User romanov - MathOverflow

Metrizable dual space

2011-07-30T21:42:06Z

I've got the following questions concerning the theory of locally convex spaces :

Let $X$ be a locally convex metrizable space, what is the necessary and sufficient condition to have its dual $X^*$ metrizable?

Is it possible that $X^*$ is the F-space when $X$ is a locally convex non-complete metrizable space which is not a normed space?

Thank you in advance for the answer.

Positive operators - norm equality

2011-06-16T20:40:06Z

I hope that somebody can help me with the following problem:

Let $A$ be a positive operator on $\mathbf{B}(\mathcal{H})$, ( $\mathcal{H}$ is a Hilbert space) with its spectral measure $E$. Show that for every Borel set $\mathbf{B}$ from the domain of $E(\cdot)$ the following equality holds $$f(\| AE(\mathbf{B})\|) = \| f(A)E(\mathbf{B})\|, $$ where $f$ is an arbitrary continuous increasing function such that $f(0)=0$. Is it also true when $f(0) \geq 0$?

I have no idea how to solve the main part. The answer for the second part is probably negative, because if I take e.g. $f(x)=x^2+1$, then

$$\| (A^2+I)E(\mathbf{B}) \| \leq \|AE(\mathbf{B})\|^2 +1$$ and the equality does not hold for every $A$.

Answer by Romanov for Positive operators - norm equality

2011-06-17T00:42:42Z

Following above steps for an arbitrary $\mathbf{B}$ we get that $$ f \left(\| M_{\phi}E(\mathbf{B})\| \right)= f \left( \sup_{x \in \mathbf{B}} \ \phi(x) \right) = \sup_{x \in \mathbf{B}} \ f(\phi(x)) = \| f(M_{\phi})E(\mathbf{B})\|.$$ Since unitary operators preserve the norm the above equality is true for an arbitrary positive operator $A$. Moreover, now it is clear that with $f(0) \geq 0$ the property holds as well. We can even generalize it for a positive $\tau$-measurable operator, where $\tau$ is a faithful normal semi-finite trace on some semi-finite von Neumann algebra. Do you agree with my answer?

trace measurable operators

2011-06-15T21:43:04Z

Hello,

I have a question about trace measurable operators and I think it's not a hard one. However, I'm quite confused because I cannot prove it.

Let $\mathcal{M}$ be a semi-finite von Neumann algebra with a faithful normal semi-finite trace $\tau$. Let $T$ be a $\tau$- measurable operator (densely defined closed (possibly unbounded) operator affiliated with $\mathcal{M}$ such that $$ \forall_{\varepsilon >0} \ \exists_{E - \text{a projection in} \text{M}} \ \mbox{Range}(E) \subset D(T) \ \& \ \tau(1-E) \leq \varepsilon.)$$

Let $E_{(s,\infty)}(|T|)$ be a spectral projection of $|T|$ corresponding to the interval $(s, \infty)$, $s \geq 0$.

How do we know that $\| |T|E_{(s,\infty)}(|T|) \| > s$ or $\| |T|E_{[0,s]}(|T|) \| \leq s$.

Thank you in advance for any help.

Partial order - Unbounded normal operators affiliated with von Neumann algebra.

2011-06-14T18:34:10Z

Hello, I have a question which is related to a partial order in a set of self-adjoint operators.

Let $\mathcal{M}$ be a semifinite von Neumann algebra with a faithful semi-finite normal trace $\tau$. Let $T$ and $S$ be two self-adjoint operators (possibly unbounded) $\tau$-measurable (here probably the assumption that they are affiliated with $\mathcal{M}$ is enough) such that $0 \leq T \leq S$ i.e. $S-T$ is positive. How to get that $$E_{(s, \infty)}(|T|) \preceq E_{(s, \infty)}(|S|), \ \ s \geq 0,$$ where $E_I(|T|)$ (resp. $E_I(|S|)$) stands for a spectral projection of $T$ (resp. $S$) corresponding to the interval $I$ and $\preceq$ means sub-equivalence relation in Murray-von Neumann sense.

I am looking also for some good references which describe the relation between $U|T|$ the elements of the polar decomposition of closed densely defined (possibly unbounded) operator $T$ affiliated with some von Neumann algebra $\mathcal{M}$. I mean that $U$ and each spectral projection of $|T|$ are in this von Neumann algebra. Probably, I can find this in Takesaki vol 2 or vol 3.

I will be really grateful for any help.

Thank you, VdM

Comment by Romanov

2011-07-30T22:43:14Z

Thank you very much.

Comment by Romanov

2011-07-08T16:32:24Z

This property is called "a tracial property" so I think if a map does not satisfy this property we cannot call it a trace. I am wondering is it possible to classify this category by the property of its trace (analogy to the type of Von Neumann algebras with a "special" trace i.e. finite, semifinite)?

Comment by Romanov

2011-07-04T17:55:04Z

Apart from the matrix form of the Toeplitz operators it is very convenient to consider their "so called" defining functions (or the symbols). With this function we have a nice form of the spectrum of chosen Toeplitz operator. For the references I can recommend "Basic Classes of Linear Operators" by Gohberg, Goldberg and Kaashoek. Look also at Laurent Operators which are in some sense generalised Toeplitz operators, but they have some nicer properties i.e. they form a commutative Banach algebra.

Comment by Romanov

2011-06-17T19:44:45Z

Ok, but if I modify it a little bit i.e. I take $\tau$-measurable positive operator $A$ and von Neumann algebra generated by the spectral projection of $A$. By the bi-commutant theorem the spectral projections of $f(A)$ will be there as well. Thus for any projection from this von Neumann algebra we will get this equality I mean $$ f( \| AE\|) = \| f(A)E \|$$. Here $A$ can be even unbounded, but \|AE\| makes sense since $A$ is $\tau$-measurable.

Comment by Romanov

2011-06-17T17:13:44Z

Nice alternative, thank you for that!

Comment by Romanov

2011-06-16T23:15:28Z

I am not pretty sure what do you mean. Did you mean to use the spectral theorem and take $M_{\phi}$ a multiplication operator on some $L^2(\mu)$ unitarly equivalent to our operator $A$. For the spectral measure of $M_{\phi}$ ( which is $\textbf{1}_{\phi^{-1}(\cdot)}$) choose $\mathbf{B}$ to be the whole space to get the identity i.e. $E(\mathbf{X})=I$. Now $f(\| M_{\phi} \|) = f(\sup_x \phi(x))$ and $\| f(\phi) \| = \sup_x f(\phi(x))$ and since $f$ is increasing continuous and $f(0)=0$ the equality holds. I think that for any set $\mathbf{B}$ it will be satisfied in a similar way.

Comment by Romanov

2011-06-16T21:33:04Z

Thank you for that, this is a really nice application of the spectral theorem.

Comment by Romanov

2011-06-16T17:00:09Z

I definitely agree. It was the key point! Thank you very much!

Comment by Romanov

2011-06-16T16:34:10Z

Thank you very much it was really easy.

Comment by Romanov

2011-06-16T15:03:54Z

Another property of $s$-numbers is that $\mu_t(f(T))= f(\mu_t(T))$ for increasing continuous $f$ on $[0,\infty) with f(0) \geq 0$ $\tau(T) = \int_{0}^{\infty}\mu_t(T) dt$ for positive $\tau$ measurable $T$ we have $$\tau(f(S)) = \int_{0}^{\infty} f(\mu_t(S)) dt \leq \int_{0}^{\infty} f(\mu_t(T)) dt = \int_{0}^{\infty} \mu_t(f(T))= \tau(f(T)).$$ So this is not a good point.

Comment by Romanov

2011-06-16T15:03:33Z

The first part of my question is in particular a part of this problem in Takesaki. Because $\mu_t(T) \leq \mu_s(T)$ iff $\lambda_s(T)= \tau(E_{(s,\infty)}(|T|)) \leq \tau(E_{(s,\infty)}(|S|))=\lambda_s(S)$ iff $E_{(s,\infty)}(|T|) \preceq E_{(s,\infty)}(|S|)$.

Comment by Romanov

2011-06-16T14:23:06Z

Ok, I didn't know that because this is a part of the problem IX.2.7 in Takesaki vol. 2 and this problem concerns "measurable operators". I can't see this from the definition of the spectral projection.

Comment by Romanov

2011-06-14T18:52:57Z

Sorry, my mistake I mean the relation between $U|T|$ and von Neumann algebra $\mathcal{M}$ i.e. that $U$ and each spectral projection of $|T|$ are in this von Neumann algebra. I know it suffices to show that $U$ and $\textbf{1}(|T|)$ are in $\mathcal{M}$, because by virtue of Double Commutant Theorem the spectral projection of $f(|T|)$ are there since $\textbf{1}(|T|)$ is. I am looking for some good references for the theory of the operators affiliated with some von Neumann algebra.