TITLE: Boundedness of operator on Hilbert space
QUESTION [6 upvotes]: I have the following question: let $\mathcal{H}$ be a Hilbert space and $\{\varphi_{i}\}_{i \in \mathbb{N}}$ be an orthonormal basis. Furthermore let $T: \mathcal{H} \rightarrow \mathcal{H}$ be an operator. If there exists a constant $K > 0$ such that $\|T \varphi_{i} \| \leq K$, $\forall i$, is then $T$ bounded? If yes, what is the argument of showing this? Thanks in advance.
Haro

REPLY [9 votes]: In $l^2(\mathbb Z)$, Let $T({\bf e}_n) = {1 \over \sqrt{n}} \sum_{i=1}^n {\bf e}_i$, where 
${\bf e}_i$ denotes the $i$th unit coordinate vector. Then $||T({\bf e}_n)|| = 1$ for each $n$. Let $v_n = \sum_{i=1}^n {\bf e}_i$. Then $||v_n|| = \sqrt{n}$, and the $j$th component of $T(v_n)$  is $\sum_{i=j}^n i^{-{1 \over 2}}$. 
For $j \leq {n \over 2}$, this 
is at least $\sum_{i={n \over 2}}^n i^{-{1 \over 2}} > C\sqrt{n}$. Since the first ${n \over 2}$ entries of $T(v_n)$ are at least $C\sqrt{n}$, we have $||T(v_n)|| \geq Cn$. Since $||v_n|| = \sqrt{n}$ this operator must be unbounded.