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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.


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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.

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Thanks for this. I was thinking about whether one could still find a counterexample without the assumption of completeness (so that the orthonormal basis might also be a Hamel basis). Your example would work in any infinite-dimensional inner product space, so the answer is yes. – Nate Eldredge Dec 1 '11 at 17:50
I agree that this is a good example. Showing that $T$ exists still requires an argument like Nate's, since it cannot be explicitly defined off of the linear span of $\{e_n\}$. – Jonas Meyer Dec 1 '11 at 21:37
Just extend the ${\bf e}_i$ to a Hamel basis for $l^2({\mathbb Z})$ and then let $T(v) = 0$ on remaining basis vectors $v$. No need for Baire. – Zarrax Dec 1 '11 at 23:07
@Zarrax: That is what I meant, and I'm sorry I was very unclear. Baire is just one way to see that $\{e_i\}$ is not already a Hamel basis. – Jonas Meyer Dec 2 '11 at 0:12

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.

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[This comment is directed to a general audience.] Another way to see that $\{\varphi_i\}$ doesn't span $H$ is to note that $\sum\limits_i \frac{1}{i}\varphi_i$ is in $H$ and not in the span of $\{\varphi_i\}$. This came up in the question An orthonormal set cannot be a basis in an infinite dimension vector space?, and in particular in Andrey Rekalo's answer. – Jonas Meyer Dec 1 '11 at 16:14
Could you explain the reason it is not bounded? I think actually for any $f= \sum_{j \in J}c_j \psi_j$, $T(f) = |c_j| ||x|| \leq ||x|| ||f|| $ – user124697 Feb 25 '14 at 10:27
@user124697: There's a typo in your comment, but you can indeed say that $\|Tf\| \le \|x\| \sum_j |c_j|$. The sum $\sum_j |c_j|$ has only finitely many nonzero terms, so it's finite, but you can't conclude that it's bounded by $\|f\|$. I'm editing to add a few more details to my argument that $T$ is not in fact bounded. – Nate Eldredge Feb 25 '14 at 15:27
Oh, I see. Thank you! – user124697 Feb 25 '14 at 18:40

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