Linear operator on a hilbert space

Question

Let $\{e_{n}\}_{n = 1}^{\infty}$ be an orthonormal basis of a Hilbert space $H$. Suppose $T: H \rightarrow H$ is a linear operator. For each $x \in H$, then $x = \sum_{n}\langle x, e_{n}\rangle e_{n}$ where the convergence in the norm topology of $H$. Must $Tx = \sum_{n}\langle x, e_{n}\rangle Te_{n}$?

Answer 1

@Vladmir asks if $T$ is bounded or closed, since this does not hold in general for linear operators. If the domain of your operator $T$ is all of $H$ and it is closed, then it is also bounded. For bounded linear operators, it is certainly true. That is because these operators are continuous and $$Tx = T\left(\lim_{N\to\infty} \sum_{n=0}^N \langle x, e_n \rangle e_n\right)= \lim_{N\to\infty} T\left( \sum_{n=0}^N \langle x, e_n \rangle e_n\right) = \lim_{N\to\infty} \sum_{n=0}^N \langle x, e_n \rangle Te_n = \sum_{n=0}^\infty \langle x, e_n \rangle Te_n$$