# Where is the $\Vert \cdot \Vert_{\infty}$ norm Fréchet differentiable on $c_0$?

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?

• 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 $h$. Nov 21, 2021 at 14:30
• @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. Nov 21, 2021 at 15:37
• 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. Nov 21, 2021 at 18:06

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?
• 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)
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.
• 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.