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Say that we have Hilbert spaces $\mathcal{H}_1$ and $\mathcal{H}_2$ and an operator $T : \mathcal{H}_1\to \mathcal{H}_2$ such that for any orthonormal sequence $\{e_n\}_{n=1}^{\infty}\subset \mathcal{H}_1$, $$\lim_{n\to \infty} \|T(e_n)\| = 0$$ It's not obvious to me that $T$ is a bounded operator, even in the case where $\mathcal{H}_1$ is separable, although I'm pretty sure that it must be true given that we can prove that $T$ is compact, which therefore implies that $T$ is bounded (although proving that $T$ is compact seems like overkill). Is there a way to show the boundedness of $T$ without first showing the compactness of $T$? Thanks for your time.

Michael L.
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    If an operator T is unbounded, then there should be an orthonormal sequence ${e_n}_n$ for which $|Te_n| > n$. Indeed, having constructed $e_1, \ldots, e_k$, we still must have that $T$ is unbounded on the orthocomplement of the span of the $e_1, \ldots, e_k$. – Bartosz Malman Apr 15 '17 at 09:21
  • Why should such a sequence be orthonormal? As far as I can tell, we only have that it lies on the unit sphere. – Michael L. Apr 15 '17 at 13:03
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    After choosing $e_i$, for $i = 1, \ldots, k$ such that $|Te_i| > i$, you can consider the linear span $S_k$ of the $k$ already chosen vectors. Then consider the orthocomplement $M_k = S_k^\perp$. By definition, every vector in $M_k$ is orthogonal to the vectors $e_1, \ldots, e_k$. Since $T$ is unbounded on the whole space, it is also unbounded on $M_k$ and hence there is a unit norm vector $e_{k+1} \in M_k$ such that $|Te_{k+1}|> k+1$. – Bartosz Malman Apr 15 '17 at 13:32
  • I understand. And $T$ is unbounded on the orthocomplement of $\mathrm{span}({e_1, \ldots, e_k})$ because it's unbounded on $\mathrm{span}({e_1})^{\mathrm{T}}$ by the reverse triangle inequality, and then on $\mathrm{span}({e_1, \ldots, e_k})^{\mathrm{T}}$ by induction. – Michael L. Apr 15 '17 at 14:17

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As Malman explained, the boundedness of $T$ follows by contradiction. Moreover, it suffices to assume that $T$ takes orthonormal sequences to bounded sequences. Suppose $T$ is unbounded. First pick $e_1$ such that $\|Te_1\|>1$.

After some orthonormal vectors $e_1,\dots,e_n$ with $\|Te_j\|>j$ have been selected, consider the spaces $M=\operatorname{span}(e_1,\dots,e_n)$ and $N=M^\perp$. Since $M$ is finite-dimensional, $T$ is bounded on $M$. If $T$ was also bounded on $N$, it would be bounded on $M+N$, i.e., all of $\mathcal H_1$, by the triangle inequality. Hence, $T$ is unbounded on $N$, which allows us to choose $e_{n+1}\in N$ with $\|Te_{n+1}\|>n+1$, and thus the process continues indefinitely.

The result is an orthonormal sequence $\{e_n\}$ such that $\|Te_n\|\to \infty$, contradicting the assumption.