What is the norm of an operator $T:c_0\to\ell^\infty$?

Question

I have the operator $T:c_0\to\ell^\infty$ defined by $$T(x)=\left(\sum_{j\geq1}a_{ij}x_j\right)_{i=1}^\infty$$ where $\sup_{i\geq1}\sum_{j\geq1}|a_{ij}|<\infty$ such that $T(x)\in\ell^\infty$. I know that $T$ is a bounded linear operator because $$\|T(x)\|_\infty\leq\left(\sup_{i\geq1}\sum_{j\geq1}|a_{ij}|\right)\|x\|_\infty$$ and I've been asked to show that $$\|T\|=\sup_{i\geq1}\sum_{j\geq1}|a_{ij}|$$ however, I'm unable to see how this can happen if $x\in{c}_0$. The definition for $\|T\|$ I'm using is $$\|T\|\equiv\sup_{x\in{c}_0:\,x\neq0}\frac{\|T(x)\|_\infty}{\|x\|_\infty}$$ which, using the definition $\|x\|_\infty=\sup_{j}|x_j|$, I can write as $$\|T\|=\sup_{x\in{c}_0:\,x\neq0}\left(\sup_{i\geq1}\sum_{j\geq1}|a_{ij}|\frac{|x_j|}{\sup_k|x_k|}\right)$$ so ultimately, to get the answer I've been given, I should have $$\sup_{x\in{c}_0:\,x\neq0}\left(\frac{|x_j|}{\sup_k|x_k|}\right)=1$$ for each $j\in\mathbb{N}$, right? But how can this happen if $x_j\to0$ when $j\to\infty$ by definition of $x\in{c}_0$? There seems to be no problem to get the result if $x\in\ell^\infty$, but I can't see how this happens too in this case.

Answer 1

Fix $\epsilon>0$.

Define $u= \sup_{i \geq 1} \sum_{j=1}^{\infty} |a_{ij}|$. Choose $N \in \Bbb N$ such that $\sum_{j=1}^{\infty} |a_{Nj}| > u-\epsilon/2$. Then choose $M \in \Bbb N$ such that $\sum_{j=M+1}^{\infty} |a_{Nj}|<\epsilon/2$. Then it is apparent that $\sum_{j=1}^M |a_{Nj}| > u-\epsilon$.

Define $x \in c_0$ as follows: $$x_j =\begin{cases} 1, \;\;\;\;\;\;\;\;a_{Nj} \geq 0, \;j \leq M \\ -1, \;\;\;\;\;\;a_{Nj}<0, \; j \leq M \\ 0, \;\;\;\;\;\;\;\;\;j>M \end{cases} $$

Then you can see that $\|x\|_{\infty}=1$ and $\sum_{j=1}^{\infty} a_{Nj}x_j = \sum_{j=1}^M|a_{Nj}|$. And thus we see that $\|Tx\|_{\infty} \geq \big| \sum_{j=1}^{\infty} a_{Nj}x_j \big| = \sum_{j=1}^M |a_{Nj}| > u-\epsilon$.

Since $\epsilon$ was arbitrary, the result follows.

Answer 2

The fact that $x_j\to0$ does not have any effect. Since $x_j\to0$, there is a $k\in\mathbb N$ such that $|x_k|=\|x\|_\infty$ (this is not true in general in $\ell^\infty$.) This gives the equality, even with $\max$ instead of $\sup$.