Let $H$ a separable Hilbert Space with a ortonormal base $\{e_n: n \in\mathbb{N}\}$. Prove that $\exists ! T\in \mathcal{B}(H): T(e_n)=d_ne_n.$

Question

Let a contable and bounded subset $D=\{d_n:n\in \mathbb{N}\}\subset \mathbb{C}$. Show that exists a unique $T\in \mathcal{B}(H)$ such that $$T(e_n)=d_n e_n,\forall n\in\mathbb{N}. $$

I know that if we had a normal operator $T\in \mathcal{B}(H)$, the autovectors form a ortonormal base from $H$. I think that this exercise it is the reciprocal of this fact, but I am not getting a way to prove the existence of such an operator.

Answer 1

I'll assume $\{e_n\}_n$ is a countable orthonormal basis of your (separable) Hilbert space. Then we can write $$x= \sum_n \langle x, e_n\rangle e_n$$ where this sum converges in the norm-topology. Define

$$Tx := \sum_n \langle x,e_n\rangle d_n e_n$$

provided the sum on the right converges. Work left for you:

(1) This sum converges in the norm topology.

(2) The operator $T$ is the one you are looking for.

(3) $T$ is unique.

Answer 2

You're thinking in the opposite direction. Given an orthonormal basis (really you can get away with maximal linear independent set) and a bounded linear operator, defining the operator on the basis is enough to uniquely identify it.

Define $T$ on the basis such that $Te_n = d_n e_n$. Then show that $T$ extends to the whole space (boundedness is important here), then show that $T$ is unique.