Let $H$ be an infinite dimensional separable complex Hilbert space.
Let $T \in B(H)$ be a unilateral weighted shift,
i.e. there exists an orthonormal basis $(e_n)_{n\in \mathbb{N}}$ for $H$ and a sequence $(w_n)_{n\in \mathbb{N}} \in \ell^{\infty}$ such that $Te_n= w_n e_{n+1}$.
Q: Write $\|T \|$ in terms of the sequence of weights $(w_n)_{n\in \mathbb{N}}$ for $T$.
Attempt:
So we have $\|T e_n\|= \| w_n e_{n+1} \|$.
$\|Te_n \|\leq | w_n| \| e_{n+1} \|$
i.e. $\|T \|\leq | w_n| $
On the other hand,
$\|T \|= \sup\{|Tx| : \| x\|=1 \} \geq |Te_n| = |w_n e_{n+1}| = |w_n| $
Thus, $\|T\| = |w_n|$.
Could you let me know if my work is correct?
Thank you!
Best Answer
You are right about your second inequality $||T|| \geq |w_n|$, but don't forget that it's true for all $n \in \mathbb{N}$. In the first part of your proof you have a logic mistake -- the equality $||T e_n|| = |w_n|||e_{n+1}||$ (and it's indeed an equality) implies only that $||T|| \geq |w_n|$ :) yes, again, because the norm of an operator is such number that the inequality $||Tx|| \leq ||T||||x||$ holds for all $x \in H$. Now that you now the answer from a comment, you should be able to prove that the last property of the norm is satisfied for it. Combine this with your second inequality which is true for all $n \in \mathbb{N}$ and you'll get the result.