T is not compact operator

Question

I want to show that if $T$ is a bounded operator between two Hilbert spaces and $T$ is not compact then there exists an orthonormal sequence $y_{n}$ and an $R>0$ such that $\forall n\in \mathbb{N}\,\,\,\|T(y_{n})\|\geq R$. I understand that there exists a sequence $y_{n}\in B[0,1]$ satisfying the above condition (because there exists a sequence doesn't contain any convergent subsequence), but i can't find an orthonormal one

Answer 1

Assume that the property does not hold. This means that for each orthonormal sequence $(e_n)$, there exists a subsequence $(e_{n_k})$ such that $\lVert Te_{n_k}\rVert\to 0$.

Claim. If $(v_n)$ is an orthonormal subsequence, then $\lVert Tv_n\rVert\to 0$.

Indeed, suppose it is not true. There for some sequence $n_k\uparrow\infty$ and some positive $\delta$, we have $\lVert T_{v_{n_k}}\rVert>\delta$. The sequence $(v_{n_k})$ is orthonormal, hence there is a further subsequence $(v_{n'_k})$ such that $\lVert v_{n'_k}\rVert\to 0$, a contradiction.

But there still is a gap: we have to show that $Tv_n\to 0$ if $v_n\to 0$ weakly, and for the moment we only showed it for a small class of weakly convergent sequences (the orthonormal ones). This gap is filled here by Haskell Curry's answer.