Fréchet Derivative of a functional in $L^2$

Question

I need to check if the functional $$ (f(x))(t):=\int_{0}^{1}k(s,t)\cdot \cos(x(t))dt $$ is Fréchet differentiable on $L^2([0,1])$, where $k:[0,1]\times [0,1]\rightarrow \mathbb{R}$ is continuous. A function $f$ is Fréchet differentiable if and only if there is a bounded linear operator $T$ and a function h such that $$ f(x+u)=f(x)+T(u)+h(u)\text{, where }\lim\limits_{\Vert u\Vert \rightarrow 0}\frac{h(u)}{\Vert u\Vert}=0$$ My intuition would be that $f$ is not Fréchet differentiable. For $x,u\in L^2([0,1])$ using Taylor we get $$f(x+u)(s)=\int_{0}^{1}k(s,t)\cdot \cos((x+u)(t))dt=\int_{0}^{1}k(s,t)\cdot\left[\cos(x(t))-u(t)\cdot \sin(x(t))+\sigma(u(t)^2)\right]dt$$ where $\sigma(u^2) $ denotes the error term. If we compute this term we get $$f(x+u)(s)=f(x(s))+ \int_{0}^{1}-k(s,t)\cdot u(t)\cdot\sin(x(t))dt+\int_{0}^{1}k(s,t)\cdot\sigma (u(t)^2)dt$$ Define $T(u):=\int_{0}^{1}-k(s,t)\cdot u(t)\cdot\sin(x(t))dt$ and $h(u):=f(x+u)-f(x)-T(u)$. Clearly $T$ is linear and bounded, hence we have to check if $\lim\limits_{\Vert u\Vert \rightarrow 0}\frac{h(u)}{\Vert u\Vert}=0 $, that is $$\lim\limits_{\Vert u\Vert \rightarrow 0}\frac{\int_{0}^{1}k(s,t)\left[\cos((x+u)(t))-\cos(x(t))+u(t)\cdot \sin(x(t))\right] dt}{\Vert u\Vert} = 0$$ I assume that the term blows up, but I have not been able to prove or disprove my assumption. Please any help with this question will be appreciated, thanks.

Answer 1

Using Taylor-Lagrange inequality, for $a, b$ real numbers : $|\cos(a+b) - \cos(a) - b \sin(a) | \leq (M/2) b^2$ Thus, the ratio you consider is bounded by : $$\int_0^1 |k(s, t)| M/2 |u(t)| dt \leq M/2 \| k(s,.) \|_2 \|u\|_2$$ (the second inequality if Holder)