Fréchet derivative of a functional.

Question

I need to show that the functional defined by $$\int_{0}^{1} \cos(u(x)) dx$$ is not Fréchet differentiable in $L^2((0,1))$. A hint is that the requested property of the remainder for a Fréchet derivative is already violated for step functions.

I was looking for similar questions and see that under a Taylor expansion, one can introduce a first order aproximation of $\cos(\cdot)$ and then take the second order residual as a step function. For example, If we consider the functional defined as $$y(x)=\sin(u(x)), \quad u \in L^2(\Omega)$$ then we have $$\sin(0+h(x))=\sin(0)+\cos(0)h(x)+\int_{0}^{1}[\cos(0+sh(x))-\cos(0)]ds$$ for $h\in L^p(0,1)$ given and $s \in (0,1)$. As far I know, that expansion is a Taylor expansion in the integral form, but I dont see why we can put those terms in the integral, because I know that a expansion of this kind wold be of the form: $$f(a+x)=f(a)+f'(a)x+\int_{0}^{1}\frac{f''(a+sx)}{2}x^2ds.$$

Is the last expression wrong? And, would it be usefull a Taylor expansion to show the non-differentiability of the first functional given? A important fact is that in the $\sin(\cdot)$ example, taking $h$ as the step function $$h(x)=\begin{cases}1,&x\in [0,\epsilon]\\0,&x\in[\epsilon,1]\end{cases}$$ it's possible to show the non-differentiability of $\sin(\cdot)$ as a funtional, and I was thinking in a similar argument.

Please any help with this question will be aprecciated. Thanks.

Answer 1

Suppose that the functional has a Frechet derivative at the constant function $0$. Then we would have for $u\in L^2([0,1])$: $$ \int_0^1\cos(u(x))\,dx=1+\langle\phi,u\rangle+o(\|u\|_2), $$ where $\phi$ is an element of the dual of $ L^2([0,1])$. By the Riesz representation theorem, there exists $f\in L^2([0,1])$ such that $\langle\phi,u\rangle=\int_0^1f(x)\,u(x)\,dx$. Take now $u$ the characteristic function of $[0,a]$, $0. Then the above equation becomes $$ a\cos1+1-a=1+ \int_0^af(x)\,dx+o(\sqrt a). $$ Divide everything by $a$ and let $a\to0$ and get a contradiction (the LHS is constant and the RHS blows up).