# Bounded linear operator from orthonormal sequence

I'm working on the following problem:

Suppose $$\{e_n\}$$ is a complete orthonormal sequence in a Hilbert space $$H$$. Let $$\{c_n\}$$ be a sequence of complex numbers.

1. Prove that there is a bounded linear operator $$T$$ on $$H$$ such that $$Te_n=c_ne_n$$ for each $$n$$, if and only if $$\{c_n\}$$ is bounded.
2. Determine $$\|T\|$$ when $$\{c_n\}$$ is bounded.
3. Determine $$T^*$$, the adjoint of $$T$$, when $$\{c_n\}$$ is bounded.

What I've done so far: I know that if $$\{e_n\}$$ is a complete orthonormal sequence then any $$x\in H$$ can be written as $$\sum_{1}^\infty\langle x,e_n\rangle e_n$$ and $$\|x\|^2=\sum_1^\infty|\langle x,e_n\rangle|^2$$. Then if $$T$$ is linear then $$Tx=T(\sum\langle x,e_n\rangle e_n)=\sum\langle x,e_n\rangle c_ne_n$$. But I'm not sure how to proceed.

Any help on this problem would be greatly appreciated!

## 1 Answer

Suppose $$(c_n)$$ is bounded. $$\|\sum c_n\langle x, e_n \rangle e_n\|^{2}=\sum |c_n\langle x, e_n \rangle |^{2} \leq M \sum |\langle x, e_n \rangle |^{2}=M^{2}\|x\|^{2}$$ where $$M =\sup_m |c_n|$$. Hence $$T$$ is bounded with$$\|T\| \leq \sup_m |c_n|$$.

Also $$\|Te_n\| =|c_n|$$ and this implies (by definition of norm of an operator) that $$\|T\| \geq |c_n|$$. This is true for each $$n$$ so we conclude that $$\|T\|=\sup_m |c_n|$$.

Conversely, $$Te_n=c_n e_n$$ show that $$|c_n| \leq \|T\|$$ for each $$n$$ so boundedness of $$T$$ implies that $$(c_n)$$ is bounded.

$$T^{*}$$ is given by $$T^{*}x=\sum \overline {c_n} \langle x, e_n \rangle e_n$$ and I will let you verify it using definitiom of the adjoint.