# Math Help - Bounded linear operators of a Hilber space

1. ## Bounded linear operators of a Hilber space

I have two questions.

Let $T \in \mathcal{B}(H)$, a bounded linear operator of a Hilbert space $H$, and let $T^{\ast}$ be the adjoint of $T$.

Firstly, is it true that $\|TT^{\ast}\|^n = \|(TT^{\ast})^n\|$?

Secondly, I want to show that for all $x \in H$, $\sum_iT_ix \rightarrow Tx$ for some $T \in \mathcal{B}(H)$. Is it enough to show that $lim_{n \rightarrow \infty}\|\sum_i^nT_ix\| \leq \|x\|$? As then surely this means that the limit exists...

2. Originally Posted by Swlabr
I have two questions.

Let $T \in \mathcal{B}(H)$, a bounded linear operator of a Hilbert space $H$, and let $T^{\ast}$ be the adjoint of $T$.

Firstly, is it true that $\|TT^{\ast}\|^n = \|(TT^{\ast})^n\|$?
Yes, that is true, but I think that you'll need something like the spectral theorem in order to prove it. In fact, $TT^*$ is a positive operator, so its norm is equal to its spectral radius $\rho(TT^*)$, and $\rho((T^*T)^n) = (\rho(T^*T))^n$.

Originally Posted by Swlabr
Secondly, I want to show that for all $x \in H$, $\sum_iT_ix \rightarrow Tx$ for some $T \in \mathcal{B}(H)$. Is it enough to show that $lim_{n \rightarrow \infty}\|\sum_i^nT_ix\| \leq 1$? As then surely this means that the limit exists...
That sounds unlikely to me, unless you know more about the sequence of operators $T_i$.

3. Originally Posted by Opalg
Yes, that is true, but I think that you'll need something like the spectral theorem in order to prove it. In fact, $TT^*$ is a positive operator, so its norm is equal to its spectral radius $\rho(TT^*)$, and $\rho((T^*T)^n) = (\rho(T^*T))^n$.
Does this hold then as

$\|TT^{\ast}\| = \rho(TT^{\ast}) = \displaystyle \lim_{n\rightarrow \infty}\|(TT^{\ast})^n\|^{\frac{1}{n}}$

so then we just take the limit? (It is the limit bit I'm confused by. Sorry for what could be a really silly question - analysis really isn't my forte!)

Originally Posted by Opalg
That sounds unlikely to me, unless you know more about the sequence of operators $T_i$.
How then would I show that a sum of operators converges to some mysterious operator $T$? What is the general technique?

4. Originally Posted by Swlabr
Firstly, is it true that $\|TT^{\ast}\|^n = \|(TT^{\ast})^n\|$?
The way that this relates to spectral radius is that $\rho(TT^*) = \sup\{\lambda:\lambda\in\sigma(TT^*)\}$ (sigma denotes the spectrum). But $\sigma((TT^*)^n) = (\sigma(TT^*))^n$ by the spectral mapping theorem. So $\|(TT^*)^n\| = \rho((TT^*)^n) = (\rho(TT^*))^n = \|TT^*\|^n$.

Originally Posted by Swlabr
Secondly, I want to show that for all $x \in H$, $\sum_iT_ix \rightarrow Tx$ for some $T \in \mathcal{B}(H)$.
If you know that $\textstyle\lim_{n \rightarrow \infty}\sum_{i=1}^n\|T_ix\| \leqslant \|x\|$ then the series $\textstyle\sum_iT_ix$ is absolutely convergent and hence convergent (because the closed ball of radius $\|x\|$ in H is complete) to some element that we can call $Tx$, with $\|Tx\|\leqslant\|x\|$. The operator $T$ defined in this way is easily seen to be linear, and is bounded. So $T\in B(H)$ and $\textstyle\sum T_i\to T$ in the strong operator topology, which is exactly what you want.

But if you only know that $\textstyle \lim_{n \rightarrow \infty}\bigl\|\sum_{i=1}^nT_ix\bigr\| \leqslant \|x\|$ then that is not enough to ensure convergence of $\textstyle\sum T_i x$. For example, let $\{e_i\}$ be an orthonormal basis for H, and define $T_i$ by $T_i(x) = \tfrac12\langle x,e_1\rangle(e_i-e_{i+1})$. Then $\textstyle\sum_{i=1}^nT_ix = \tfrac12\langle x,e_1\rangle(e_1-e_{n+1})$, which has norm $\leqslant\|x\|$. But $\textstyle\sum_{i=1}^nT_ie_1 = \tfrac12(e_1-e_{n+1})$, which does not converge.

5. Originally Posted by Opalg
The way that this relates to spectral radius is that $\rho(TT^*) = \sup\{\lambda:\lambda\in\sigma(TT^*)\}$ (sigma denotes the spectrum). But $\sigma((TT^*)^n) = (\sigma(TT^*))^n$ by the spectral mapping theorem. So $\|(TT^*)^n\| = \rho((TT^*)^n) = (\rho(TT^*))^n = \|TT^*\|^n$.
That makes sense, thanks!

Originally Posted by Opalg
If you know that $\textstyle\lim_{n \rightarrow \infty}\sum_{i=1}^n\|T_ix\| \leqslant \|x\|$ then the series $\textstyle\sum_iT_ix$ is absolutely convergent and hence convergent (because the closed ball of radius $\|x\|$ in H is complete) to some element that we can call $Tx$, with $\|Tx\|\leqslant\|x\|$. The operator $T$ defined in this way is easily seen to be linear, and is bounded. So $T\in B(H)$ and $\textstyle\sum T_i\to T$ in the strong operator topology, which is exactly what you want.

But if you only know that $\textstyle \lim_{n \rightarrow \infty}\bigl\|\sum_{i=1}^nT_ix\bigr\| \leqslant \|x\|$ then that is not enough to ensure convergence of $\textstyle\sum T_i x$. For example, let $\{e_i\}$ be an orthonormal basis for H, and define $T_i$ by $T_i(x) = \tfrac12\langle x,e_1\rangle(e_i-e_{i+1})$. Then $\textstyle\sum_{i=1}^nT_ix = \tfrac12\langle x,e_1\rangle(e_1-e_{n+1})$, which has norm $\leqslant\|x\|$. But $\textstyle\sum_{i=1}^nT_ie_1 = \tfrac12(e_1-e_{n+1})$, which does not converge.
I can't seem to get my norm within the summation sign. However, I have that it is bounded (as $\lim_{n \rightarrow \infty}\|\sum_i^nT_ix\| \leq \|x\|$) and I can also show that it is monotone. Would this, then, be enough? Or would I also have to show that my sequence is Cauchy?

6. Originally Posted by Swlabr
I can't seem to get my norm within the summation sign. However, I have that it is bounded (as $\lim_{n \rightarrow \infty}\|\sum_i^nT_ix\| \leq \|x\|$) and I can also show that it is monotone. Would this, then, be enough? Or would I also have to show that my sequence is Cauchy?
I'm not sure if this is what you want, but there is a theorem that a bounded increasing sequence of selfadjoint operators has a least upper bound, which is the strong limit of the sequence. This result is given in Appendix II of Dixmier's book Les algèbres d'opérateurs dans l'espace Hilbertien. (Dixmier's book is available in an English translation under the title of Von Neumann algebras. The English translation has the added bonus of an extended preface written by me. )

The key thing here is that the sequence should be increasing (or decreasing and bounded below). For the series $\textstyle\sum_i T_i$ that is of course equivalent to each operator $T_i$ being positive. I'm not sure if that is what you mean by saying that your sum is monotone.

Strong convergence means exactly that $\textstyle\sum_i T_ix\to Tx$ in norm.

7. Originally Posted by Opalg
The English translation has the added bonus of an extended preface written by me. )
Haha, I can't wait to contribute a book so I can do this!

8. Originally Posted by Opalg
I'm not sure if this is what you want, but there is a theorem that a bounded increasing sequence of selfadjoint operators has a least upper bound, which is the strong limit of the sequence. This result is given in Appendix II of Dixmier's book Les algèbres d'opérateurs dans l'espace Hilbertien. (Dixmier's book is available in an English translation under the title of Von Neumann algebras. The English translation has the added bonus of an extended preface written by me. )

The key thing here is that the sequence should be increasing (or decreasing and bounded below). For the series $\textstyle\sum_i T_i$ that is of course equivalent to each operator $T_i$ being positive. I'm not sure if that is what you mean by saying that your sum is monotone.

Strong convergence means exactly that $\textstyle\sum_i T_ix\to Tx$ in norm.
Hmm...I think the question is tough, but perhaps not quite so tough at to warrant a trip to the library over Easter Weekend. Anyway, a quick search reveals that my uni lacks this book!

Okay, I have, after many dead ends, proven (in a very elementary way) that $\sqrt{\sum_i^n\|T_ix\|^2} \leq \|x\|$, so I believe this gives us absolute convergence, and so convergence? I am, however, suspicious, as it is too elementary...