Define $F:C[0,1]\rightarrow \mathbb{R}$ as $F(u)=\int_{0}^{1} (u(x))^2 dx$.
We have verified that directional derivative of $F$ at $u$ in direction $g$ is $D_gF(u)=\int_0^12u(x)g(x)dx$.
Now there is a claim that the differential of $F$ at $u$ is the same map $(F'(u))(v)= \int_0^12u(x)v(x)dx$, under both $d_1$ and $d_\infty$.
I use definition of differentiation, after several steps, I get
$\frac{|F(u+h)-F(u)-F'(u)(h)|}{\|h\|} = \frac{\int_0^1 (h(x))^2dx}{\|h\|}$
But downstairs is either a sup (in $d_\infty$) or an integral (in $d_1$), and I don't know how to continue to bound the expression to $0$ (as $h\rightarrow0$) in either case..
Note that \begin{align} \|h\|_2 \le \|h\|_\infty^\frac{1}{2}\|h\|_1^\frac{1}{2} \end{align} then it follows that \begin{align} \frac{\int^1_0 h(x)^2\ dx}{\|h\|_\infty} \le \|h\|_1 \end{align} and likewise \begin{align} \frac{\int^1_0 h(x)^2\ dx}{\|h\|_1} \le \|h\|_\infty. \end{align}
It remains to prove the first inequality. Observe \begin{align} \int^1_0 h(x)^2\ dx \le \|h\|_\infty \int^1_0 |h(x)|\ dx . \end{align}