New answers tagged

4

The deeper way to do this is to use Riesz Lemma to show that the unit ball is compact if and only if $\dim E<\infty$, and conclude that the identity operator is not compact, and thus not a limit of finite-rank operators. But there is a much straightforward way. If $E$ is infinite-dimensional, no finite-rank operator can be invertible. This implies that $\|...


0

I expanded my original answer to the question you linked, hopefully the updated version should clear up the confusion regarding point 1. As for point 2, the space was first investigated due to its role as a trace of $H^1$-functions, where the quotient construction is a natural abstract characterization of the space. The other characterization was not ...


1

Here is a heat kernel approach: Consider the heat semigroup $(e^{t\Delta})$. It is well known (and can be seen quite easily using the Fourier transform) that $$ e^{t\Delta}f(x)=(4\pi t)^{-n/2}\int_{\mathbb{R}^n}e^{-\frac{|x-y|^2}{4t}}f(y)\,dy. $$ Direct computation shows that $\|e^{t\Delta}f\|_p\leq \|f\|_p$. In fact, it suffices to show this for $p=2$ and $...


1

You define $\psi'(x)=\varphi((1-s(\psi))x(1-s(\psi)))$. Taking $\psi=\omega$, you have $$ \omega(x)=\langle \pi_\omega(x)\,\Omega,\Omega\rangle,\ \text{ for some } \Omega\in\mathfrak H_\omega. $$ Let $\xi=u\Omega$. Then $$ \omega(x)=\langle u^*\pi_\omega(x)u\Omega,\Omega\rangle =\langle \pi_\varphi(x)u\Omega,u\Omega\rangle =\langle \pi_\varphi(x)\xi,\xi\...


0

Let $$ H_k=\overline{\operatorname{span}}\{e_{k,j}:\ j\in\mathbb Z\}. $$ Then $H=\bigoplus_kH_k$. More importantly, each $H_k$ is invariant for $T$, so $T=\bigoplus_k T_k$, where $T_k=T|_{H_k}$. Then $\sigma(T)=\overline{\bigcup_k\sigma(T_k)}$. On $H_k$, we have $$ T\sum_j x_je_{k,j}=\sum_j \mu^kx_je_{k,j+1}. $$ That is, $T_k=\mu^k\,B$, where $B$ is the ...


2

Considering the two topological spaces $(U,\|{\cdot}\|)$ and $(U,w)$, we have that the former is compact and the latter is Hausdorff. Since we know that the identity map $$ (U,\|{\cdot}\|) \to (U,w) $$ is continuous, it must therefore be a homeomorphism by a well know Theorem of General Topology. Thus the weak- and norm-convergent sequences are the same.


1

Indeed it is true. However, I would argue differently. Pick some subsequence $(x_{n_k})_{k}$. By compactness it admits a convergent subsequence. However, as the strong and the weak limit coincide, it comverges strongly to $x$. This implies that every subsequence of $(x_n)_n$ admits a subsequence converging strongly to $x$ and thus the whole sequence ...


0

In the context of QIT, the inner-product over $\mathcal L(\mathcal H)$ is defined by $$ \langle \rho_1,\rho_2 \rangle_{\mathcal L(\mathcal H)} = \operatorname{Tr}(\rho_1 \rho_2^*). $$ With that in mind, we want to show that for any $\rho \in \mathcal L(\mathcal H)$ and $\sigma \in \mathcal {L(K)},$ your formulas for $\Phi$ and $\Phi^*$ are such that $$ \...


1

If $M_*$ is separable, just take $\psi=\sum_n2^{-n}\phi_n$, where the $\phi_n$ form a dense subset of the normal state space. Otherwise such a state might not exist.


1

If $M$ is finite but it is not a factor, you can construct two different traces. As the modular group of a trace is trivial, you get two distinct normal states with the same modular group. If I remember correctly, the condition $\sigma^\varphi_t=\sigma^\psi_t$ is equivalent to the existence of invertible positive $h\eta M_\psi\cap M_\varphi$ with $\psi=h\,\...


1

@user854214 has already settled this question in the comments with the reference to Pal's paper but it might perhaps be useful to see an example illustrating why must the definition of a regular operator in the context of Hilbert-modules be so much more intricate than the definition for Hilbert spaces. Recall that if $A$ is a C*-algebra and $E$ and $F$ ...


1

The identity mapping from $D$ to $E_1=D$ has a closed graph but $D$ might not be closed.


1

Hint: $T$ is abounded operator iff $(a_n) \in \ell_{s}$ where $s=\frac {pq} {p-q}$ and the operator norm is $\|T\|=(\sum |a_n|^{s})^{1/s}$. This can be proved using Holder's inequality and condition for equality in Holder's inequality: choose $x_n$ to be of the form $\pm |a_n|^{\alpha}$ for a suitable $\alpha$ to see that the norm is exactly $\|(a_n)\|_s$. ...


1

Yes. If $\|Az\|=\|Bz\|$ for all $z\in D(A)$ (where $X$ is a space with scalar product $(.,.)$ - assumed from the question) then we can put $z=x-y$ and we get $\|Az\|^2=\|Ax\|^2-2(Ax,Ay)+\|Ay\|^2=\|Bx\|^2-2(Bx,By)+\|By\|^2=\|Bz\|^2$. Using our assumption again (i.e. $\|Ax\|=\|Bx\|$ and $\|Ay\|=\|By\|$), we arrive to $(Ax,Ay)=(Bx,By)$ for every $x,y\in D(A)$.


2

Note that the finite rank maps are dense in the space of compact operators on a Hilbert space. Furhter note that the finite rank maps are spanned by the rank $1$ maps. These can all be written as $v\otimes w^*$, which denotes the map $x\mapsto (x,w)\cdot v$. Now suppose $(1)$ holds, then for any $v\in H$ there is a sequence $a_n(h_n)\to v$ with $a_n\in A$, $...


0

As $\xi_1\otimes\xi_2$ is cyclic for $\mathcal{M}_1'\overline\otimes\mathcal{M}_2'$ and we have $\mathcal{M}_1'\overline\otimes\mathcal{M}_2'\subseteq (\mathcal{M}_1\overline\otimes\mathcal{M}_2)'$, therefore $\xi_1\otimes\xi_2$ is cyclic for $(\mathcal{M}_1\overline\otimes\mathcal{M}_2)'$. Hence $\xi_1\otimes\xi_2$ is separating for $\mathcal{M}_1\overline\...


1

$\newcommand{\dim}{\text{dim}}\newcommand{\codim}{\text{codim}}\newcommand{\Ind}{\text{Ind}}\newcommand{\Ker}{\text{Ker}}\newcommand{\Ran}{\text{Ran}}$Notice that $\Ker(A\oplus B) = \Ker(A)\oplus \Ker(B)$, hence a finite dimensional space, and $\Ran(A\oplus B) = \Ran(A)\oplus \Ran(B)$, hence a finite co-dimensional space. Therefore $A\oplus B$ is ...


1

I get it now. The point is, if $y\in M_{n}(B)$ contains nonzero constants among its elements, the ideal generated by $y$ is the whole $M_{n}(B)$. Suppose $y\in M_{I(n)}(B)$ is nonzero, by the assumption, there is a sufficiently large $m$ such that $\pi_m\otimes 1_{I(n)}(y)\neq 0$, which means $\psi_m^n(y)$ has nonzero constant elements. Then $h_{m-1}\circ ......


1

I just want to point out another possible positive answer that people may be interested in. Claim: suppose $V_1, V_2 \in R^{n\times r}$ where $r \leq n$ and $V_1^{\star}V_1 =V_2^{\star}V_2 = I_{r}.$ If $\|V_1V_1^{\star} - V_2V_2^{\star}\| \leq \epsilon$, then there exists a rotation matrix $H \in R^{r\times r}$ where $H^{\star}H = I_{r}$ and $$ \|V_1 H - V_2\...


1

Based on the comments above, it's easy to see that $$ ||\phi||_{op} \le \int^1_0 |a(t)|dt $$ and for equality you can consider the sign function $f = sgn(a)$, then for $a(t) \neq 0$ $$ ||\phi||_{op} \ge |\phi(f)|= \left|\int^1_0 sign(a) \cdot a(t)dt\right| $$ but the product, $sign(a) \cdot a(t) = |a(t)|$. Finally $$ ||\phi||_{op} \ge \int^1_0 |a(t)|dt $$ ...


3

It might help to separate $T$ into the composition of two simpler maps. Note that the map $s:C[0,1] \to \mathbb{R}^n$ defined by $s(f) = (f(t_1),...,f(t_n))$ has norm one (using the $l_\infty$ norm on $\mathbb{R}^n$) and is surjective. Furthermore, for any $y$ with $\|y\|_\infty =1$ it is easy to construct (by interpolation & extrapolation) some $f \in C[...


4

WLOG, one can assume that $a_k\ne0$ for all $k$ (if not, just drop all the zero-values, and in the case that all are zero, the problem is trivial). Take $f(x)$ a continuous function with these properties: $f(t_k)=\frac{|a_k|}{a_k}$. $|f(x)|\le 1$ for all $x\in[0,1]$. For example, you can take the polygonal through the points $\{(t_k,f(t_k)),\,1\le k\le n\}\...


2

Yes. Let $\phi:A\hookrightarrow A\rtimes G$ denote the canonical embedding, and let $\tau_A=\tau\circ\phi$, so that $\tau_A$ is the restriction in question. As a composition of two positive linear maps, we have that $\tau_A$ is positive and linear. To show that $\tau_A$ is a state, let $(u_\lambda)$ be an approximate unit for $A$. Then $(\phi(u_\lambda))$...


1

I might be wrong, but I don't see how you could use that $\lambda$ is an approximate eigenvalue to conclude that it is an eigenvalue. What you can use, due to $T$ being normal, is the continuous funcional calculus. Take $f$ to be continuous on $\sigma(T)$, with $f(\lambda)=1$ and $f=0$ on $\sigma(T)\setminus\{\lambda\}$. Then $P=f(T)$ is a nonzero projection,...


2

It's a general fact about the Weyl algebra $k[x, D]$ (where $D = \partial_x$) that every element can be written this way. The proof is just by induction; you apply the canonical commutation relation $D x - x D = 1$ a lot to move $x$'s past $D$'s. As a simple example, $$(xD)^2 = x(Dx)D = x(xD + 1)D = x^2 D^2 + xD.$$ In general, it suffices to show that $D^i x^...


2

A quick read of the proof did not show any error, although I would need to take a good look into it. The result is indeed correct, so I hope there is no major error. For the second question, there is indeed a more efficient proof of this identity. \begin{align*} \sum_{k=0}^nk^j&=\sum_{k=0}^n\sum_{i=0}^j S(j,i)(k)_i\\&=\sum_{i=0}^j i!S(j,i)\sum_{k=0}^...


2

One one speaks about things generated by a set (linear subspaces, subgroups, subrings, and so on) there is usually a category in the background and then the generated object is supposed to be the smallest "acceptable" thing containing the set of generators. If the background category is that of rings, you would only take polynomials without ...


4

The algebra generated by $T$ is usually defined to be the intersection of all algebras containing $T$. If you work out this definition, you will find that the result is one of the algebras you mention, depending on whether you require algebras to have a unit. More genreally, if $A$ is an algebra and $S \subseteq A$ is a subset, then the subalgebra generated ...


0

Here is an alternative method of proof. By the polar decomposition one may write $ T=UP, $ where $U$ is a partial isometry which also satisfies, $ U^*T=P. $ This implies that $U$ is an isometric isomorphism from $\text{Ran}(P)$ to $\text{Ran}(T)$ and, in particular, $$\text{dim}(\text{Ran}(P)) = \text{dim}(\text{Ran}(T)).$$ PS: Thanks to user &...


1

Given an abelian subalgebra $N$ of a von Neumann algebra $M$, such as the center of $M$, one can always write $N=L^\infty (X)$, for some measure space $X$. Furthermore one can decompose $M$ as a "direct integral" $$ M=\int_X^\oplus M_x\,dx $$ (sort of a continuous direct sum) of von Neumann algebras indexed by $X$. When $N$ is the center of $M$ ...


1

Hint: If $\ P\ $ is a square root of $\ T^*T\ $, then $\ T^*T=P^2\ $.


3

Indeed the key point is that $p$-norm is a continuous function as stated in the comments. Although, a small argument is needed: Fix $x=(x_1,...,x_n)\in \mathbb{R^n}$ (i assume that your $p$ is $>1$). Then, by Hölder's Inequality used in the vectors $x,y\in \mathbb{R^n}$ where $y=(1,...,1)$ we have \begin{align} \sum_{k=1}^{n}|x_k|&=\sum_{k=1}^{n}|x_k|\...


1

$$\|Px\|^2\le \|Px\|^2+\|Px-x\|^2=\|x\|^2$$ so $\|Px\|\le\|x\|$ which implies that $\|P\|\le 1$.


1

Use the following statement: If $A$ is a unital $C^*$-algebra and $x\in A$ there are unitaries $u_1,...,u_4$ and complex numbers $a_1,...,a_4$ with $x=\sum_i a_i u_i$. So: $$\varphi(x^*x)=\varphi\left(\sum_i(a_iu_i)^*\sum_j(a_ju_j)\right)= \sum_{ij} \overline{a_i}a_j\varphi(u_i^*u_j)$$ now note that $\varphi(u_i^*u_j)= \varphi\left(u_i(u_i^*u_j)u_i^*\right)...


3

Note that: $$\|(E-E_i)h\|^2 = \|Eh\|^2+\|E_ih\|^2 - \langle Eh, E_ih\rangle - \langle E_ih,Eh\rangle$$ As you have seen, $E_i\to E$ in WOT implies $\|E_ih\|^2\to \|Eh\|^2$, this is the only step that uses $E_i$ and $E$ being projections. From the definition WOT convergence it follows that $\langle Eh, E_ih\rangle\to\langle Eh,Eh\rangle = \|Eh\|^2$ and that $\...


1

Just promoting my comments to an answer because it gives the constant that appears in the question; Ruy gives a (better) method that works for more general kernels, but the kernel here is nice (it is continuous, bounded, $L^1$, ...) so we can apply generalised Minkowski as follows: \begin{align}\left\| \int_{0}^x K(r) f(x-r)dr\right\|_{L^p([0,1])} &\le \...


2

In this answer, we will write $$ D^+_{B}F(A) := \lim_{\epsilon \to 0^+} \frac{F(A+\epsilon B) - F(A)}{\epsilon} $$ whenever the limit exists. Then the following lemma will be useful: Lemma. We have $$ D^+_{B}\exp(A) = \int_{0}^{1} e^{sA}Be^{(1-s)A} \, \mathrm{d}s. $$ Proof. We have $$ D^+_{B}\exp(A) = \sum_{n=1}^{\infty} \frac{1}{n!} D^+_{B}(A^n) = \sum_{n=...


2

By the spectral Theorem you are allowed to think that $\mathcal H=L^2(X,\mu)$, where $(X,\mu)$ is a $\sigma$-finite measure space, and that your operator (which I am going to call) $T$ is given by $$ T(\xi)|_x = \phi(x)\xi(x), \quad \forall \xi\in \mathcal H, \quad \forall x\in X, \tag 1 $$ where $\phi$ is some measurable real valued function on $X$, ...


1

Note that in principle you don't know that $f$ is differentiable. This is probably not a big deal, as differentiable functions are dense. But it complicates the proof. In your use of integration by parts, if you do integration by parts twice while exchanging the roles of the two functions, so indeed you get nothing. Here the argument becomes much easier if ...


1

If you are satisfied with a weaker bound $\frac1{\sqrt{n}(n-1)!}$ then we can use Cauchy-Schwarz on the $\|T^nf\|_\infty$ estimate: \begin{align} \|T^nf\|_2 &\le \|T^nf\|_\infty \\ &\le \frac1{(n-1)!} \sup_{x\in[0,1]}\int_0^1 |x-r|^{n-1}|f(r)|\,dr\\ &\le \frac1{(n-1)!} \sqrt{\sup_{x\in[0,1]}\int_0^1 |x-r|^{2n-1}\,dr} \sqrt{\int_0^1 |f(r)|^2\,dr}\\...


1

Okay, I guess in my way we will not be able to finally get $\frac{1}{n!}$, but a little bit worse bound $\frac{1}{\sqrt{n}(n-1)!}$, but it does not affect the general decaying I need in the end, so it is okay. The following is my proof: The $n-$fold iterate of $T$ is given by the formula $$(T^{n}f)(x)=\dfrac{1}{(n-1)!}\int_{0}^{x}(x-r)^{n-1}f(r)dr.$$ This ...


3

Given $k\in L^2([0, 1]\times[0, 1])$, one may define an operator $A_k$ on $L^2([0, 1])$ by $$ A_k(f)|_x = \int_0^1 k(x, y)f(y)\, dy, \quad \forall f\in L^2([0, 1]), \quad \forall x\in [0,1]. $$ This is called an integral operator and $k$ is said to be its integral kernel. Besides being bounded, $A_k$ is Hilbert-Schmidt, and its Hilbert-Schmidt norm $\...


3

You wrote as if $B$ is the finite-dimensional one, so I'll stick with that. Let $\gamma$ be a C$^*$-norm on $A\otimes M_n(\mathbb C)$ (one certainly exists, because we can represent $A\subset B(H)$ and then $A\otimes M_n(\mathbb C)$ can be represented in $B(H\otimes\mathbb C^n)$). For any $k$, the map $a\longmapsto \gamma(a\otimes E_{kk})$ defines a C$^*$-...


2

Note that the bound will have to scale with $\|X\|$, because the map $X\mapsto U_1XU_1^*-U_2XU_2^*$ is linear. Assuming you are using the (Euclidean) operator norm, we have \begin{align*} \|U_1XU_1^*-U_2XU_2^*\|&\leq\|U_1XU_1^*-U_1XU_2^*\|+\|U_1XU_2^*-U_2XU_2^*\|\\ &\leq\|U_1X\|\|U_1^*-U_2^*\|+\|U_1-U_2\|\|XU_2^*\|\\ &\leq 2\|X\|\varepsilon, \end{...


0

If $x \in X, x \ne 0$, then $u := \frac1{\|x\|_X}x \in X$ and $\|u\|_X = \frac1{\|x\|_X}\|x\|_X = 1$. And $$\|T(u)\|_Y = \left\|T\left(\frac x{\|x\|_X}\right)\right\|_Y = \frac{\left\|T\left(x\right)\right\|_Y}{\|x\|_X}$$ since $T$ is linear and norms commute with positive multipliers. So every value in $$S_1 := \left\{\frac{\|T(x)\|_Y}{\|x\|_X}\ \middle|\ x ...


2

The answer is negative and here is a counter-example. Let us begin with the following: Lemma. If $T$ and $S$ are positive operators on a Hilbert space, with $0\leq T\leq S\leq 1$, and if $\xi $ is any vector such that $T\xi =\xi $, then $S\xi =\xi $, as well. Proof. We have $$ \|\xi \|^2 = \langle \xi , \xi \rangle = \langle T\xi , \xi \rangle \leq \...


4

This is not true. Consider $A=M_2(\mathbb C)$, and $$B=\left\{\begin{bmatrix}b&0\\0&0\end{bmatrix}:b\in\mathbb C\right\}.$$ Note that $$\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}\begin{bmatrix}b&0\\0&0\end{bmatrix}=\begin{bmatrix}a_{11}b&0\\a_{21}b&0\end{bmatrix},$$ so clearly $AB$ cannot be all of $A$.


1

Yes. Let $y$ be vector in the codomain. $A$ is bijective so there is a vector $x$ in the domain such that $Ax = y$. We have $$\langle A^{-1}y,y\rangle = \langle A^{-1}Ax,Ax\rangle = \langle x,Ax\rangle \ge 0.$$ Therefore, $A^{-1}$ is positive.


1

Certainly true for square matrices. $$A x = \lambda_j x$$ with all $\lambda_j >0$ Then $$A^{-1} A x = \lambda_j A^{-1} x$$ or $$\frac{1}{\lambda_j} x= A^{-1} x.$$ $$\lambda_j >0 \iff \frac{1}{\lambda_j} >0.$$ Making use of the property that $A>0 \iff$ all eigenvalues are positive.


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