I want to prove that the dual of $c_0=\{(x_n)_{n\in\mathbb N}\subset\mathbb R : \lim x_n=0 \text{ and }\|x\|_\infty=\sup\limits_n|x_n|\}$ is $l^1$ .

So I defined the following map, $$T:l^1\rightarrow(c_0)^*$$ $$ x_n\mapsto f_x$$ where $f_x(a)=\sum\limits_{n=0}^\infty x_na_n$. In a next step I showed that tis map is well defined.

Now I want to show that T is an isometry and thats where I am not sure how to do it.

Could someone help me? Thanks

  • $\begingroup$ The definition of an isometry is distance preservation, so you need for $a,b\in c_0$, $\|a-b\|_{\infty}=\|T_x(a)-T_x(b)\|=\sum_{n=1}^{\infty}|a_n-b_n||x_n|$. $\endgroup$ – Ellya Mar 30 '14 at 19:04
  • $\begingroup$ @ellya I know what the definition is :D but I dont see how to obtain the equality. $\endgroup$ – Thorben Mar 30 '14 at 19:06
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    $\begingroup$ Consider $a \in c_0$ given by $a_n = \operatorname{sign} x_n$ for $n \leqslant N$ and $a_n = 0$ for $n > N$. $\endgroup$ – Daniel Fischer Mar 30 '14 at 19:08
  • $\begingroup$ no worries, just checking, is what you've found similar to what I have above? $\endgroup$ – Ellya Mar 30 '14 at 19:08

Now choose $x_n=\frac{1}{2^n}$, $ \forall n$, so $\|T_x(a)-T_x(b)\|=\sum_{n=1}^{\infty}|a_n-b_n||x_n|\leq\sum_{n=1}^{\infty}\frac{\|a-b\|_{\infty}}{2^n}=\|a-b\|_{\infty}$, so we just need to show the other way now and you're done.


The map $T$ is linear, hence we have to show that $\lVert f_x\rVert_{(c_0)^*}=\lVert x\rVert_1$.

Notice that $$|f_x(a)|\leqslant\sum_{j=1}^{+\infty}|a_j||x_j|\leqslant \sup_{j\geqslant 1}|a_j|\cdot \lVert x\rVert_1=\lVert a\rVert_\infty \cdot \lVert x\rVert_1,$$ hence $\lVert f_x\rVert\leqslant \lVert x\rVert_1$.

In order to show the converse inequality, consider for a fixed integer $N$ the sequence $a:=(\epsilon(1),\dots,\epsilon(N),0,\dots)$, an element of $c_0$, where $\epsilon(j)=1$ if $x_j\geqslant 0$ and $-1$ otherwise. The norm of $a$ is $1$ and we have $$f_x(a)=\sum_{j=1}^N|x_j|.$$


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