A confusion about Brezis’ exercise 5.10

convex-analysisderivativesfunctional-analysishilbert-spaces

I'm trying to solve an exercise in Brezis' Functional Analysis

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With a small change of wording,

Let $(H, \langle \cdot, \cdot \rangle)$ be a real Hilbert space. Let $F: H \rightarrow \mathbb{R}$ be a convex and continuously differentiable. Let $K \subset H$ be convex and let $u \in H$. Show that the following properties are equivalent:

(i) $F(u) \leq F(v) \quad \forall v \in K$,

(ii) $\langle \nabla F (u), v-u \rangle \geq 0 \quad \forall v \in K$.

It seems I got a "counter-example". Let $H = \mathbb R$ and $F(x) = x^2$ for all $x \in \mathbb R$. Then $\nabla F(x)= 2x$. Let $u=-1$ and $K=\{2\}$. We have $F(-1) < F(2)$, but $\langle \nabla F (-1), 2- (-1) \rangle =-6<0$.

Could you elaborate on where I got wrong?

Best Answer

There's nothing wrong with your counterexample.

Even though I was unable to find errata for this book, I'm quite certain that it should require $u \in K$ instead of only $u \in H$.

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