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I am looking to either prove or disprove that if a linear operator is bounded on an orthonormal basis of a (separable) Hilbert space, the operator is itself bounded.

Expanding each $x$ according to the orthonormal basis $\left\{e_{j}\right\}$ and using $\left\|Te_j \right\| \leq K, \forall j \in \mathbb{N} $ what I have got so far is

\begin{align} \left\|Tx \right\|^2 = \left\| \sum_{i=1}^\infty \langle x,e_i \rangle Te_i \right\|^2 &= \sum_{i=1}^\infty \sum_{j=1}^\infty \langle x,e_j \rangle \langle x,e_j \rangle \langle Te_i, Te_j \rangle \\ &\leq K^2 \left( \sum_{i=1}^\infty \langle x, e_i \rangle \right)^2 \end{align}

but we would like to have is $\left\|T_x \right\|^2 \leq C \left\|x\right|^2$, where $\left\| x \right\|^2 = \sum_{i=1}^\infty | \langle x, e_i \rangle|^2 $ by Bessel's equality. I'm wondering then, is there another way to prove the statement or is it perhaps that the statement is untrue?

Thank you in advance.

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    $\begingroup$ As indicated by Jochen below, the answer is no. However, if $A : H\to H$ and you replace $\|Ae_n\|$ being bounded by the requirement that each of the linear functionals $x\mapsto\langle Ax,e_n\rangle$ is bounded (which means that $e_n\in dom\,A^*$), then $A$ is bounded since then $A^*$ is densely defined. This is equivalent to $A$ being closable. But as $dom\,A = H$, $A$ is bounded by the closed graph theorem. $\endgroup$ – amsmath Oct 10 '17 at 14:14
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No. Extend $\{e_i: i\in\mathbb N\}$ to a HAMEL basis $B$ such that $\|b\|\le 1$ for all $b\in B$ and define $f:B\to \mathbb R$ so that $f(e_i)=1$ but $f$ is unbounded and extend it to a linear map on the whole space. This extension isn't continuous.

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  • $\begingroup$ But does there exist an extension that is continuous? $\endgroup$ – mr_e_man Dec 5 '19 at 2:15
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We have a orthonormal sequence $(e_n)$ such that $$ x =\sum_{n=1}^\infty \langle x,e_n\rangle e_n \quad \forall x\in H $$ and the existence of $K>0$ such that $$ \|Te_n\|_H \le K \quad \forall n. $$ Moreover, $T:H\to H$ is linear, and it remains to prove continuity. As the counter-example in the other answer shows this is not possible. I will show that under an additional assumption we can get continuity.

Now let us take $x$ such that $$ x = \sum_{n=1}^N \langle x,e_n\rangle e_n $$ for some (finite) $N$. By linearity $$ Tx= \sum_{n=1}^N \langle x,e_n\rangle T e_n $$ and by orthonormality for $y\in H$ $$ |\langle Tx,y\rangle| \le \sum_{n=1}^N \langle x,e_n\rangle \langle Te_n,y\rangle\le \left(\sum_{n=1}^N |\langle x,e_n\rangle|^2\right)^{1/2} \left(\sum_{n=1}^N |\langle Te_n,y\rangle|^2\right)^{1/2}\\ \le \|x\| \left(\sum_{n=1}^N |\langle Te_n,y\rangle|^2\right)^{1/2}. $$ This shows by using this identity twice $$ \|Tx\|^2 \le \|x\| \left(\sum_{n=1}^N |\langle Te_n,Tx\rangle|^2 \right)^{1/2}\\ \le \|x\|^2 \left(\sum_{n,m=1}^N |\langle Te_n,Te_m\rangle|^2 \right)^{1/2}. $$ Now if in addition to the original assumptions we would have $$ \sum_{n,m=1}^\infty |\langle Te_n,Te_m\rangle|^2<\infty $$ then this proves that $T$ is bounded on the dense subspace $span(e_n,n\in \mathbb N)$. Hence it can be uniquely extended to a continuous operator on the whole space.

My assumption above yields to a so-called Hilbert-Schmidt operator. The assumption is much too strong, as the example $T=I$ shows.

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  • $\begingroup$ If you make the assumption that $T$ is Hilbert-Schmidt, you don't need to go through all these steps. The operator would be automatically compact, hence bounded. $\endgroup$ – JohnK Oct 11 '17 at 5:31
  • $\begingroup$ I do not assume Hilbert-Schmidt. I used an assumption on the $Te_n$'s which make $T$ continuous, but then also Hilbert-Schmidt. $\endgroup$ – daw Oct 11 '17 at 18:31

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