How to show that $\vert \vert T \vert \vert = \sqrt{c}$ where $c:=\sum\limits_{j=1}^{\infty}\sum\limits_{k=1}^{\infty}\vert t_{jk}\vert^{2}$

Question

Define $c:=\sum\limits_{j=1}^{\infty}\sum\limits_{k=1}^{\infty}\vert t_{jk}\vert^{2} <\infty$

and $T:\ell^{2} \to \ell^{2}$ where $(Tx)_{j}=\sum\limits_{k=1}^{\infty}t_{jk}x_{k}$ for all $j \in \mathbb N$.

I have already shown that $(Tx)_{j}$ is absolutely convergent also that $Tx\in \ell ^{2}$ for any $x \in \ell^{2}$ and lastly that $T$ is a bounded linear operator namely, $\vert \vert T \vert \vert \leq \sqrt{c}$. But now I am asked to show that $\vert \vert T \vert \vert =\sqrt{c}$ but I have no idea on how to go about this. Any ideas?

Answer 1

Let $t$ be diagonal: $t_{ii}=1$ for $i=1\dots n$, $t_{ii}=0$ for all other $i$. Then $c=n$. But $\|T\|\le 1< \sqrt c$ for $n>1$.

Answer 2

To elaborate on "the greater picture" here, the condition $c=c(T)<\infty$ is equivalent to $T$ being a Hilbert-Schmidt operator / Schatten-$p$-operator for $p=2$, cf. Prop.16.10 in the book "Introduction to Functional Analysis" by Meise & Vogt (1997).

In this case $\|T\|_2$ (the Hilbert-Schmidt / Schatten-2-norm) is finite and equal to $\sqrt c$; but it is also well-known that $$\|T\|_\infty\leq\|T\|_2=\sqrt{c}$$ with $\|\cdot\|_\infty$ being the usual operator norm.

While daw gave a simple example for $\|T\|_\infty<\sqrt{c}$, if the norm in your original question was the Hilbert-Schmidt / Schatten-2-norm then the statement you have to prove is actually correct.