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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.

  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 $n$.2012-05-29
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    Did you mean $T(x) = \sum c_n x_n$, or did you mean $T(x) = \sum c_n x_n e_n$?2012-05-29
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    @copper.hat: What is $e_n$? Note that $\{ x_n\}$ is an orthornormal basis.2012-05-29
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    @ChrisEagle: Oh, you're right. I think I misinterpreted the question. Hold on, I will fix it.2012-05-29
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    @ChrisEagle: Thanks, I missed that point.2012-05-29
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    @jb88: Consider the finite dimensional case first.2012-05-29
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    @copper.hat For all three parts?2012-05-29
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    @jb88: I meant for Part 1. Originally I meant Part 2 as well, but I must have misread the question.2012-05-29
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    Look at the Open Mapping Theorem for Part 2.2012-05-29
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    copper.hat is correct that you could use the Open Mapping Theorem, but it is not necessary to do so here because you can explicitly write down the conditions for the inverse to exist, and what it is, and check directly that it is bounded.2012-05-29
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    @JonasMeyer: I'm having a bad 'question-reading' day. I interpreted the question as for any $T$.2012-05-29
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    @copper.hat: It might be me, especially given that the third question suddenly seems to jump to a general statement. Perhaps I was hasty to assume that "$T$" is the same $T$ in #2. Oh well.2012-05-29
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    I think for Part 3 you need to conclude that $V$ is finite dimensional?2012-05-29
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    Yes, T in part 2 is the same T, whereas part 3 is a general question.2012-05-29

1 Answers 1

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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.

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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.2012-05-29
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    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?2012-05-29
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    @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.2012-05-29
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    @jb88: I have updated 3.2012-05-29
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    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.2012-05-29
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    @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 $\begin{bmatrix}c_1&0\\0&c_2\end{bmatrix}$?2012-05-29
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    @jb88: Look at the finite dimensional case to see what the inverse is.2012-05-29
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    Ah, so $T^{-1}(x_n) = \frac{1}{c_n}x_n$, which is bounded since $\{\frac{1}{c_n}\}$ is.2012-05-29
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    @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.2012-05-29
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    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.2012-05-29
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    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.2012-05-29
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    Ok, well I'll see if anyone else can help me with part 3, but I will eventually accept your answer either way.2012-05-29
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    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.2012-05-29
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    @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.)2012-05-29
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    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.2012-05-29
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    @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.2012-05-29
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    @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 $|3a_1|^2+|4a_2|^2\leq |4a_1|^2+|4a_2|^2$.2012-05-29
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    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|\}$2012-05-29
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    @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.2012-05-29
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    Alright, thanks a lot.2012-05-29
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    Use the finite, Luke.2012-05-30