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Xn , . . ) : |xn | ≤ 2−n }. Let f (x) = − ∞ ∞ n=1 √ −n 2 + xn if x ∈ C, elsewhere. 2 The subgradient 37 Then f is convex and its restriction to the set C is a continuous function. n−2 An easy calculation shows that f (0; en ) = −2 2 . Now suppose x∗ ∈ ∂f (0). Then f (2−n en ) ≥ f (0) + x∗ , 2−n en , ∀n ∈ N, whence (1 − √ n 2)2 2 ≥ x∗ , en , ∀n ∈ N. Thus f has all directional derivatives at 0, but ∂f (0) = ∅. 13 below shows. 9 Let x ∈ dom f , x∗ ∈ ∂f (x), u∗ in the normal cone to dom f at x ( u∗ , x − u ≤ 0, ∀u ∈ dom f ).

If x ∈ int dom f , then R+ (dom f −x) = X. 14. If X is ﬁnite dimensional, the previous result can be reﬁned (same proof) since ∂f (x) = ∅ ∀x ∈ ri dom f . In inﬁnite dimensions it can be useless, since dom f could possibly have no interior points. 13). 12 we immediately get the following result providing an estimate from above of the norm of the elements in ∂f . 16 Let f ∈ Γ (X) be Lipschitz with constant k in an open set V x. Then x∗ ≤ k, ∀x∗ ∈ ∂f (x). As a last remark we observe that the subdiﬀerential keeps a fundamental property of the derivative of a convex function.

Consider the line segment S = {(x, f (x) − ε) + t(d, α) : 0 ≤ t ≤ 1}. S is a compact convex set disjoint from epi f . Thus there are y ∗ ∈ X ∗ , r ∈ R such that y ∗ , y + rf (y) > y ∗ , x + td + r(f (x) − ε + tα), ∗ ∀y ∈ dom f , ∀t ∈ [0, 1]. As usual, r > 0. Dividing by r and setting x∗ = − yr , we get x∗ , d ≥ α − ε, (with the choice of y = x, t = 1), and if v ∈ X is such that x + v ∈ dom f , setting y = x + v and t = 0, f (x + v) − f (x) + ε ≥ x∗ , v , which means x∗ ∈ ∂ε f (x). The last two facts provide sup{ x∗ , d : x∗ ∈ ∂ε f (x)} ≥ α − ε, and this ends the proof.