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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.

  3. 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.

rt93
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1 Answers1

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  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.

Community
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Jonas Meyer
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  • 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
  • @jb88: If you think of $T$ as a really big diagonal matrix, you might find it easier to see the inverse. What is the inverse of $\begin{bmatrix}c_1&0\0&c_2\end{bmatrix}$? – Jonas Meyer May 29 '12 at 17:45
  • @jb88: Look at the finite dimensional case to see what the inverse is. – copper.hat May 29 '12 at 17:45
  • Ah, so $T^{-1}(x_n) = \frac{1}{c_n}x_n$, which is bounded since ${\frac{1}{c_n}}$ is. – rt93 May 29 '12 at 17:52
  • @jb88: Yes. The problem is, for such a solution to be complete, you would have to argue that $T$ is not invertible if ${1/c_n}$ is not bounded, which is equivalent to $0$ being in the closure of ${c_n}$. You can show that $T$ is not surjective in such a case. This can be done directly, but admittedly, it might be easier to cite a big theorem like the open mapping theorem. – Jonas Meyer May 29 '12 at 17:55
  • Would there be another way to do part 3? I'm not sure if I'm allowed to use the results from that second question you listed. My teacher mentioned something about a proof by contradiction by considering a sequence ${x_n}$ in the unit sphere such that $||T(x_n)|| \rightarrow \infty$ and then considering a convergent subsequence of that. – rt93 May 29 '12 at 17:56
  • I don't know what your teacher had in mind, and don't have further comments at the moment, but I hope someone else has some good ideas. – Jonas Meyer May 29 '12 at 17:58
  • Ok, well I'll see if anyone else can help me with part 3, but I will eventually accept your answer either way. – rt93 May 29 '12 at 18:07
  • Sorry for all the questions, but I still can't get part 1 (even though it should be obvious). The problem is I don't know which definition of the operator norm to use, and I don't know where to apply Parseval's identity. – rt93 May 29 '12 at 18:22
  • @jb88: What is the norm of the matrix $\begin{bmatrix}c_1&0\0&c_2\end{bmatrix}$ when considered as an operator on the Hilbert space $\mathbb C^2$ with orthonormal basis ${(1,0),(0,1)}$? Considering what $T$ does to $x_n$ shows that $|T|\geq |c_n|$ for each $n$. How about an upper bound? (I can give another hint if that isn't clear.) – Jonas Meyer May 29 '12 at 18:29
  • As for which definition to use, $|T|=\inf{K>0:|Tv|\leq K|v|\forall v\in V}$ or $|T|=\sup{|Tv|:|v|\leq 1}$ or $|T|=\sup\left{\frac{|Tv|}{|v|}:v\neq 0\right}$, or any other definition that is equivalent should work. – Jonas Meyer May 29 '12 at 18:38
  • @JonasMeyer Thanks for the extra hint. I'm not sure what the upper bound for $||T||$ would be. It must be some combination of the $c$'s I guess, but I can't see what. – rt93 May 29 '12 at 18:45
  • @jb88: Did you have an answer or guess for the 2-by-2 case? What is the norm of $\begin{bmatrix}3&0\0&4\end{bmatrix}$? Of $\begin{bmatrix}-9&0\0&7i\end{bmatrix}$? Hint: Note that $|3a_1|^2+|4a_2|^2\leq |4a_1|^2+|4a_2|^2$. – Jonas Meyer May 29 '12 at 18:48
  • Ok, I think I see now. The upper bound for the norm is just the "largest" absolute value of the numbers in the matrix, so it would be 4 for the first matrix and 9 for the second one. So $||T||$ is the maximum of ${|c_1|, |c_2|}$ – rt93 May 29 '12 at 18:55
  • @jb88: Exactly. So when you have a bounded sequence and its corresponding "diagonal" operator, the same idea applies, except that you have to replace maximum with supremum. – Jonas Meyer May 29 '12 at 18:57
  • Alright, thanks a lot. – rt93 May 29 '12 at 19:11
  • Use the finite, Luke. – copper.hat May 30 '12 at 06:58