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0

Yeaaahh, I got it! :D Great thanks to David C. Ullrich!!! Counterexample Given the Hilbert space $\mathbb{C}^4$. Regard the matrices: $$N:=\begin{pmatrix}1&0\\0&-1\end{pmatrix}\oplus\begin{pmatrix}0&0\\0&0\end{pmatrix}\quad N':=\begin{pmatrix}0&0\\0&0\end{pmatrix}\oplus\begin{pmatrix}1&0\\0&-1\end{pmatrix}$$ Then they are ...


0

1) Yes. 2) CP on a Hilbert space does not make sense. You need $X\subset B (H) $ a C$^*$-algebra, and $Y\subset B (K) $ on the codomain. There, you have to convince yourself that entry wise sot convergence in $M_n (Y) $ implies sot convergence in $M_n (Y) $. Note that $\langle T_jx,x\rangle \to\langle Tx,x\rangle$ is wot convergence, not sot.


1

From $a^*ab^*b=0$ we get $$0=a^*ab^*ba^*a=(ba^*a)^*ba^*a, $$ so $ba^*a =0$. But then $ba^*ab^*=0$, which implies $ab^*=ba^*=0$ The others are similar.


0

Given the Hilbert space $\ell^2(\mathbb{N}_0)$. Consider the left shift: $$L_0:\mathcal{D}(L_0)\subseteq\ell^2(\mathbb{N}_0)\to\ell^2(\mathbb{N}_0):\quad L_0:=L$$ For nondense domain: $$\mathcal{D}(L_0):=\ell^2(\mathbb{N}):=\{\varphi\in\ell^2(\mathbb{N}_0):x_0=0\}$$ But it has an inverse: $$0\not\in\sigma(L_0):\quad RL_0=1_0\quad L_0R=1$$ Concluding ...


1

To prove what you want, it suffices to prove that if $|\lambda| \neq 1$, then $(A-\lambda)$ is bijective. I was able to prove injectivity - surjectivity seems hard to just brute force, so I will post this and let someone else come up with an elegant answer :) Injectivity: Suppose $Ax = \lambda x$, and $\lambda, x\neq 0$, then $$ \frac{1}{3}x_1 = \lambda x_0 ...


1

The claim is not true for merely continuous and bounded $g$. For instance, take $$ g(v):=\min( \sqrt{\|v\|},1). $$ Fix $x$ such that $\Pi x \ne \hat\Pi x$. Then for $s>0$ such that $\|s\hat\Pi x\|\le1$ and $\|s\Pi x\|\le1$ $$ \|g(s\hat\Pi x)-g(s\Pi x) \|= \sqrt s\left|\sqrt{\|\hat\Pi x\|} - \sqrt{\Pi x}\right|, $$ hence for $s\searrow 0$ the quantity $$ ...


2

Let's prove it. Suppose $k$ is in the kernel of $A$ and $x_0$ is a specific solution, so that $Ax_0=b$. We have $$A(x_0+k)=Ax_0+Ak=Ax_0=b.$$ Conversely, suppose that $x_0$ and $y_0$ are two solutions. Then $$A(x_0-y_0)=0$$ implying that $x_0-y_0=:k$ is in the kernel. Therefore any two solutions differ by an element of the kernel.


1

With a clear domain definition, and $\|\frac{1}{\Delta t}\{U_{\alpha}(\Delta t)-1_\alpha\}\varphi_{\alpha}-H_{\alpha}\varphi_{\alpha}\|=\|\frac{1}{\Delta t}\int_{0}^{\Delta t}(U_{\alpha}(t)-1_\alpha)H_{\alpha}\varphi_\alpha dt\|$, can you now better establish convergence?


1

Since $p$ is infinite, there exists $r\leq p$ with $r\ne p$ and $r$ equivalent to $p$. That is, there exists $v$ with $v^*v=r$, $vv^*=p$. Note that $v=pvr$, so in particular $(q-p)v=0$, and $v(q-p)=0$. Now let $s=r+q-p$. Then $s$ is a proper subprojection of $q$. Let $w=v+q-p$. Then $$ w^*w=(v^*+q-p)(v+q-p)=v^*v+q-p=r+q-p=s, $$ $$ ww^*=vv^*+q-p=p+q-p=q. ...


0

Concerning 1, yes. This is an example where abstract nonsense is the most useful: An inverse limit is a categorical limit in the category of $C^*$-algebras, and those are unique up to unique isomorphism. If you check that $A_0$ fulfills the universal property, that means automatically that $A_0 \cong \lim{A_I}$. A proof goes like that: Assume there's a ...


0

Left Shift Denote for shorthand: $$1_0(0,x_1,\ldots):=(0,x_1,\ldots)$$ For the modulus: $$L^*L=RL=1_0\implies|L|=1_0$$ For the argument: $$L=U|L|\implies U=1_0,1,\ldots$$ So it admits one. Right Shift For the modulus: $$R^*R=LR=1\implies|R|=1$$ For the argument: $$R=U|R|\implies U=R$$ So it admits none. Reference For much more details: Polar ...


1

Unilateral shifts (including weighted shifts) are never normal.


1

This was answered in the negative in this post at MathOverflow. See the first answer there and Bill Johnson's comment to the second answer.


1

As @DanielFischer said, you can first show $I-Z$ is invertible if $\|Z\|<1$; then consider $I:Y\rightarrow Y$ and $Z=(A_0-A)A_0^{-1}:Y\rightarrow Y$. The $Z$ norm can arbitrary small by your choice of $\varepsilon$. Now you can rewrite $A^{-1}$ as $A^{-1}=A_0^{-1}(I-(A_0-A)A_0^{-1})^{-1}$.


2

First for the eigenvalues: By the spectral theorem (and since $A$ is a positive (hence self-adjoint) operator), there is a measure space $(X, M, \mu)$ and some (measurable) function $g : X \to [0,\infty)$, such that $A$ is unitarily equivalent to the multiplication operator $$ M_g : L^2(\mu) \to L^2 (\mu), h \mapsto g\cdot h. $$ Thus, it suffices to prove ...


1

I don't see how to understand $A \wedge B$. Is it the operator that takes $x \wedge y$ to $Ax \wedge By$, or the operator that takes $x\wedge y = -y\wedge x$ to $-Ay\wedge Bx = Bx \wedge Ay$? So it only makes sense if $A = B$. Also, we don't have $A \wedge A = 0$, we only have $x\wedge x = 0$. And in fact some people prefer the notation $\Lambda^{(2)} ...


1

Let $\mathcal S$ be an orthonormal basis of eigenvectors for $|A|$ (this exists, since $|A|$ is trace-class and thus compact). Let $\mathcal T$ be any other orthonormal basis. For any $\sigma\in\mathcal S$, we have $$ |A|\sigma=s_\sigma(A)\,\sigma $$ and $\sum_\sigma s_\sigma(A)=\sum_\sigma\langle |A|\sigma,\sigma\rangle<\infty$. Then, for any $B\in ...


1

Yes. The functional calculus preserves approximation by polynomials. So, from $$ T^n=\begin{bmatrix}A^n&0\\0&B^n\end{bmatrix}, $$ you get that $$ p(T)=\begin{bmatrix}p(A)&0\\0&p(B)\end{bmatrix} $$ for any polynomial $p$. Now using a sequence of polynomials that converges uniformly to $f$, you get that $$ ...


0

I can explain the first question, the author said "if $prox_f$ were a contraction", which is not guaranteed. Banach fixed-point theorem tells us repeatedly applying contraction would find a (here, unique) fixed point. So if $prox_f$ is a contraction, then by repeatedly applying proximal operator will result a fixed point.


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According to wikipedia, an operator is a function whose domain and codomain are both vector spaces or modules. Since $\mathbb{R}, \mathbb{Q}, \mathbb{C}$ are all (one-dimensional) vector spaces, many familiar functions are also operators. However, a general function might be from a domain that is not a vector space, and hence not be an operator, e.g. ...


1

It is not true, and it is easy to construct counterexamples. For instance, take $X = \Bbb R^2$ with the usual norm $\Vert y \Vert = \langle y, y \rangle^{1/2}$, where $\langle \cdot, \cdot \rangle$ denotes the standard real inner product, viz. $\langle (w_1, w_2), (v_1, v_2) \rangle = w_1v_1 + w_2 v_2$; let $T$ be the matrix $T = \begin{bmatrix} a & 0 ...


3

Invertible and nonivertible operators may have arbitrary norms: just note that $\|cT\|=|c|\|T\|$ and $T$ is ivertible if and only if $cT$ is invertible (provided $c \neq 0$)


0

Assuming that for finite dimensional Hilbert spaces $K$ and $H$, I can write any $T$ as a linear combination of these products, then by Goldstine's Theorem, these linear combinations are $w^{*}$ dense in $B(K)$.


2

It is the whole of $B(H)$. This simply follows from Goldstine's theorem and the dualities between $K(H)$, $K(H)^*$ and $B(H)$ (try to write down the corresponding duality brackets yourself). You may replace here $H$ with any reflexive Banach space that has the bounded approximation property.


1

Although $0 \le A \le B$ implies $\overline{{\cal R}(A)} \subseteq \overline{{\cal R}(B)}$, it's not true without the closures. For a counterexample, take $L^2[0,1]$. Let $A$ be multiplication by $x$ (i.e. $A f(x) = x f(x)$), and $B = A + u u^*$ where $u(t) = t^{1/4}$ and $u^*$ is the corresponding linear functional, i.e. $$ B f(x) = A f(x) + u^*(f) u(x) = ...


0

Firstly I just wanted to clarify what you meant by "twice absolutely continuous", my assumption is that means "twice absolutely continuously differentiable". If that is the case, then the fact that the question is posed on $\Bbb R$ is important, of course $f\in L^2(\Bbb R)$ is a strong assumption. Assume $Hf=-f''+x^2f\in L^2(\Bbb R)$, this tells us there ...


1

The implication $0\leq A\leq B$ $\implies $ $\mathcal RA\subset\mathcal RB$ can be proven as follows. From $0\leq A\leq B$, we easily see that if $Bx=0$, then $$ 0\leq\langle Ax,x\rangle\leq\langle Bx,x\rangle=0, $$ so $A^{1/2}x=0$, and thus $Ax=0$. In other words, $\ker B\subset\ker A$. Then $$ \overline{\mathcal RA}=(\ker A)^\perp\subset(\ker ...


0

Construction Suppose one has: $$\dim\mathcal{H}\geq\dim\mathcal{K}$$ Remember the unitaries: $$\#S=\dim\mathcal{H}:\quad U_\mathcal{H}(\varphi):=(\langle \sigma_s,\varphi\rangle)_{s\in S}\in\ell^2(S)$$ $$\#T=\dim\mathcal{H}:\quad U_\mathcal{K}(\psi):=(\langle \tau_t,\varphi\rangle)_{t\in T}\in\ell^2(T)$$ Regard an embedding: $$\iota:T\hookrightarrow ...


0

I think I got it. As a result of the first isomorphism theorem, the dual space of a direct sum is the direct sum of the annihilators. So answer 1 is a wrong answer.


2

The result is false in this generality. Fix some $z\in S$ and define $O\mu=\mu_c+\mu_d(S)\delta_z$, where $\mu_d$ and $\mu_c=\mu-\mu_d$ are the discrete and continuous parts of $\mu$, respectively. Then $O$ satisfies your conditions, but cannot possibly be given by a kernel, if there is a non-discrete probability measure $\nu$ on $S$. Why not? If ...


2

The uniform bounded principle does work for any collection of operators, see wiki or any book on functional analysis. Note that convergent nets may not be bounded (in contrary to convergent sequences), see the post to which David Mitra is referring.


2

First part: Assuming the series on right does converge to R, then R will be linear operator. For ex. For any matrix $A$, $$exp(A) := \sum_{n=0} ^{\infty} \frac{A^n} {n! } $$ is a linear operator as $exp(A)(k u + t v) =k ~exp(A)u +t~ exp(A)v $ for real number $k, t$. Regarding second part, series will have all the "nice" properties like series of real or ...


0

Maybe you are interested in the Functional Calculus ( of different types) https://en.wikipedia.org/wiki/Functional_calculus ?


0

Existence For complex sums: $$\sum_\sigma\langle A\sigma,\sigma\rangle\in\mathbb{C}\iff\sum_\sigma|\langle A\sigma,\sigma\rangle|<\infty$$ By polar decomposition: $$|\langle A\sigma,\sigma\rangle|\leq\||A|^{1/2}\sigma\|\cdot\||A|^{1/2}J^*\sigma\|$$ For partial isometries: $$\tau^{(\prime)}:=J^*\sigma^{(\prime)}:\quad\sigma\perp\sigma'\implies ...


0

Sorry, but I'm not comfortable with nets. I can answer part a) however. Consider a weakly open basic neighbourhood $\mathcal{U}$ of $0$. That is, for some $y_1, y_2, \ldots, y_m \in H$ and $\alpha_1, \alpha_2, \ldots, \alpha_m \in (0, \infty)$, we have $$\mathcal{U} = \lbrace x \in H : \forall k = 1 \ldots m, |\langle x, y_k \rangle| < \alpha_k ...


0

Let me elaborate on the comment of Ian. Indeed you can characterize the solvability by the spectrum of the operator. As you wrote, the equation is has a unique solution for every $b$, iff $I-L$ is invertible. This is equivalent to $0 \not\in \sigma(I-L)$ and to $1 \not\in \sigma(L)$. And there are many examples for such $L$.


0

Take $L(x) = 2 x$ for $x \in \Bbb{R}$ $$x = 2 x + b \Rightarrow x = -b \text{ is the unique solution}$$


0

Meanwhile I got it... Limit Projection By orthogonality one has: $$\varphi\in\mathcal{R}^\perp\implies\varphi\in\mathcal{R}_\lambda^\perp\quad(\lambda\in\Lambda)$$ So one obtains: $$\|P_\lambda\varphi-0\|=\|0-0\|=0\stackrel{\lambda}{\to}0$$ By monotony one has: ...


3

You can approximate the constant function with Gaussians $\phi_{\sigma^2}(x)=\frac{1}{\sqrt{2\pi \sigma^2}} \exp(-x^2/2\sigma^2)$. If I calculated correctly you have $\|\phi_{\sigma^2}\|^2= \frac{1}{2\sqrt{\pi \sigma^2}}$ and since $A\phi_{\sigma^2} = \phi_{\sigma^2+1}$ (the convolution of two Gaussians is Gaussian with the sum of the variances) you get ...


3

Yes, it is true that $\|A\|=1$. The operator $A$ is sometimes called a Fourier multiplier or simply a multiplier because it acts by pointwise multiplication of the Fourier transform. (Alternatively, multipliers are the same as convolution operators). So anyway, we can prove the following: Let $m\in L^\infty(\mathbb{R})$ be a bounded and measurable ...


3

Have you tried to check $2\times2$ matrices $$ A = \begin{pmatrix} a&b\\0&c\end{pmatrix} $$ What can you say for $$ B = A^*A-AA^* $$


1

Thr natural thing is to prove the contrapositive. If $T $ is not bounded, there exists a sequence $\{x_n\}_X $ with $\|x_n\|=1$ and $\|Tx_n\|>n^2$. Then $x_n/n $ is a sequence that converges to zero with its image through $T $ unbounded. Conversely, if $x_n\to0$ with $\{Tx_n\} $ unbounded, then $T $ is unbounded.


1

Consider the pushforward: $$E_\eta:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H}):\quad E_\eta(A):=E(\eta^{-1}A)$$ Their domain agree as:* $$\int|\vartheta\circ\eta|^2\mathrm{d}\nu_\varphi(\lambda)=\int|\vartheta|^2\mathrm{d}\nu^\eta_\varphi$$ Denote for shorthand: ...


1

Eigenvalue problems were first seriously considered in separation of variables problems arising out of Partial Differential Equations. The separation parameter was the eignevalue, and general eigenfunction analysis came out of Fourier's method. The matrix methods came out of this method, which is the opposite direction of abstraction that one would naturally ...


2

Nets that converge do not need to be bounded. Even when you use the uniform boundedness principle, you see that the operators applied to individual elements need not be bounded. For an example of this, imagine a net that is indexed by two copies of $\mathbb{N}$ cascaded. (So every element of one copy is greater than the other.) I believe this is typically ...


0

Let me leave out technicalities.. Closed Operators Given a Banach space $E$. Consider a closed operator: $$T:\mathcal{D}(T)\subseteq E\to E:\quad T=\overline{T}$$ Denote its resolvent: $$R(\lambda):=(\lambda-T)^{-1}\in\mathcal{B}(E)$$ Then one can construct: $$\eta\in\mathcal{H}(\mathbb{C}):\quad\eta(T):=\oint\eta(\lambda)R(\lambda)\mathrm{d}\lambda$$ ...


1

In $M_2(\mathbb C)$, the normal operators are precisely the unitarily diagonalizable ones. So, any non-selfadjoint normal operator is, as TrialAndError mentioned, of the form $$ N=U^*\,\begin{bmatrix}\lambda&0\\0&\mu\end{bmatrix}\,U $$ with at least one of $\lambda,\mu$ not real (if both were real, $N=N^*$). The general $2\times 2$ unitary is of ...


1

In general, they are not weakly closed (not even strongly closed). To see this, consider the Hilbert space $\ell^2([0,1])$ and the operator $N=M_{\rm id}$. Then the (bounded) Borel functional calculus will yield all operators $M_f$ (multiplication with $f$, where $f$ is Borel measurable and bounded. But now consider the set $A:=\{I\subset [0,1]\,\mid\,I ...


1

By density it is enough to consider smooth functions. The arguments work for general Sobolev functions as well, but there is less to worry about with smooth functions. Consider first $i_{k,l}$ for $0\leq l\leq k$. Take a function $u\in C^\infty$. Its squared norm is $$ \|u\|_{H^k}^2=\sum_{|\alpha|\leq k}\|\partial^\alpha u\|_{L^2}. $$ Now $$ ...


0

Let $\phi$ be any $C^\infty$ function with compact support in $(0,1)$, like $$ \phi(x)=e^{-\tfrac{1}{x(1-x)}}\text{ if } x\in(0,1),\quad\phi(x)=0\text{ otherwise}. $$ Note. The support of $\phi$ is $[0,1]$, so it does not have compact support in $(0,1)$, as commented by Jonas Meyer. Change it to $$ \phi(x)=e^{-\tfrac{1}{(x-a)(b-x)}}\text{ if } ...



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