Let $C([0,1])=\{f|f:[0,1]\to \Bbb R$ is continuous$\}$ with norm $||f||_{\infty}=\sup_{x\in[0,1]}|f(x)|$.
Prove that $L:C([0,1])\to C([0,1])$, define by $L(f)(x)=\int_0^x f^2(t)dt$ is Fréchet differentiable.
I have to prove that $$ \dfrac {L (f+h) (x) -L (f) (x) + Ah} {|| h ||} = 0 $$ when $ || h || \to 0 $, for some $ A $ array. Some help to do the exercise, to see which matrix is the one that works for me. Thank you.
There are a couple of mistakes in what you said: You need to find a continuous linear operator $A \colon C[0,1] \to C[0,1]$ such that $$\lim_{h \to 0} \dfrac{L(f+h)-L(f)+Ah}{\|h\|_\infty} = 0$$ which means that, in formal terms, for every $\varepsilon>0$ you can find a $\delta>0$ with the property that for any $h \in C[0,1]$ with $\|h\|_\infty<\delta$ we have $$\left\| \dfrac{L(f+h)-L(f)+Ah}{\|h\|_\infty} \right\|_\infty<\varepsilon. \tag{$*$}$$ Now observe that the computation \begin{align} L(f+h)(x) - L(f)(x) &= \int_0^x (f(t)+h(t))^2dt - \int_0^x f(t)^2dt \\ &= \int_0^x (2f(t)h(t)+h(t)^2)dt \\ &=2\int_0^x f(t)h(t)dt+L(h)(x) \end{align} suggests us to define $A$ by the formula $$(Ah)(x) = -2\int_0^x f(t)h(t)dt.$$ Can you prove that this operator works? (That is, prove that $A$ is linear, continuous, and that for every $\varepsilon>0$ you can find a $\delta>0$ such that $(*)$ holds for any $h \in C[0,1]$ with $\|h\|_\infty<\delta$).