# If $(\lambda_n)_{n=1}^\infty$ is a bounded sequence, then there is a bounded linear operator $A$ on a Hilbert space $H$ such that $Ae_n=\lambda_n e_n$

If $(\lambda_n)_{n=1}^\infty$ is a bounded sequence, then there is a bounded linear operator $A$ on a Hilbert space $H$ such that $Ae_n=\lambda_n e_n$ for all $n\in \mathbb{N}$.

Let ${e_n}$ be a complete orthonormal sequence in $H$ and $\lambda_n\in \mathbb{C}$

I've proved the converse of this, but I'm stuck on this direction now. Any solutions or hints are greatly appreciated.

If $\{e_n\}$ is an orthonormal basis for a Hilbert space $H$, then $$H = \left\{ \sum_{n=1}^\infty \alpha_n e_n : \sum_{n=1}^\infty |\alpha_n|^2 < \infty \right\}.$$

For $f=\sum \alpha_n e_n \in H$ we formally define the operation: $$Af = \sum_{n=1}^\infty \lambda_n \alpha_n e_n.$$

This operator is linear, and it is a well defined map from $H \to H$ provided $$\sum_{n=1}^\infty |\lambda_n|^2 |\alpha_n|^2 <\infty.$$

However, we were already told that the sequence is bounded, let's say by $M$. Thus $$\sum_{n=1}^\infty |\lambda_n|^2 |\alpha_n|^2 \le M \sum_{n=1}^\infty |\alpha_n|^2 = M^2\|f\|^2.$$

We conclude that $A:H \to H$ is a well defined linear map, and $\|A\| <M$, so it is bounded.