Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
add comment

2 Answers

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.

share|improve this answer
    
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
add comment

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.

share|improve this answer
2  
[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 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 at 15:27
    
Oh, I see. Thank you! –  user124697 Feb 25 at 18:40
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.