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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 $Tx = \sum_{n}\langle x, e_{n}\rangle Te_{n}$?

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    $\begingroup$ Is $T$ assumed to be bounded? closed? What is the domain of $T$? $\endgroup$ – Vladimir Jul 24 '14 at 20:00
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    $\begingroup$ Assuming T is continuous, write the infinite sum as a limit of finite sums, and use continuity and linearity. $\endgroup$ – Nate Eldredge Jul 24 '14 at 20:07
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@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 $$Tx = T\left(\lim_{N\to\infty} \sum_{n=0}^N \langle x, e_n \rangle e_n\right)= \lim_{N\to\infty} T\left( \sum_{n=0}^N \langle x, e_n \rangle e_n\right) = \lim_{N\to\infty} \sum_{n=0}^N \langle x, e_n \rangle Te_n = \sum_{n=0}^\infty \langle x, e_n \rangle Te_n$$

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