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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.

JohnK
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    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. – amsmath Oct 10 '17 at 14:14

3 Answers3

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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.

Jochen
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  • But does there exist an extension that is continuous? – mr_e_man Dec 05 '19 at 02:15
  • @mr_e_man: Consider the sequence $x_n = \sum_{i=1}^n \frac{1}{i} e_i$. This sequence converges in $H$ by Bessel's inequality and completeness (since $\sum_{i=1}^n \frac{1}{i^2} < \infty$). But $f(x_n) = \sum_{i=1}^n \frac{1}{i}$ which diverges, and therefore $f$ cannot be continuous. – Nate Eldredge Dec 03 '21 at 20:24
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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.

daw
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  • 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. – JohnK Oct 11 '17 at 05:31
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    I do not assume Hilbert-Schmidt. I used an assumption on the $Te_n$'s which make $T$ continuous, but then also Hilbert-Schmidt. – daw Oct 11 '17 at 18:31
  • @JohnK Slightly beside the point but that implication you claimed in your comment is incorrect; a linear operator $T:H\to H$ can satisfy $\sum_j|Te_j|^2<\infty$ (or $\sum_{j,k}|\langle Te_j,Te_k\rangle|^2<\infty$) for some orthonormal basis $(e_j)_j$ of $H$ without being bounded – Frederik vom Ende Mar 17 '25 at 20:37
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I'm not saying it is, but this could be (in fact should be!) a homework question.

Here's an explicit counterexample: for each $n = 1,2,3,\ldots$, define

$$T(e_n) = u_n = n^{-1/2} (e_1 + e_2 + \cdots + e_n).$$

Then $\|u_n\| = 1$, and we have defined a linear operator $T$ on the finite linear span of the orthonormal basis $(e_n)$, with $\|T(e_n)\|=1$.

Now consider

$$T(u_n) = n^{-1/2} \sum_{j=1}^n a(j,n)e_j$$

for some coefficients $a(j,n) \geq 0$, which we must estimate. Thus

$$ \|T(u_n)\| = n^{-1/2} \left( \sum_{j=1}^n |a(j,n)|^2 \right)^{1/2}.$$

If you carefully consider what happens when $j \leq n/2$ and do some estimates, you should be able to show (using integration estimates, or just elementary algebra and induction - another great exercise!) that $a(j,n) \geq a(n/2, n) \geq An^{1/2}$ for some fixed $A>0$ and all even $n$, say.

Putting it back in gives

$$ \|T(u_n)\| \geq An^{-1/2} \left( \sum_{j=1}^{n/2} n \right)^{1/2} \geq Bn^{1/2} $$

for all even $n$, and some fixed $B>0$ (actually for odd $n$ also if you think about it).

Thus we have an explicit operator $T$ and explicit vectors $u_n$ such that $\|u_n\| = 1$ but $\|T(u_n)\| \rightarrow +\infty$, which shows that $T$ is unbounded.

Note: we've shown that $T$ is unbounded on a particular dense but non-closed linear subspace, the set of all finite linear combinations of $(e_j)$. Thus, obviously $T$ cannot be extended to a bounded linear operator on the whole space. We have not given an explicit example of an unbounded operator on the whole space. In fact, such examples are impossible to give: some nonconstructive axioms such as the axiom of choice are required (to produce Hamel bases, etc.), because it is consistent with ZF set theory that all linear operators on the whole Hilbert space (or, more generally, a Banach space, or even a Frechet space) are bounded! (And thus, any unbounded linear operator on the whole space is necessarily nonconstructive within "ordinary" mathematics. So in some sense, it's actually a matter of opinion whether unbounded linear operators on the whole space even exist!)

Zen Harper
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