Having a normed linear space $S=C^0[0,1]$ of continuous functions $f:[0,1] \rightarrow \Bbb R%$, with sup norm: $\|f\|=\sup_{\space x \in [0,1]}|f(x)|$,
prove that $F(f)=\|f\|$ is nowhere differentiable in $S$,
that is, for all $f_0 \in S$, there is no linear operator $A:S \rightarrow \Bbb R$ such that : $$\lim_{\space \|h\| \rightarrow 0} \frac{\|F(f_0-h)-F(f_0)-A(h)\|}{\|h\|}=0$$ where $h \in S$.
Could someone state the proof, or link to it?
On
I got the hint from the book Functional Analysis and Infinite Dimensional Geometry by Marián Fabian. The following proposition is the crucial one:
Let $X$ be a Banach space and $x\in S_{X}$. $\|\cdot\|$ is Frechet differentiable if and only if $\lim_{n}\|f_{n}-g_{n}\|=0$ whenever $f_{n},g_{n}\in S_{X^{\ast}}$ satisfy $\lim_{n}f_{n}(x)=\lim_{n}g_{n}(x)=1$.
Now $X=C[0,1]$. For $x\ne 0$, choose some $t_{0}\in[0,1]$ such that $x(t_{0})=1$. Also choose distinct $t_{n}\in[0,1]$, $t_{n}\ne 1$ and that $x(t_{n})\rightarrow 1$. The Dirac measures $\delta_{t_{n}}$ are such that $\delta_{n}(x)=x(t_{n})$. Choosing $f_{n}$ the even subsequence of $\delta_{n}$ and $g_{n}$ the odd one, and use the fact that $\|f_{n}-g_{n}\|=\text{the total variation of }f_{n}-g_{n}=2$. Note that $(C[0,1])^{\ast}$ is the space of complex Borel measures.
Suppose that it is differentiable at $0 \in S$. Let the derivative be $A: S \to \mathbb{R}$. Then for any function $f \in C^0[0,1]$ with $\|f\|=1$ we need to have $$ 0=\lim_{h \to 0^+ } \frac{\lvert \|h f \| - h A(f) \rvert}{\lvert h \rvert } = \lvert \| f \| - A(f) \rvert $$ So $A(f) = \|f\|$. But we could apply the same argument to $-f$, then we find $A(-f) = \|f\|$. Since $A$ is linear, we must have $A(f) = 0$ for all $f \in C^0[0,1]$. This leads to a contradiction.
Note that you don't really need any properties of the particular space $S$ or its norm. This proves that for any normed space, the norm isn't differentiable at $0$.