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I am in trouble trying to solve the following:

Let $X = \ell^1$ with the canonical norm and let $f \colon \ell^1 \ni x\mapsto \Vert x \Vert$. Then $f$ is Gateaux differentiable at a point $x = (x_1, x_2, \ldots )$ if and only if $x_i \ne 0$ for every $i \in \mathbb N$ with Gateaux differential given by
$$ \Phi_G f(x) = \left( \frac{x_1}{\vert x_1 \vert}, \ldots , \frac{x_n}{\vert x_n \vert}, \ldots \right ) $$

Well, I show you what I've done.

1 If for some $i$ we have $x_i = 0$ then I am quite sure the limit $$ \lim_{t \to 0} \frac{f(x + te_i) - f(x)}{t} $$ does not exist (in an analogous way to the one variable function $x \mapsto \vert x \vert$) but I don't manage to prove it rigorously (how can I evaluate the left and the right limits?).

2 Secondly, I do not know how to show that $f$ is Gateaux differentiable if $x_i \ne 0$ for every $i$. How can I do?

3 Eventually, here there is a self-posed question: is the function $f$ Fréchet differentiable at some point $x$? My guess is "no", since Gateaux differential is non-linear but I have some doubts...

I thank you for any help.

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I have some questions: 1) Do you mean that $f(x)$ is the l1 norm of $x$ or the absolut value of each component? 2) The Gateaux derivative needs two arguments: $\Phi_G(x)$ is the derivative at the point $x$ but I think that you need the 'direction', isn't it? –  guacho May 24 '13 at 14:38
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1 Answer

up vote 1 down vote accepted

Fix some $v\in \ell^1$ and note that $$\tag{1}\frac{\|x+tv\|-\|x\|}{t}=\frac{\sum_{i=1}^\infty(|x_i+tv_i|-|x_i|)}{t}$$

It follows from $(1)$ that $$\tag{2}\lim_{t\to 0}\frac{\|x+tv\|-\|x\|}{t}=\sum_{i=1}^\infty \lim_{t\to 0}\frac{(|x_i+tv_i|-|x_i|)}{t}$$

Now the question is: In what points the function $|\cdot|:\mathbb{R}\to\mathbb{R}$ is differentiable? and what is its derivative in those points?

For your third question: the fact that Gateaux differential is non-linear has nothing to do with $f$ being or not Frechet differentialbe. Try to verify if in the points where $f$ is Gateaux differentiable the derivative is continuous.

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