Let $(e_n) = (e_n)_{n \in \mathbb{N}}$ be an orthonormal basis for a Hilbert space $H$. Let $(a_n)$ be an arbitrary sequence of complex numbers. In multiple textbooks, I see an operator $T:H \to H$ defined by $$ Te_n = a_n e_n $$

Thus $$ Tx = \sum_{n=1}^{\infty} a_n \langle x,e_n \rangle e_n $$

Is this operator well-defined?

I can see that if $(a_n)$ is a bounded sequence, then the operator is bounded: $$ \| Tx \|^2 = \sum_{n=1}^{\infty} |a_n|^2 |\langle x,e_n \rangle|^2 \leq \sup_n |a_n|^2 \sum_{n=1}^{\infty} |\langle x,e_n \rangle|^2 = \|x\|^2 $$

But if $(a_n)$ is an arbitrary sequence of complex numbers, I don't see why $\| Tx \|$ should be finite, i.e., I don't see why the sum $\sum_{n=1}^{\infty} a_n \langle x,e_n \rangle e_n$ should converge.


1 Answer 1


It is not well-defined. Take $x=\sum \frac 1n e_n$ which is a convergent series. Then the series defining $Tx$ converges iff $\sum \frac 1n a_ne_n$ converges iff $\sum \frac 1 {n^{2}}|a_n|^{2} <\infty$. Take $a_n =n$ to see that this can fail.


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