I'm trying to solve below exercise in Brezis' Functional Analysis
Let $(H, \langle \cdot, \cdot \rangle)$ be a real Hilbert space and $\Vert \cdot\Vert $ its induced norm. Let $a: H \times H \rightarrow \mathbb{R}$ be a continuous bilinear form such that $$ a(v, v) \geq 0 \quad \forall v \in H $$ Prove that the function $F: H \to \mathbb R, v \mapsto a(v, v)$ is convex, of class $C^1$, and determine its differential.
The solution by the author is
The convexity inequality $a(t u+(1-t) v, t u+(1-t) v) \leq t a(u, u)+$ $(1-t) a(v, v)$ is equivalent to $t(1-t) a(u-v, u-v) \geq 0$. Consider the operator $A \in \mathcal{L}(H)$ defined by $a(u, v)=(A u, v)$ for all $u, v \in H$. Then $F^{\prime}(u)=A u+A^{\star} u$, since we have $$ F(u+h)-F(u)=\left(A u+A^{\star} u, h\right)+a(h, h) . $$
I would like to verify that the explicit form of $F'(u):H \to \mathbb R$ is indeed $F'(u)[v] = a(u, v)+ a(v, u)$ for all $v\in H$. Could you have a check on my attempt?
Because $a$ is continuous, there is $C>0$ such that $\Vert a(u, v)\Vert \le C\Vert u\Vert \Vert v\Vert $ for all $u, v\in H$. Then $$ \begin{align} & \lim_{v \to u} \frac{\Vert F(v)- F(u) - F'(u)[v-u]\Vert }{\Vert v-u\Vert } \\ = & \lim_{v \to u} \frac{\Vert a(v, v)- a(u, u) - (a(u, v-u)+ a(v-u, u))\Vert }{\Vert v-u\Vert } \\ = & \lim_{v \to u} \frac{\Vert a(v-u, v-u)\Vert }{\Vert v-u\Vert } \\ \le & \lim_{v \to u} C \Vert v-u\Vert =0. \end{align} $$
This completes the proof.