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).