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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?

jd27
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Dagon
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2 Answers2

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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 $0$.

Demophilus
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  • What about the points that aren't $0$? – Jonas Meyer Nov 06 '17 at 21:45
  • Ah I missed a part it seems. I only read "there is no linear operator $A:S \rightarrow \Bbb R$ such that for all $f_0 \in S$: $$\lim_{\space |h| \rightarrow 0} \frac{|F(f_0-h)-F(f_0)-A(h)|}{|h|}=0$$ where $h \in S$.", so I thought it sufficient to prove that isn't differentiable for a $f_0 \in S$. This is of course different from proving that $| \cdot |$ is nowhere differentiable. – Demophilus Nov 06 '17 at 21:49
  • Yeah, good point, they stated it wrong. – Jonas Meyer Nov 06 '17 at 21:50
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    Sorry, You are right. I've edited it. – Dagon Nov 06 '17 at 22:00
  • @Demophilus Could you explain how exactly you achieved $A(f)=|f|$? I'm particularly interested in the $hA(f)=A(h)$ equality. – Dagon Nov 07 '17 at 22:05
  • What step confuses you? I proved that $\lvert |f| - A(f) \rvert = 0$ so we must have $|f| =A(f)$. – Demophilus Nov 07 '17 at 22:07
  • I see now that I probably wrote the previous comment based on a wrong assumption, but what still confuses me, is the expression following "we need to have". How exactly $0=\lim_{h \to 0^+ } \frac{\lvert |h f | - h A(f) \rvert}{\lvert h \rvert }$ is equivalent to $\lim_{\space |h| \rightarrow 0} \frac{|F(h)-A(h)|}{|h|}=0$? I'm mostly curious about how the A(f) superposition came to be. – Dagon Nov 07 '17 at 22:33
  • They're absolutely not equivalent. The second statement implies the first statement. You should try to prove this yourself, it's not that hard. What do you mean by $A(f)$ superposition? – Demophilus Nov 07 '17 at 22:36
  • Sorry, I meant composition. – Dagon Nov 07 '17 at 22:45
  • Then I still don't know what you mean, the map $A:S \to \mathbb{R}$ maps every function $f \in S$ to a real number. There's no composition going on here, $A(f)$ just means $A$ evaluated in $f$. – Demophilus Nov 07 '17 at 22:48
  • Let us continue this discussion in chat. – Dagon Nov 07 '17 at 22:50
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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.

user284331
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