Hilbert space operator that takes all orthonormal sequences to strong null-sequences 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.

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. 
