I'm trying to solve below exercise
Let $(H, \langle \cdot, \cdot \rangle)$ be a real Hilbert space. Let $F:H \to \mathbb R$ be convex and differentiable. Let $K$ be a convex subset of $H$ and $u \in K$. Then $F(u) \le F(v)$ for all $v \in K$ IFF $\langle \nabla F(u), v-u) \ge 0$ for all $v \in K$.
Could you verify my below attempt?
Because $F$ is convex, $F(x) - F(u) \ge \langle \nabla F(u), x-u)$ for all $x \in H$. Then $\langle \nabla F(u), v-u) \ge 0$ implies $F(u) \le F(v)$. Let's prove the reverse. Assume there exists $v \in K$ such that $\langle \nabla F(u), v-u) < 0$. We have $$ \langle \nabla F(u), v-u) = \lim_{t \to 0} \frac{F(u + t(v-u)) - F(u)}{t}. $$
This implies there is $t_0 \in (0, 1)$ such that $F(u + t_0 (v-u)) < F(u)$. This is equivalent to $F(t_0 v + (1-t_0)u) < F(u)$. Because $u,v \in K$ and $K$ is convex, we get $v_0 :=t_0 v + (1-t_0)u \in K$. So there exists $v_0 \in K$ such that $F(v_0) < F(u)$. This completes the proof.