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As an exercise for my analysis class, I have to find the Fréchet derivative of $F : [0,1] \times \mathcal{C}([0,1]) \rightarrow R : (x,f) \mapsto f(x)$, in $(x_0,f_0)$, where $f_0$ is differentiable in $x_0$.

I tried proceding as follows, let $L(x,f) = f'(x_0).x + L'(x,f)$: $\begin{align*}&\lim_{(x,f) \rightarrow 0} \frac{|F(x_0 + x,f_0+f) - F(x_0,f_0) - L(x,f)|}{\lVert (x,f)\rVert}\\& = \lim_{(x,f) \rightarrow 0} \frac{|(f_0+f)(x_0+x) - f_0(x_0) - L(x,f)|}{\lVert (x,f)\rVert}\\ & =\lim_{(x,f) \rightarrow 0} \frac{|(f_0+f)(x_0+x) - f_0(x_0) - f_0'(x_0).x - L'(x,f)|}{\lVert (x,f)\rVert}\\ &=\lim_{(x,f) \rightarrow 0} \frac{|f(x_0+x)+ f_0(x_0+x) - f_0(x_0) - f_0'(x_0).x - L'(x,f)|}{\lVert (x,f)\rVert}\\ &\leq \lim_{(x,f) \rightarrow 0} \frac{|f(x_0+x)- L'(x+f)| + |f_0(x_0+x) - f_0(x_0) - f_0'(x_0).x|}{\lVert (x,f)\rVert}\\ &=\lim_{(x,f) \rightarrow 0} \frac{|f(x_0+x)- L'(x+f)|}{\lVert(x,f)\rVert} + \lim_{(x,f) \rightarrow 0}\frac{|f_0(x_0+x) - f_0(x_0) - f_0'(x_0).x|}{\lVert (x,f)\rVert}\\ &= \lim_{(x,f) \rightarrow 0} \frac{|f(x_0+x)- L'(x+f)|}{\lVert(x,f)\rVert} + 0 \end{align*}$

So I have to find some continuous linear function $L'(x,f)$, such that $\lim_{(x,f) \rightarrow 0} \frac{|f(x_0+x)- L'(x+f)|}{\lVert(x,f)\rVert} = 0.$

Setting $L' = f(x_0+x)$ would do the trick, but this function is not linear, hence I cannot use it. Could anyone give me a pointer?

My guess is that it has to do with the fact that $F$ is continuous. (No full solutions please).

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    @richard, $C[0,1]$ is endowed with the sup norm.2012-11-02

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I don't think $F$ is differentiable at $(x_0,f_0)$, because once it is differentiable, $L'(x,f)$ must be of the form $cx+\int f d\mu$ for some $c\in\mathbb{R}$ and signed Borel measure $\mu$ on $[0,1]$. However, given $c$ and $\mu$, it is not hard to show that $\limsup_{(x,f)\to 0}\frac{|f(x_0+x)-cx-\int f d\mu|}{\|(x,f)\|}\ge \max(1,|c|).$

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    @DimitriSurinx: By introducing measures, I only want to show the following fact. Given any \delta>0, any $x_0\in[0,1]$ and any bounded linear functional $l:C[0,1]\to\mathbb{R}$, there exists $|x|\le \delta$ and $f\in C[0,1]$, such that $x+x_0\in[0,1]$, $l(f)=0$ and f(x+x_0)=∥f∥>0. With this fact, you may obtain the $\limsup$ inequality in my answer where $\int fd\mu$ is replaced with $l(f)$.2012-11-02