I am having a bit of difficulty with the following homework problem.

Let $\{x_n\}$ be an orthonormal basis in a Hilbert space $V$ over $\mathbb{C}$ and let $\{c_n\}_{n \in \mathbb{N}}$ be a fixed bounded sequence of complex numbers. Consider the bounded linear operator $T: V \to V$ defined by $T(x_n) = c_nx_n$.

There are numerous parts to the question, but below are the ones I am having trouble with

  1. Find the adjoint operator $T^*$ and its norm $||T^*||$
  2. If T is invertible, is its inverse continuous?
  3. Show that any linear operator on a normed space is continuous if the unit sphere is compact.
  1. I have managed to find $T^*$. As for the norm, I know that $||T^*|| = ||T||$. But is there an explicit value for $||T||$ that can be found? I can't think of a way to find $||T||$ explicitly since we don't know what the norm on $V$ is.

  2. I am not really sure how to do this one. Firstly, I know that a linear operator is continuous iff it is bounded, so I need to show that a linear operator $T: V \to V$ is bounded if the unit sphere $\{x \in V : ||x|| = 1\}$ is compact. I have been told to assume that $T$ is unbounded and try to get a contradiction. If T is unbounded then $||T|| = \sup_{||x|| = 1}\{||Tx||\} = \infty$. I don't know what to do from here.

  • $\begingroup$ $T$ is clearly not invertible in general, for example if $c_n=0$ for some $n$. $\endgroup$ May 29 '12 at 17:08
  • $\begingroup$ Did you mean $T(x) = \sum c_n x_n$, or did you mean $T(x) = \sum c_n x_n e_n$? $\endgroup$
    – copper.hat
    May 29 '12 at 17:08
  • $\begingroup$ @copper.hat: What is $e_n$? Note that $\{ x_n\}$ is an orthornormal basis. $\endgroup$ May 29 '12 at 17:09
  • $\begingroup$ @ChrisEagle: Oh, you're right. I think I misinterpreted the question. Hold on, I will fix it. $\endgroup$
    – rt93
    May 29 '12 at 17:10
  • $\begingroup$ @ChrisEagle: Thanks, I missed that point. $\endgroup$
    – copper.hat
    May 29 '12 at 17:11
  1. We do know the norm on $V$, because we know that $\{x_n\}$ is an orthonormal basis. That means that each $v\in V$ can be written as $v=\sum_n a_nx_n$ with $a_n=\langle v,x_n\rangle$ and $\|v\|^2=\sum_n\|a_n\|^2$. Using this fact, you should be able to find the norm of $\|T\|$ in terms of the sequence $\{c_n\}$.

  2. This is typically false. If $c_n=0$ for some $n$, the map is not injective. If $0$ is in the closure of $\{c_n\}$, then the map is not surjective. The sum you mention would converge if the sequence $\left\{\frac{1}{c_n}\right\}$ is bounded, so that would be a good condition to focus on. You may also find it useful to note that a bijective bounded linear operator on a Hilbert space automatically has a bounded inverse.

  3. You could combine the facts that “Every linear mapping on a finite dimensional space is continuous” and the Characterization of normed vector spaces of finite dimension in terms of compactness of the unit sphere.

  • $\begingroup$ 2. was based on a previous version of the question, which was edited out while I was writing, but it will hopefully still help. $\endgroup$ May 29 '12 at 17:18
  • $\begingroup$ Thanks Jonas. So regarding the updated version of part 2, if T is invertible, then the inverse is bounded since T is bounded. Could you elaborate on your hint for part 3 at all? Am I supposed to prove by contradiction that T is continuous? $\endgroup$
    – rt93
    May 29 '12 at 17:30
  • $\begingroup$ @jb88: Regarding 2, yes, this is a general fact, which as copper.hat mentions follows from the open mapping theorem. However, if $T$ is still the same $T$ from above, then you can be very explicit about it as I mentioned in a comment. And as I mentioned above, boundedness of $\{1/c_n\}$ is key. Regarding 3, I've temporarily removed that because I had misread/misthought. $\endgroup$ May 29 '12 at 17:36
  • $\begingroup$ @jb88: I have updated 3. $\endgroup$ May 29 '12 at 17:42
  • $\begingroup$ I see. So if $c_n \neq 0$ for all $n$ and if that sequence is bounded, then $T$ will be bijective. You mentioned that you can explicitly write down the inverse and check that it is bounded -- I am not sure how to do this. $\endgroup$
    – rt93
    May 29 '12 at 17:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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