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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.

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  • $\begingroup$ math.stackexchange.com/questions/2508079/… $\endgroup$
    – user284331
    Commented Dec 24, 2019 at 17:43
  • $\begingroup$ 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 $|x_k|$. $\endgroup$ Commented Dec 24, 2019 at 17:45
  • $\begingroup$ 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. $\endgroup$
    – Thomas
    Commented Dec 24, 2019 at 17:54

1 Answer 1

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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.

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