How to show this diagonal operator is bounded?

We define the diagonal operator on a separable Hilbert space as $$Tx=\sum\limits_{n=1}^\infty \lambda_n\langle x,e_n\rangle e_n$$ where ${e_n}$ is an orthonormal basis and $\lambda_n$ is a sequence in $l^\infty$. I want to show the operator T is bounded, but I run into a slight problem at the end with an inequality. Since we have a separable Hilbert space with an orthonormal basis, every element can be expressed as $x=\sum\limits_{n=1}^\infty\langle x,e_n\rangle e_n$ What I want to show is $||Tx||\leq M||x||$ for some $M>0$ We can define $M$ to be the absolute value bound of $\lambda_n$ since the sequence is bounded. Starting with $||Tx||=||\sum\limits_{n=1}^\infty \lambda_n\langle x,e_n\rangle e_n||$ and applying the triangle inequality we get $||Tx||\leq M\sum\limits_{n=1}^\infty||\langle x,e_n\rangle e_n||$ which is not what we want because $||x||=||\sum\limits_{n=1}^\infty\langle x,e_n\rangle e_n||$ which is quite different than $\sum\limits_{n=1}^\infty||\langle x,e_n\rangle e_n||$

Solution 1:

Do not use triangle inequlaity. Instead, use orthogonality. $\|\sum a_ne_n\|^{2}=\sum |a_n|^{2}$ and the proof of $\|T|| \leq M$ becomes obvious if you use this (for both $\|Tx\|^{2}$ and $\|x\|^{2}$). [$\sum |\lambda_n \langle x, e_n \rangle|^{2} \leq M^{2} \sum | \langle x, e_n \rangle|^{2}=M^{2}\|x\|^{2}$]. Note also that $Te_n=\lambda_n e_n$ so $\|T\|\geq |\lambda_n|$ for each $n$. Hence, $\|T\|\geq \sup_n|\lambda_n|$ proving that $\|T\|=\sup_n|\lambda_n|$