# Bounded linear operator on a Hilbert space

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.

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$T$ is clearly not invertible in general, for example if $c_n=0$ for some $n$. – Chris Eagle May 29 '12 at 17:08
Did you mean $T(x) = \sum c_n x_n$, or did you mean $T(x) = \sum c_n x_n e_n$? – copper.hat May 29 '12 at 17:08
@copper.hat: What is $e_n$? Note that $\{ x_n\}$ is an orthornormal basis. – Chris Eagle May 29 '12 at 17:09
@ChrisEagle: Oh, you're right. I think I misinterpreted the question. Hold on, I will fix it. – rt93 May 29 '12 at 17:10
@ChrisEagle: Thanks, I missed that point. – 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.

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2. was based on a previous version of the question, which was edited out while I was writing, but it will hopefully still help. – Jonas Meyer May 29 '12 at 17:18
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? – rt93 May 29 '12 at 17:30
@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. – Jonas Meyer May 29 '12 at 17:36
@jb88: I have updated 3. – Jonas Meyer May 29 '12 at 17:42
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. – rt93 May 29 '12 at 17:43