Boundedness assumption in the definition of Hilbert-Schmidt operators

Question

A Hilbert Schmidt operator $T$ on a Hilbert space $H$ is typically defined as a bounded linear operator satisfying $\sum ||Te^i||^2<\infty$ with respect to some (and hence any) orthonormal basis ${e^i}$. Is the specification that $T$ is bounded really necessary, i.e. does it not follow from the Hilbert-Schmidt condition? If not, are there examples of unbounded operators satisfying the Hilbert-Schmidt condition (with respect to some orthonormal basis, as this might not be basis-independent in the unbounded setting)?

Answer 1

_{(The following proposition is a version of Chapter II.§1, Exercise 3 in Conway's "A Course in Functional Analysis", 2nd ed.)}

If you restrict the domain of $T$ properly, then you can get a result of the form "Hilbert-Schmidt implies bounded":

Proposition. Let an orthonormal basis $(e_j)_j$ of $H_1$ be given, and define $D:={\rm span}\{e_j:j\}$. If a linear operator $T:D\to H_2$ satisfies $(\sum_j\|Te_j\|^2)^{1/2}=:\|T\|_{\rm HS}<\infty$, then $T$ is bounded with $\|T\|_{\rm op}\leq\|T\|_{\rm HS}$, and thus it admits a bounded linear extension $\widehat T:H_1\to H_2$ of the same operator norm.

Proof. For all $x\in D$, using the basis expansion in Hilbert spaces (which is finite(!) here by definition of $D$), the triangle inequality, Cauchy-Schwarz on $\ell^2$, and Parseval's identity we estimate \begin{align*} \|Tx\|&=\Big\|\sum_jT\langle x,e_j\rangle e_j\Big\|\\ &\leq\sum_j\|Te_j\||\langle x,e_j\rangle|\\ &\leq\Big( \sum_j\|Te_j\|^2 \Big)^{1/2}\Big(\sum_j|\langle x,e_j\rangle|^2\Big)^{1/2}=\|T\|_{\rm HS}\|x\|\,. \end{align*} In particular, because $D$ contains an orthonormal basis of $H_1$ it is a dense subspace meaning the BLT theorem guarantees the existence of a unique bounded extension $\widehat T$ of $T$ of same operator norm. $\square$

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However, you may have noticed the deliberate formulation of the proposition and the way $D$ is defined. This is on purpose; indeed, if you do not constrain $D$ the implication you are interested in can fail:

Example. A linear operator $T:H_1\to H_2$ which satisfies $\sum_j\|Te_j\|^2<\infty$ for some orthonormal basis $(e_j)_{j\in\mathbb N}$ of $H_1$ but which is not bounded.

_{(The following construction follows the idea of this answer)}

Starting from $(e_j)_j$ and using Zorn's lemma we can complete it to a Hamel basis $\mathcal B$ of $H_1$. In fact, $\mathcal B$ is uncountable meaning the same is true for $\mathcal B_0:=\mathcal B\setminus(e_j)_{j\in\mathbb N}$. Thus we can find a countable sequence $(x_n)_{n\in\mathbb N}$ in $\mathcal B_0$. Now define a linear functional $f:H_1\to\mathbb C$ as follows:

• $f(x_n)=\|x_n\|n$ for all $n\in\mathbb N$
• $f(x)=0$ for all $x\in\mathcal B\setminus(x_n)_n$. In particular, this means $f(e_j)=0$ for all $j$

Because $f$ is defined on a basis and it is assumed to be linear, it is uniquely defined on all of ${\rm span}\,\mathcal B=H_1$. Notably, $f$ is unbounded because $\lim_{n\to\infty}f(\frac{x_n}{\|x_n\|})=\lim_{n\to\infty}n=\infty$.

Next, let any $y\in H_2\setminus\{0\}$ be given and define $T:H_1\to H_2$ via $Tx:=f(x)y$. Then $T$ is well-defined and linear (because $f$ is), and $$ \sum_j\|Te_j\|^2=\sum_j|f(e_j)|^2\|y\|^2=0<\infty $$ by definition of $f$—so $T$ has the Hilbert-Schmidt property for the orthonormal basis $(e_j)_j$—but $T$ is not bounded because $f$ is not bounded. $\square$

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In plain words, the problem is that requiring $T$ to have the Hilbert-Schmidt property for just one orthonormal basis $(e_j)_j$ leaves too much ambiguity as to how $T$ acts outside of ${\rm span}\{e_j:j\}$. In fact, the above construction continues to hold if $T$ is Hilbert-Schmidt with respect to countably many orthonormal bases. But once you require it for every orthonormal basis it is of course true, i.e., any linear map $T:H_1\to H_2$ which satisfies $\sum_j\|Te_j\|^2<\infty$ for all orthonormal bases $(e_j)_j$ of $H_1$ has to be bounded (and hence compact). The simplest argument is by contraposition: if $T$ is unbounded, then one can find a sequence of pairwise orthogonal vectors $(e_j)_{j\in\mathbb N}$ with $\|e_j\|=1$ such that $\|Te_j\|\geq j$. Hence $\sum_j\|Te_j\|^2=\infty$ and we are done.