Notation: Let $X$ and $Y$ be Banach spaces. Assume that $T:X\to Y$ is a linear operator with norm $$\|T\|:=\sup_{\|x\|=1}\|Tx\|.$$ Recall that $$\ell^2=\{(x_n)_{n=1}^\infty: \sum_{n=1}^\infty \|x_n\|^2<\infty \}.$$
In this post, OP asked the following question:
Question: Is it always true that operator norm attain supremum?
Omnomnomnom gives the following answer:
For example: take $T:\ell^2 \to \ell^2$ to be given by $$ T[(x_n)_{n=1}^\infty] = \left(\frac{n-1}{n}x_n\right)_{n=1}^\infty $$ We have $\|T\| = 1$, but there is no non-zero $x \in \ell^2$ with $\|Tx\| = \|x\|$.
I have trouble showing $\|T\|=1.$
I can show that $\|Tx\|\neq \|x\|$ for all $x\in\ell^2.$ Indeed, for any $(x_n)_{n=1}^\infty\in \ell^2,$ we have $$\|T(x_n)_{n=1}^\infty\| = \sqrt{\sum_{n=1}^\infty \bigg(\frac{n-1}{n}\bigg)^2 \|x_n\|^2} < \sqrt{\sum_{n=1}^\infty \|x_n\|^2} =\|(x_n)_{n=1}^\infty\|.$$
By similar reasoning, I can also show that $\|T\|\leq 1.$ However, I fail to show that $\|T\| =1.$
Any hint would be appreciated.
Let $\delta_m = (\delta_m(n))_{n = 1}^\infty$ be the sequence defined by $\delta_m(n) = 1$ if $m = n$ and $\delta_m(n) = 0$ otherwise. So $\delta_m$ is the sequence $0,0,\dots,0,1,0,\dots$ with the $1$ in the $m$-th position. We then have
$$ T\delta_m = 0,0,\dots,0,\frac{m-1}{m},0,\dots$$
So as you can see, $||\delta_m|| = 1$ and $||T\delta_m|| = (m - 1)/m$.
You've already seen that $||T|| \le 1$. This shows that
$$ ||T|| \ge \frac{m-1}{m}, \quad \forall m \ge 1,$$
and it follows that $||T|| = 1$.