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At what points $x\neq 0$ is the mapping $x\mapsto \Vert x\Vert_{\infty}$ Fréchet differentiable on $c_0:=\lbrace (x_n)_{n\in\mathbb{N}} \subset \mathbb{C}:x_n\rightarrow 0\rbrace$? A function $f$ is Fréchet differentiable in $x$ if there exists a bounded linear operator $T$ such that $$\lim\limits_{h \rightarrow 0}\frac{f(x+h\cdot v)-f(x)}{h}=Tv\text{ , uniformly in } v\in\overline{B_1(0)}$$ Right now I have concluded that $\Vert T\Vert$ is bounded by $1$ since $$\frac{\Vert x_n+h\cdot v\Vert_{\infty}-\Vert x_n\Vert_{\infty}}{h}\leq \frac{\Vert x_n+ hv-x_n\Vert_{\infty}}{|h|} =\Vert v\Vert_{\infty}\leq 1\forall (x_n)_{n\in\mathbb{N}}\subset c_0$$ Is there a trick I could use to derive the points where $\Vert \cdot \Vert_{\infty}$ is Fréchet differentiable?

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  • $\begingroup$ An operator whose the norm is finite is always bounded. Your reasoning is wrong for $h < 0$. Using the reversed triangle inequality, you will get that the slope is bounded by $\|v\|_{\infty}$ regardless of the sign of $h$. $\endgroup$
    – halbaroth
    Nov 21, 2021 at 14:30
  • $\begingroup$ @halbaroth Right, but because of $v\in B_1(0)$ it is also bounded by 1. Nevertheless, I find it hard to imaging what the derivative of $\Vert x\Vert_{\infty}=sup_{n\in\mathbb{N}}|x_n|$ should look like in general. $\endgroup$
    – stats19
    Nov 21, 2021 at 15:37
  • $\begingroup$ I didn't say that your conclusion was wrong but your reasoning was ;) Do you want a nice closed-form expression for the differential of $x \to \|x\|_{\infty}$? It is not what you were asking for in the first place. $\endgroup$
    – halbaroth
    Nov 21, 2021 at 18:06

2 Answers 2

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Let $x$ be a sequence in $c_0$ with the property that there exists a unique $n_x$ such that $|x_n|< |x_{n_x}|$ for all $n\neq n_x$. Then for $h\in c_0$ with $\|h\|_\infty$ sufficiently small, it follows that $\|x+h\|_\infty=|x(n_x)+h(n_x)|$. Of course, $\|x\|_\infty=|x(n_x)|.$ Hence, \begin{align} &\frac{|f(x+h)-f(x)-A(x)(h)|}{\|h\|_\infty} \\ &=\frac{|\|x+h\|_\infty-\|x\|_\infty-A(x)(h)|}{\|h\|_\infty}\\ &=\frac{\lvert|x(n_x)+h(n_x)|-(|x(n_x)|+A(x)(h))\rvert}{\|h\|_\infty}. \end{align} Can you figure out what $A(x)(h):c_0 \to \mathbb{C}$ should be?

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  • $\begingroup$ One should note that this proof uses the particular properties of $c_0$. The proof would be false for the $\|\cdot\|_\infty$-norm on $c$ (the space of converging sequences) $\endgroup$
    – daw
    May 8 at 14:00
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The mapping is not Fréchet differentiable. One can show that the candidate for the derivative $A$ is $A(v)=0 \ \ \forall v\in \overline{B_1(0)}$. To see this let $n_x\in\mathbb{N}$ be such that $\Vert x\Vert_{\infty} = |x_{n_x}|>0.$ Because $x_n\in c_0$ there is $N\in\mathbb{N}$ such that $|x_n|\leq 0.5\cdot|x_{n_x}|\ \ \forall n\geq N.$ Consider $v:=(0,\dots,0,1,0,\dots) $, where the N-th component is $1$. Then $$|x_N+h|\leq |x_n|+h< |x_{n_x}| \ \ \forall h<0.5\cdot |x_{n_x}|$$ $$\implies\left| \frac{\Vert x+h\cdot v\Vert_{\infty}-\Vert x\Vert_{\infty}}{h}-Av\right| =\left|\frac{\vert x_{n_x}\vert-\vert x_{n_x}\vert}{h}\right|=0 $$ This shows that $A(v)=0$ is the possible derivative. Now if we consider $\tilde{v}=\frac{x_{n_x}}{|x_{n_x}|}\cdot (0,\dots,0,1,0\dots)$ with $1$ at the $n_x-th$ spot, we see that $\tilde{v}\in \overline{B_1(0)}$, but the term equals $1\neq 0$ . Hence, the mapping is not Fréchet differentiable.

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  • $\begingroup$ This answer is wrong. You calculated $A(v)=0$ for one particular choice of $v$. See the answer of ProfOak for points, where the norm is differentiable. $\endgroup$
    – daw
    May 8 at 13:57

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