User romanov - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T11:34:36Z http://mathoverflow.net/feeds/user/15777 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/71687/metrizable-dual-space Metrizable dual space Romanov 2011-07-30T21:42:06Z 2011-07-30T22:08:48Z <p>I've got the following questions concerning the theory of locally convex spaces :</p> <p>Let $X$ be a locally convex metrizable space, what is the necessary and sufficient condition to have its dual $X^*$ metrizable? </p> <p>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?</p> <p>Thank you in advance for the answer.</p> http://mathoverflow.net/questions/67999/positive-operators-norm-equality Positive operators - norm equality Romanov 2011-06-16T20:40:06Z 2011-06-17T17:10:39Z <p>I hope that somebody can help me with the following problem:</p> <p>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 $f(0) \geq 0$?</p> <p>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</p> <p>$$\| (A^2+I)E(\mathbf{B}) \| \leq \|AE(\mathbf{B})\|^2 +1$$ and the equality does not hold for every $A$.</p> http://mathoverflow.net/questions/67999/positive-operators-norm-equality/68019#68019 Answer by Romanov for Positive operators - norm equality Romanov 2011-06-17T00:42:42Z 2011-06-17T00:42:42Z <p>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?</p> http://mathoverflow.net/questions/67898/trace-measurable-operators trace measurable operators Romanov 2011-06-15T21:43:04Z 2011-06-16T20:35:56Z <p>Hello,</p> <p>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.</p> <p>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 $$\forall_{\varepsilon >0} \ \exists_{E - \text{a projection in} \text{M}} \ \mbox{Range}(E) \subset D(T) \ \&amp; \ \tau(1-E) \leq \varepsilon.)$$</p> <p>Let $E_{(s,\infty)}(|T|)$ be a spectral projection of $|T|$ corresponding to the interval $(s, \infty)$, $s \geq 0$.</p> <p>How do we know that $\| |T|E_{(s,\infty)}(|T|) \| > s$ or $\| |T|E_{[0,s]}(|T|) \| \leq s$.</p> <p>Thank you in advance for any help.</p> http://mathoverflow.net/questions/67791/partial-order-unbounded-normal-operators-affiliated-with-von-neumann-algebra Partial order - Unbounded normal operators affiliated with von Neumann algebra. Romanov 2011-06-14T18:34:10Z 2011-06-16T16:52:06Z <p>Hello, I have a question which is related to a partial order in a set of self-adjoint operators.</p> <p>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.</p> <p>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. </p> <p>I will be really grateful for any help.</p> <p>Thank you, VdM</p> http://mathoverflow.net/questions/71687/metrizable-dual-space/71688#71688 Comment by Romanov Romanov 2011-07-30T22:43:14Z 2011-07-30T22:43:14Z Thank you very much. http://mathoverflow.net/questions/69671/tracexytraceyx-in-full-generality Comment by Romanov Romanov 2011-07-08T16:32:24Z 2011-07-08T16:32:24Z This property is called &quot;a tracial property&quot; 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 &quot;special&quot; trace i.e. finite, semifinite)? http://mathoverflow.net/questions/69419/spectra-of-a-symmetric-toeplitz-operator Comment by Romanov Romanov 2011-07-04T17:55:04Z 2011-07-04T17:55:04Z Apart from the matrix form of the Toeplitz operators it is very convenient to consider their &quot;so called&quot; 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 &quot;Basic Classes of Linear Operators&quot; 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. http://mathoverflow.net/questions/67999/positive-operators-norm-equality/68019#68019 Comment by Romanov Romanov 2011-06-17T19:44:45Z 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. http://mathoverflow.net/questions/67999/positive-operators-norm-equality/68072#68072 Comment by Romanov Romanov 2011-06-17T17:13:44Z 2011-06-17T17:13:44Z Nice alternative, thank you for that! http://mathoverflow.net/questions/67999/positive-operators-norm-equality Comment by Romanov Romanov 2011-06-16T23:15:28Z 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. http://mathoverflow.net/questions/67898/trace-measurable-operators/67998#67998 Comment by Romanov Romanov 2011-06-16T21:33:04Z 2011-06-16T21:33:04Z Thank you for that, this is a really nice application of the spectral theorem. http://mathoverflow.net/questions/67791/partial-order-unbounded-normal-operators-affiliated-with-von-neumann-algebra/67974#67974 Comment by Romanov Romanov 2011-06-16T17:00:09Z 2011-06-16T17:00:09Z I definitely agree. It was the key point! Thank you very much! http://mathoverflow.net/questions/67898/trace-measurable-operators/67967#67967 Comment by Romanov Romanov 2011-06-16T16:34:10Z 2011-06-16T16:34:10Z Thank you very much it was really easy. http://mathoverflow.net/questions/67791/partial-order-unbounded-normal-operators-affiliated-with-von-neumann-algebra Comment by Romanov Romanov 2011-06-16T15:03:54Z 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. http://mathoverflow.net/questions/67791/partial-order-unbounded-normal-operators-affiliated-with-von-neumann-algebra Comment by Romanov Romanov 2011-06-16T15:03:33Z 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 $E_{(s,\infty)}(|T|) \preceq E_{(s,\infty)}(|S|)$. http://mathoverflow.net/questions/67898/trace-measurable-operators Comment by Romanov Romanov 2011-06-16T14:23:06Z 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 &quot;measurable operators&quot;. I can't see this from the definition of the spectral projection. http://mathoverflow.net/questions/67791/partial-order-unbounded-normal-operators-affiliated-with-von-neumann-algebra Comment by Romanov Romanov 2011-06-14T18:52:57Z 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.