraints for the regular function h is equivalent to the fact that one or both of the radii of curvature on R K (A) are equal to α or 0. Since R K (A) is a strictly convex regular surface, its curvature radii are not zero. As a consequence, on all points of R K (A) at least one of the curvature radii is equal to α. Consider now the surjective application from ω to R K (-A) given by

If this application is not injective, R K (-A) is not smooth since at least one point of this surface has a non empty subdiferrential. We conclude that the previous application is an admissible parametrization of R K (-A). It is now straightforward to compute that on all points of R K (-A), at least one of the curvature radii is equal to 0. Again, this fact contradicts the regularity of R K (-A) which concludes the proof. Remark 3.F. If we assume additionally that the lines of curvature on R K (A) of the body K have no torsion, it is possible to show that R K (-A) is a


Sharath Ram Kumar
email: sharath.ramkumar@cnrsatcreate.sg

Arvind Easwaran
email: arvinde@ntu.edu.sg

Benoit Delinchant
email: benoit.delinchant@g2elab.grenoble-inp.fr

Remy Rigo-Mariani
email: remy.rigo-mariani@g2elab.grenoble-inp.fr

Behavioural Cloning based RL Agents for District Energy Management

Keywords: CCS CONCEPTS, Computing methodologies → Intelligent agents Reinforcement Learning, Imitation Learning, District Energy Management

In this work, we discuss a method to incorporate domain knowledge into a Reinforcement Learning (RL) environment through the process of behavioral cloning, in the context of a district energy management system. Prior knowledge, encoded into heuristic rule-based programs, is used to initialize a policy network for an RL agent, after which an RL algorithm is used to improve on this by optimizing against a reward function. The key ideas are implemented in the CityLearn framework, where the resulting controller is used to manage the electrical energy storage for 5 buildings in a district. We demonstrate that the resulting agents offer measurable performance gains compared to existing baselines, offering a 3.8% improvement over the underlying rule-based controller, and a 20% improvement over a pure RL based controller. We also illustrate the possibility of using imitation learning to develop agents with desirable characteristics without explicit reward shaping.

INTRODUCTION

Demand Response at the building level is widely considered as a promising avenue to improve the cost and energy efficiency of power grids. The field has grown in popularity in recent years due the growing penetration of distributed energy resources in the energy mix, such as solar panels and battery systems. However, the development of control systems for distributed building energy management remains a challenge due to the complex nature of buildings, which necessitates the use of detailed models and requires significant engineering effort [START_REF] Medina | Demand response and distribution grid operations: Opportunities and challenges[END_REF]. Reinforcement Learning (RL) techniques are an active area of research in this context, since they can be used to build model-free controllers which learn by directly interacting with the target environment. Specifically, recent developments in the field of Deep RL, which combine traditional RL algorithms with deep neural networks, have proven effective in various continuous control tasks across diverse fields. RL solutions are typically not used in critical tasks in practice today due to the inherent unpredictability and the curse of dimensionality [START_REF] Thi Nguyen | Deep reinforcement learning for multiagent systems: A review of challenges, solutions, and applications[END_REF].

The CityLearn environment is a computer simulation framework based on the popular OpenAI Gym paradigm, and was developed to drive research in the field of Deep RL based Building Energy Management solutions. It uses a set of real-world datasets covering electricity demand, weather, and electricity pricing data, along with detailed models of various DERs such as electrical storage systems. Many RL-based controllers have been developed using CityLearn, based on different strategies related to the architecture and level of inter-agent communication. Conventionally, the operational effectiveness is reported relative to a reference Rule-Based Controller (RBC) -however, recent work has shown the drawbacks of this approach, demonstrating that a slightly modified RBC is able to achieve higher performance than most known RL-based solutions in CityLearn [START_REF] Nweye | Real-world challenges for multi-agent reinforcement learning in grid-interactive buildings[END_REF].

Imitation learning is an active area of research in the context of deep RL, as it holds the promise for agents to exploit existing knowledge and greatly speed up the training, while mitigating the need for complex reward function design. Behavioural Cloning is one such approach, where an agent is trained (typically through supervised learning) on data collected from the actions of an expert in the same environment. The technique has been used to obtain promising results across a range of tasks related to robotic manipulation [START_REF] Osa | An algorithmic perspective on imitation learning[END_REF].

In this work, we propose a solution in which an RL-based controller can be developed utilizing a reference RBC. The approach uses the Proximal Policy optimization (PPO) learning algorithm [START_REF] Schulman | Proximal policy optimization algorithms[END_REF] in combination with behavioural cloning. The use of imitation learning results in controllers with much more predictable actions and speeds up the training process, and a subsequent training step results in a small performance improvement over the underlying RBC. Notably, the learning process does not mandate the use a complex reward function, instead using a simple metric which is empirically known to be effective in the Citylearn environment.

RELATED WORK

The CityLearn environment has been used to study the use of reinforcement learning in the context of district energy management, and the potential of such techniques to develop purely data-driven, model-free controllers has been demonstrated [START_REF] José | Reinforcement learning for demand response: A review of algorithms and modeling techniques[END_REF]. Several controller architectures have been developed to solve this environment, with the performance of the resulting agents evaluated against a benchmark rule-based controller. Vazquez-Cantelli et al proposed MARLISA, a distributed framework in which each agent learns to predict its own energy consumption while taking control actions, with the aim of improving agent performance through information sharing [START_REF] Jose | MARLISA: Multi-Agent Reinforcement Learning with Iterative Sequential Action Selection for Load Shaping of Grid-Interactive Connected Buildings[END_REF]. Centralized controllers have also proven effective in solving the environment, where a single agent oversees the states and decides the actions for all buildings. A centralized agent based on the Soft Actor-Critic algorithm was developed by Kathirgamanathan, and was shown to exhibit good performance compared to the reference rule-based controller [START_REF] Kathirgamanathan | A Centralised Soft Actor Critic Deep Reinforcement Learning Approach to District Demand Side Management through CityLearn[END_REF]. However, Nweye et al recently reported on the limitations of such approaches, and showed that an improved rule-based controller exhibited state-of-the-art performance in this environment [START_REF] Nweye | Real-world challenges for multi-agent reinforcement learning in grid-interactive buildings[END_REF].

Imitation learning, when combined with deep learning, has shown promise in a diverse set of problems. Hua recently reviewed the use of imitation and transfer learning in the field of robotics, especially for tasks such as manipulation and fine control [START_REF] Hua | Learning for a robot: Deep reinforcement learning, imitation learning, transfer learning[END_REF]. Recently, a method to use imitation learning in the context of a building energy management problem was described by Gao et al [START_REF] Gao | Online Optimal Power Scheduling of a Microgrid via Imitation Learning[END_REF], where the final agent learns to imitate the actions of a Model Predictive Controller (MPC) with a perfect forecast of the environment variables. Pezzotti et al developed MimicBot, a deep RL agent in a Fantasy Football game environment, which achieved excellent performance when RL training techniques were applied after an initial behaviour cloning step [START_REF] Pezzotti | MimicBot: Combining Imitation and Reinforcement Learning to win in Bot Bowl[END_REF]. In a more advanced scenario, Liu et al created a framework to train simulated humanoid figures utilizing motion-capture data, and showed that complex behaviours requiring multi-agent coordination could be captured using this approach [START_REF] Liu | From motor control to team play in simulated humanoid football[END_REF].

While previous work has focused on the application of deep RL and imitation learning in the context of building energy management systems (BEMS), we are not aware of any existing work where these techniques are applied in the context of a district energy management (DEM) problem. Traditional control methods used in BEMS, such as MPCs, are often not viable here due to the increased complexity caused by the presence of multiple buildings, resulting in the need to resort to heuristic approaches [START_REF] Sameti | Optimization approaches in district heating and cooling thermal network[END_REF]. As such, the combination of IL and DRL presented in this work is of significance as a practical method to transition from these heuristic rules to a data-driven controller, without losing the knowledge encoded in the former approach.

IMPLEMENTATION 3.1 CityLearn 2022 Environment

The CityLearn 2022 Environment is an OpenAI gym environment which allows for the development and simulation of RL-based controllers for distributed energy management. The environment offers the possibility of choosing from up to 28 different observations for each building, covering both global parameters such as weather and building-specific data such as hourly demand and solar generation. Each building is equipped with a electric energy storage system, which can be charged or discharged at each time step by the controller. The performance of this controller is finally evaluated against that of a naive agent, which does not use the battery, on two metrics -net electricity price and net carbon consumption. In this study, we use only the net electricity price as a comparison metric. This choice is justified since the focus of the study is on discussing the use of behaviour cloning, and not on achieving the best scores in the simulation.

In the present work, data for a period of 1 year for 5 buildings is used -the models are trained on the data for the first 10 months, and evaluated on their performance for the subsequent 2 months. A centralized architecture is chosen for the controller, such that a single agent is able to observe the states for all 5 buildings, and take actions to charge or discharge the electric storage of each building at any time step. The observation space used in our formulation, made up of 22 data points for the whole district, is outlined in Table 1. The above parameters were chosen from the set of all available information provided by the environment at every time step, omitting metrics related to the temperature, humidity and daylight savings status. This selection was done empirically, and adjusted based on the performance of the agents in the simulation. The chosen metrics are normalised based on their running mean and variance before being input to the policy network.

The reward function 𝑟 (𝑡) at time step 𝑡 is given below, where 𝑁 is the number of buildings and 𝑒 𝑖 represents the net electricity cost for building 𝑖 at time step 𝑡:

𝑟 (𝑡) = -1 * ( 𝑁 𝑖=1 max(0, 𝑒 𝑖 )) 2
The reward function used for the agent is purely based on the rectified net electricity price per time step, which is a simple metric based on the energy consumption and the electricity price. It should be emphasised that this choice was made to demonstrate the effectiveness of using behaviour cloning in encouraging positive behaviours for a given agent without explicitly rewarding them.

Algorithms

The learning process consists of two steps. First, the policy, represented by the weights of a neural network, is initialized by cloning the behaviour of a reference rule-based controller. This process is outlined in Algorithm 1. Next, the PPO algorithm is used to train this network further to generate the final policy, which typically achieves marginal gains over the RBC. The hyperparameters used in this work are listed in Table 2. It should be noted that a lower learning rate and clipping parameter were chosen for the second PPO training cycle, so that subsequent policies are not too different from the cloned policy. The RBC used in our work is outlined in Fig 1 . The rules are structured such that the battery is charged during the late nights and early mornings, and discharged to meet the peak load. The loss function L used in this implementation consists of a log-loss term which penalizes the difference between the model output and the ground truth, and a term to promote entropy in the stochastic policy. The first term addresses bias in the model output, while the latter term address variance.

RESULTS AND DISCUSSION

The trained agents are able to find a better policy than an agent which does not use electrical storage, and exhibit good performance It is visible from the training curves that the agent which starts out with a behaviour cloning prior does not exhibit a significant improvement in the training reward, especially when compared to the PPO-only agent. It is notable that such an approach still leads to small increase in the score for the final agent.

An important observation is that the actions of the agent trained using behaviour cloning are more approachable than those of the agent trained directly using PPO; for instance, the latter sometimes takes sub-optimal actions such as neglecting the energy storage system altogether for some buildings. The net energy demand for a period of 1 week, taken from the evaluation set, is shown in Fig 3. As the latter was developed manually using prior knowledge of the environment dynamics, the behavioural cloning step acts as a method to transfer existing knowledge into a neural-network based agent without constructing a complex reward function. 3 occurs as a result of the trained agent learning to make marginal changes to the actions of the RBC, in an attempt to maximize its reward.

Once the cloning step is completed, the resulting set of weights can be trained on-line in a continuous training operation. For instance, this opens up the possibility to develop functional controllers that constantly improve and adapt as new data is made available, which is highly desirable in this context. Since weight updation is a purely data-driven step, there is minimal additional effort required to implement the same.

An important caveat with the method presented here is that the performance of the final agent strongly depends on that of the initial rule-based agent. Training the clone agent using PPO only resulted in marginal gains compared to the underlying prior, since the hyperparameters chosen tend to restrict the exploration aspect of the learning process in exchange for stability. Further research is required to determine whether this limitation can be overcome by a different choice of hyperparameters.

CONCULSION AND FUTURE SCOPE

In this paper, we explore the possibility of embedding prior knowledge into a Deep RL based control system using imitation learning, in the context of the building energy management problem presented in the CityLearn environment. The resulting agents are able to translate a hard-coded set of rules into the weights of a neural network, and achieve marginal gains over the underlying policy through continuous training. The possibility of adapting the method to implement a controller that is training on-line using real-time data was discussed. It was noted that the technique allows for the training of viable agents without designing a complex reward function -however, the performance was observed to be strongly dependent on, and limited by, the performance of the initial heuristic agent. Future work may focus on achieving a significant performance improvement over the underlying controller, through hyperparameter tuning or by using a different reward function.

Figure 1 :

 1 Figure 1: Rule-Based Agent Action Sequence

  after training for about 2M steps on the training data. The scores achieved by each agent are listed in Table3-it must be noted that a lower score is desirable, since the figures represent cumulative costs. Fig 2 shows the training curves for the control PPO agent and the agent which started the PPO training process after behaviour cloning.

Figure 2 :

 2 Figure 2: Training Rewards for the Agents

Figure 3 :

 3 Figure 3: Net Electricity Consumption of the District

Figure 4 :

 4 Figure 4: Comparison between RBC Agent and Final Agent

Table 1 :

 1 State space for RL Problem Formulation

	Parameter	Observation Type
	Month	Shared
	Hour	Shared
	Direct Solar Irradiance	Shared
	Direct Solar Irradiance (6h Forecast)	Shared
	Carbon Intensity	Shared
	Electricity Pricing	Shared
	Electricity Pricing (6h Forecast)	Shared
	Non-shiftable Load	Per Building
	Solar Generation	Per Building
	Electrical Storage SOC	Per Building

1 :

 1 Initialize the rule-based policy 𝜋 𝑅 , target policy 𝜋 𝐵𝐶 𝜃 and imitation buffer 𝐵 2: Initialize environment 𝜀 and initial state 𝑠 = 𝑠 0 3: while 𝐵 is not full do 𝜋 𝑅 (𝑠 (𝑡))) to 𝐵, and update 𝑠 (𝑡 + 1) = 𝜀 (𝜋 𝑅 (𝑠 (𝑡)) 5: end while 6: Split 𝐵 into 𝐵 𝑡𝑟𝑎𝑖𝑛 , 𝐵 𝑡𝑒𝑠𝑡 7: for 𝑖 = 0, 𝑖 < 𝑛𝑢𝑚_𝑡𝑟𝑎𝑖𝑛𝑖𝑛𝑔_𝑖𝑡𝑒𝑟𝑠 do

	4: Add (𝑠 (𝑡), 8: Sample mini-batch 𝐵 𝑚𝑏 from 𝐵 𝑡𝑟𝑎𝑖𝑛
	9:	Calculate Loss 𝐵 𝑚𝑏 L (𝜋 𝑅 (𝑠 (𝑡)), 𝜋 𝐵𝐶 𝜃 (𝑠 (𝑡)))
	10:	Update 𝜃 := 𝜃 -𝛾 ∇L
	11: end for
	12: Evaluate the model using 𝐵 𝑡𝑒𝑠𝑡

Table 2 :

 2 Hyperparameters Used In This Work

	Parameter	PPO BC + PPO Description
	𝑙𝑟 𝐵𝐶	None	1𝑒 -4	Learning
				Rate	for
				Behavioural
				Cloning
	𝑒 𝐵𝐶	None	1𝑒 -3	Entropy
				Weight for
				Behavioural
				Cloning Loss
	𝑙𝑟 𝑃𝑃𝑂	3𝑒 -4	1𝑒 -4	Initial Learn-
				ing Rate for
				the PPO Al-
				gorithm
	𝐵𝑎𝑡𝑐ℎ𝑆𝑖𝑧𝑒	256	256	Mini-batch
				size during
				PPO updates
	𝐻𝑖𝑑𝑑𝑒𝑛𝐿𝑎𝑦𝑒𝑟𝐷𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑠 64,64	64,64	Dimensions
				of the policy
				network
				hidden layers
	𝑐 𝑃𝑃𝑂	0.2	0.1	PPO	Clip
				Range
	𝛾 𝑃𝑃𝑂	0.99	0.99	PPO Future
				Rewards
				Coefficient
	𝜆 𝐺𝐴𝐸	0.95	0.95	PPO GAE Co-
				efficient

Table 3 :

 3 Net Electricity Price for Different Agents

	Agent Type	Whole Year ($) Last 2 Months ($)
	Baseline (No Storage)	8277.73	1614.90
	Rule-Based Controller	6357.17	1281.82
	PPO Agent (no BC)	7681.95	1535.99
	Only BC (no PPO)	6136.06	1219.43
	BC + PPO	6113.04	1215.50

ACKNOWLEDGEMENTS

This research is supported by the National Research Foundation, Prime Minister's Office, Singapore under its Campus for Research Excellence and Technological Enterprise (CREATE) programme.