# Where is the $l_\infty$ norm Fréchet differentiable?

Suppose $$(V, \|\cdot\|_V)$$ and $$(W, \|\cdot\|_W)$$ are two Banach spaces and $$f: V \to W$$ is some function. We call a bounded linear operator $$A \in B(V, W)$$ Fréchet derivative of $$f$$ in $$x \in V$$ iff

$$\lim_{h \to 0} \frac{\|f(x + h) - f(x) - Ah\|_W}{\|h\|_V} = 0$$

We call a $$f$$ Fréchet differentiable in $$x$$ iff there exists a Fréchet derivative of $$f$$ in $$x$$.

Now suppose $$l_\infty$$ is the set of bounded real sequences equipped with "uniform convergence" norm $$\|(x_n)_{n \in \mathbb{N}}\|_\infty = \sup\{|x_n|| n \in \mathbb{N}\}$$. It is not hard to see, that $$l_\infty$$ is a Banach space.

Suppose $$f: l_\infty \to \mathbb{R}$$ is defined as $$(x_n)_{n \in \mathbb{N}} \mapsto \|(x_n)_{n \in \mathbb{N}}\|_\infty$$. Suppose $$D$$ is the set of all points of $$l_\infty$$ in which $$f$$ is Fréchet differentiable. Is there some sort of explicit description of $$D$$?

The only thing I currently know about it is that if for $$(x_n)_{n \in \mathbb{N}} \in l_\infty \exists n_0 \in \mathbb{N}$$ such that $$|x_{n_0}|> sup\{x_k| k \neq n_0\}$$, then $$(x_n)_{n \in \mathbb{N}} \in D$$ because $$(h_n)_{n \in \mathbb{N}} \mapsto sign(x_{n_0})h_{n_0}$$ is a Fréchet differentiable of $$f$$ in that point.

However, have no idea how to deal with other cases.

• math.stackexchange.com/questions/2508079/… Commented Dec 24, 2019 at 17:43
• The one case you mention is not correct; you need $|x_n|$ to be not just greater than $|x_k|$ for all $k\neq n$ but also greater than the sup of all such $|x_k|$. Commented Dec 24, 2019 at 17:45
• at the end of your question you are using $n$ both as a fixed integer and a sequence index. That does not make sense, could you please fix that? Thx. Commented Dec 24, 2019 at 17:54

If $$x=(x_n)$$ is such that $$|x_n|>\sup_{k\neq n}|x_k|$$ for some $$n$$, then $$f$$ is given by $$h\mapsto \operatorname{sign}(x_n)h_n$$ in a neighborhood of $$x$$ and so is differentiable at $$x$$, as you say. At any other point, I claim $$f$$ is not differentiable.
To prove this, suppose $$x$$ does not satisfy the condition above. This means that either there are two distinct $$m,n$$ such that $$|x_m|=|x_n|=f(x)$$, or else there exists a subsequence $$(x_{k_n})$$ such that $$|x_{k_n}|\to f(x)$$. In either case, we can find two disjoint subsets $$S,T\subset\mathbb{N}$$ such that $$\sup_{n\in S}|x_n|=\sup_{n\in T}|x_n|=f(x)$$ (in the first case take $$S=\{m\}$$ and $$T=\{n\}$$ and in the second case split $$\{k_n\}$$ into two infinite sets $$S$$ and $$T$$). Now let $$h=(h_n)$$ be the sequence such that $$h_n=\operatorname{sign}(x_n)$$ for $$n\in S$$, $$h_n=-\operatorname{sign}(x_n)$$ for $$n\in T$$, and $$h_n=0$$ otherwise. Then $$f(x+th)=f(x-th)=f(x)+t$$ for all sufficiently small $$t$$. So if $$f$$ had Frechet derivative $$A$$ at $$x$$, then it would have to satisfy $$Ah=A(-h)=1$$, which is a contradiction since $$A$$ must be linear.