Boundedness of operator on Hilbert space 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
 A: 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.
A: No, not necessarily.
By Baire category, the linear span of $\{\varphi_i\}$ is not all of $H$.  So we can extend $\{\varphi_i\}$ to a Hamel basis for $H$ by adding some more vectors $\{\psi_j\}_{j \in J}$ (using Zorn's lemma).  Fixing any nonzero $x \in H$ and any $j_0 \in J$, I can define an operator $T$ by $T \varphi_i = 0$, $T \psi_{j_0} = x$, $T \psi_j = 0$ for $j \ne j_0$.  Then $T$ certainly has the condition you request, but I claim $T$ is not bounded, i.e. not continuous.  It's a standard fact from topology that two continuous maps that agree on a subset of a space must agree everywhere.  $T$ agrees with the zero operator on the dense subpace spanned by $\{\varphi_i\}$ but is not identically zero, so it cannot be continuous.
