I am interested in determining whether the function $F:L^{\infty}(0,\infty;L^{\infty}(0,1)) \to \mathbb R$ defined by $$u \mapsto F(u)=\int_0^\infty \int_0^1 u(x,t)^2 \ \mathrm dx \, \mathrm dt $$ is differentiable/Frechét differentiable at an element $u \in L^2(0,\infty;L^2(0,1))$.
Any suggestions or assistance would be greatly appreciated.
First of all, instead of $L^{\infty}(0,\infty;L^{\infty}(0,1))$, you should write $L^{\infty}([0,\infty);L^{\infty}(0,1))$ (or maybe $L^{\infty}((0,\infty);L^{\infty}(0,1))$, depending of what you want here).
Second, for any $u\in L^{\infty}([0,\infty);L^{\infty}(0,1))$ (say), any $t\in[0,\infty)$, and any $x\in(0,1)$, the value at $x$ of the value $u(t)$ of $u$ at $t$ should be written as $u(t)(x)$, rather than $u(x,t)$.
Finally, there is no function $F$ from $L^{\infty}([0,\infty);L^{\infty}(0,1))$ (or from $L^{\infty}((0,\infty);L^{\infty}(0,1))$) to $\mathbb R$ such that $$F(u)=\int_0^{\infty} \int_0^1 u(t)(x)^2 \,dx \,dt$$ for all $u\in L^{\infty}([0,\infty);L^{\infty}(0,1))$. Indeed, if $u(t)(x)=1$ for all $t\in[0,\infty)$ and $x\in(0,1)$, then $u\in L^{\infty}([0,\infty);L^{\infty}(0,1))$ but $\int_0^{\infty} \int_0^1 u(t)(x)^2 \,dx \,dt=\infty$. (The case of $L^{\infty}((0,\infty);L^{\infty}(0,1))$ is quite similar.)
So, your question is about a nonexistent function $F$, and hence any answer about $F$ would be correct.
