Linear Supporting Form and Derivative

Nel documento Weak KAM Theorem in Lagrangian Dynamics Seventh Preliminary Version (pagine 21-25)

As is usual, if E as a vector space (over R) we will denote by E = Hom(E, R) its algebraic dual space. We will indifferently use both notations p(v) or hp, vi to denote the value of v ∈ E under the linear form p ∈ E.

Definition 1.2.1 (Supporting Linear Form). We say that the linear form p ∈ E is a supporting linear form at x0 ∈ U for the function f : U → R, defined on U ⊂ E, if we have

∀x ∈ U, f (x) − f (x0) ≥ p(x − x0) = hp, x − x0i.

We will denote by SLFx(f ) the set of supporting linear form at x for f , and by SLF(f ) the graph

SLF(f ) = ∪x∈U{x} × SLFx(f ) ⊂ U × E.

In the literature, the linear form p is also called subderiva-tive of f at x0 or even sometimes subgradient. We prefer to call it supporting linear form to avoid confusion with the notion of subdifferential that we will introduce in another chapter.

Example 1.2.2. a) If f : R → R, t 7→ |t| then SLF0(f ) = [−1, 1], for t > 0, SLFt(f ) = {1}, and for t < 0, SLFt(f ) = {−1}.

b) If g : R → R, t 7→ t3 then SLFt(g) = ∅, for everyt ∈ R.

The following Proposition is obvious.

Proposition 1.2.3. The set SLFx(f ) is a convex subset of E. Moreover, if we endow Ewith the topology of simple convergence on E (”weak topology”) then SLFx(f ) is also closed.

Here is the relation between supporting linear form and deriva-tive.

Proposition 1.2.4. Let f : U → R be a function defined on an open subset U of the normed space E.

a) If f is differentiable at some given x ∈ U then SLFx(f ) ⊂ {Df (x)}, i.e. it is either empty or equal to the singleton {Df (x)}.

b) If E = Rn, and all partial derivatives ∂f /∂xi(x), i = 1, . . . , n, at some given x ∈ U , then SLFx(f ) is either empty or reduced to the single linear form (a1, . . . , an) 7→Pn

i=1ai∂f /∂xi(x).

Proof. a) If SLFx(f ) 6= ∅, let p be a supporting linear form of f at x. If v ∈ E is fixed, for all ǫ > 0 small we have x + ǫv ∈ U and thus f (x + ǫv) − f (x) ≥ ǫp(v). Dividing by ǫ and taking the limit as ǫ goes to 0 in this last inequality, we find Df (x)(v) ≥ p(v). For linear forms this implies equality, because a linear form which is

≥ 0 everywhere has to be 0.

b) We denote by (e1, . . . , en) the canonical base in Rn. Let us consider a point x = (x1, . . . , xn) ∈ Rn where all partial deriva-tives exist. This implies that the function of one variable h 7→

f (x1, . . . , xi−1, h, xi+1, . . . , xn) is differentiable at xi, hence by part a), if p ∈ SLFx(f ), we have p(ei) = ∂f /∂xi(x). Since this is true for every i = 1, . . . , n, therefore the map p must be (a1, . . . , an) 7→

Pn

i=1ai∂f /∂xi(x).

We have not imposed any continuity in the definition of a sup-porting linear form for a function f . This is indeed the case under very mild conditions on f , as we will see presently.

Proposition 1.2.5. Let U is an open subset of the topological vector space E, and let f : U → R be a function. Suppose that f is bounded from above on a neighborhood of x0 ∈ U , then any supporting linear form of f at x0 is continuous.

Proof. Let V be a neighborhood of 0 such that V = −V , and f is defined and bounded from above by K < +∞ on x0+ V . Since V

9 is symmetrical around 0, for each v ∈ V , we have

p(v) ≤ f (x0+ v) − f (x0) ≤ 2K

−p(v) = p(−v) ≤ f (x0− v) − f (x0) ≤ 2K,

hence the linear form p is thus bounded on a nonempty open sub-set, it is therefore continuous.

As is customary, if E is a topological vector space, we will denote by E ⊂ E the topological dual space of E, namely E is the subset formed by the continuous linear forms.Of course E = E if E is finite-dimensional. If E is a normed space, with norm k·k, then E is also a normed space for the usual norm

kpk = sup{p(v) | v ∈ E, kvk ≤ 1}.

In the case of continuous map, we can improve Proposition 1.2.3.

Proposition 1.2.6. Suppose that f : U → R is a continuous function defined on the topological vector space E. If we endow E with the topology of simple convergence on E (”weak topology”), then the graph SLF(f ) is a closed subset of U × E.

The proof of this Proposition is obvious.

Exercise 1.2.7. Let f : U → R be a locally bounded function defined on the open subset U of the normed space E. (Recall that locally bounded means that each point in U has a neighborhood on which the absolute value of f is bounded)

a) Show that for every x ∈ U , we can find a constant K, and a neighborhood V such that for every y ∈ V and every p ∈ SLFy(f ) we have kpk ≤ K. [Indication: see the proof of Theorem 1.4.1]

b) If E is finite dimensional, and f is continuous, show the following continuity property: for every x ∈ U , and every neigh-borhood W of SLFx(f ) in E = E, we can find a neighborhood V of x such that for every y ∈ V we have SLFy(f ) ⊂ W .

As we will see now the notion of linear supporting form is tailored for convex functions.

Proposition 1.2.8. If the function f : U → R, defined on the convex subset U of the vector space E, admits a supporting linear form at every x ∈ U , then f is convex.

Proof. Let us suppose that x0 = y + (1 − t)z with y, z ∈ U and t ∈ [0, 1]. If p is a supporting linear form at x0, we have

f (y) − f (x0) ≥ p(y − x0) and f (z) − f (x0) ≥ p(z − x0), hence

tf (y) + (1 − t)f (z) − f (x0) ≥ p(t(y − x0) + (1 − t)(z − x0))

= p(ty + (1 − t)z − x0) = 0.

The following theorem is essentially equivalent to the Hahn-Banach Theorem.

Theorem 1.2.9. Let U be a convex open subset of the locally convex topological vector space E. If f : U → R is continuous and convex, then we can find a supporting linear form for f at each x ∈ U .

Proof. As f is continuous and convex, the set O = {(x, t) | x ∈ U , f (x) < t}

is open, non-empty, and convex in E × R. Since (x0, f (x0)) is not in O, by the Hahn-Banach Theorem, see [RV73, Theorem C, page 84] or [Rud91, Theorem, 3.4, page 59], there exists a continuous and non identically zero linear form α : E × R → R and such that

∀(x, t) ∈ O, α(x, t) > α(x0, f (x0)).

We can write α(x, t) = p0(x) + k0t, with p0: E → R a continuous linear form and k0 ∈ R. Since α(x0, t) > α(x0, f (x0)) for all t > f (x0), we see that k0 > 0. If we define ˜p0 = k−10 p0, we get

˜

p0(x)+t ≥ ˜p0(x0)+f (x0), for all t > f (x), therefore f (x)−f (x0) ≥ (−˜p0)(x − x0). The linear form −˜p0 is the supporting linear form we are looking for.

The following Proposition is a straightforward consequence of Theorem 1.2.9 and Proposition 1.2.4

11 Proposition 1.2.10. Let f : U → R be a continuous convex function defined on an open convex subset U of the normed space E. If f is differentiable at x0 then the derivative Df (x0) is the only supporting linear form of f at x0. In particular, we have

∀x ∈ U, f (x) − f (x0) ≥ Df (x0)(x − x0).

Corollary 1.2.11. Let f : U → R be a continuous convex func-tion defined on an open convex subset U of a normed space. If f is differentiable at x0, then x0 is a global minimum if and only if Df (x0) = 0.

Proof. Of course, if the derivative exists at minimum it must be 0, this is true even if f is not convex. The converse, which uses convexity, follows from the inequality f (y) − f (x0) ≥ Df (x0)(y − x0) = 0 given by Proposition 1.2.10 above.

Corollary 1.2.12. If U ⊂ Rn is open and convex and f : U → R is a convex function, then, for almost all x, the function f admits a unique supporting linear form at x.

Proof. This is a consequence of Proposition 1.2.10 above and Rade-macher’s Theorem 1.1.10.

Exercise 1.2.13. Let U be an open and convex subset of Rn. Suppose that f : U → R is convex and continuous. Show that if f admits a unique supporting linear formp0 at x0 then Df (x0) exists, and is equal to p0. [Indication: For each x ∈ U \ 0, choose px∈ SLFx(f ), and prove that

p(x − x0) ≤ f (x) − f (x0) ≤ px(x − x0).

Conclude using exercise 1.2.7.

Nel documento Weak KAM Theorem in Lagrangian Dynamics Seventh Preliminary Version (pagine 21-25)