How to prove $(\ell_1\ |.\|)$ is Gateaux differentiable at $x$ if and only if $x_i\not= 0$ for all $i\in\mathbf{N}$?

Question

I am in trouble to prove that $\|x\|=\sum\limits_{i=1}^{\infty}|x_i|$ of $x=(x_i)\in\ell_1$ is Gateaux differentiable at $x$ if and only if $x_i\not= 0$ for all $i\in \mathbf{N}$. I want to use this theorem

the norm $\|.\|$ is Gateaux differentiable at $x\in S_X$ if and only if there is a unique $f\in S_{X^*}$ such that $f(x)=1.$

Since $\ell_{1}^{*}=\ell_{\infty},$ so we have that foe every $f\in\ell_{1}^{*}$ there is a unique $(a_i)\in\ell_{\infty}$ such that $f(x)=\sum\limits_{i=1}^{\infty}a_ix_i$ for all $x=(x_i)\in \ell_1$

Please help me how to continue? Thanks.

Answer 1

Given $x$ such that $\sum |x_i|=1$, you are looking for $a$ with $\sup |a_i|=1$ such that $\sum x_i a_i=1$.

By the triangle inequality, $$\sum x_i a_i\le \sum |x_i a_i|\le \sum |x_i|=1$$ so equality is required at every step. This means $x_ia_i=|x_ia_i|$ for every $i$.

• If all $x_i$ are nonzero, the only choice of $a_i$ is $\operatorname{sign} x_i$. This is why $a$ is uniquely determined.
• If some $x_i$ is zero, you can choose $a_i=\pm 1$ (or in between) as you wish. So, $a$ is not uniquely determined.