Let $f: R \rightarrow R$ be a continuously differentiable function on the real numbers (if needed also infinitely many often differentiable).
Define the Operator $F : L^2([0,1]) \rightarrow R$ for $x \in L^2([0,1])$ by
$F(x)=\int_0^1 f(x(s)) ds $
We assume that $F(x)<A−B∫x(s)^2 ds\leq A$ with constants $A,B>0$.
Hence it is bounded from above but not from below
Question:
Is $F$ now Frechet differentiable?
Is $F$ Gateaux differentiable?
Known:
If we were looking at $F$ as an operator on $C([0,1])$, then it is Frechet differentiable (as shown for example at wikipedia: http://de.wikipedia.org/wiki/Fréchet-Ableitung#Integraloperator )
But they use that on this space they know that each function is bounded.
Best Answer
The operator $F$ is Frechet from $L^\infty(0,1)$ to $\mathbb R$ without further assumptions.
In order to prove differentiability from $L^2$ to $\mathbb R$, one needs growth conditions: there exists $a,b\ge0$ such that $$ |f(y)| \le a + b |y|^2 $$ and $$ |f'(y)| \le a + b |y| $$ for all $y\in \mathbb R$. This makes $F(x)$ well defined for all $x\in L^2$.
To prove Gateaux differentiability: show directional differentiability first. Convergence of difference quotients can be argued with Lebesgue dominated convergence theorem and Taylor expansion of $f$. The directional derivative is given by $$ F'(x; h) = \int_0^1 f'(x(s))h(s)ds, $$ which is clearly linear and continuous with respect to $h\in L^2$. This proves Gateaux differentiability.
To show Frechet differentiability one can use Egorov's theorem.
I am pretty sure that these conditions are also necessary (See Appell and Zabrejko: Superposition operators - which is a great source of results of this type).