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Let $\ell^{\infty}=\{x\in \mathbb{R}^{\mathbb{N}}: x\,\, \text{is bounded}\}$ and $E=\{x\in \ell^{\infty}:x_n\rightarrow 0\}$ with the norm $||\cdot||_{\infty}$ and let $f(x)=||x||_{\infty}$. How to prove that:

a) If $x\in E$ then there exists $m\in\mathbb{N}$ such that $f(x)=x_m$.

b) $f$ is differentiable at $x\in E$ if and only if $m$ is unique.

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And... what did you try to solve these? –  Did Jul 3 '13 at 20:52
    
I've tried but I can not really get it ... –  Roiner Segura Cubero Jul 3 '13 at 20:54
2  
Hmmmm... You have to give something to work on to the people who might want to answer your question. –  Did Jul 3 '13 at 20:55
    
For example, do you think that a) is correct if $x_n=-\frac1n$? –  Hagen von Eitzen Jul 3 '13 at 20:56
    
FWIT, $E$ is usually denoted by $c_0$. –  1015 Jul 3 '13 at 21:11

1 Answer 1

up vote 0 down vote accepted

a) If $x_n$ is identically zero, the claim is trivial. So select $n$ with $x_n\ne 0$. Then there exists $N$ with $|x_k|<\frac12|x_n|$ for all $k>N$. Hence $\lVert x\rVert_\infty=\max\{|x_1|,\ldots,|x_N|\}$ and there exists $m\in\{1,\ldots,N\}$ with $f(x)=|x_m|$.

b) Select $m$ with $f(x)=|x_m|$. Let $y=x$ except that $y_m=x_m$. If $m$ is unique then $f(y)<|x_m|$ by a) and especially $|x_m|>0$. Let $\epsilon=\frac12(f(x)-f(y))$. Then for all $z\ne0$ we have $f(x+hz)=|x_m+hz_m|$ for $|h|<\frac{\epsilon}{\lVert z\rVert_\infty}$. Since $h\mapsto |a+hb|$ is differentiable at $h=0$ if $a\ne 0$, we find that $f$ is differentiable at $x$.

On the other hand, assume $f(x)=|x_m|=|x_n|$ with $n\ne m$. Let $z_m=x_m$, $z_n=-x_n$ and $z_k=0$ otherwise. Then $f(x+hz)=(1+|h|)f(x)$. If $f(x)>0$, this shows that $f$ is not differentiable at $x$. And if $f(x)=0$, we consider $z_1=1$, $z_2=-1$ instead and obtain $f(x+hz)=|h|$, again not differentiable.

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