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

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?

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 $$n>1$$.
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.