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1

The claim is false in the general infinite dimensional setting. For example, let $X$ be the Banach space $l^p$, and define $A,B$ by$$A(x_1,x_2,\ldots)=(x_2,x_3,\ldots),\qquad B(x_1,x_2,\ldots)=(0,x_1,x_2,\ldots).$$Then $AB$ is the identity, but $A$ and $B$ are not invertible.


0

For the question in the title, the answer is No. Take $B : \ell^2 \to \ell^2$ be the right-shift operator $$ (x_n) \mapsto (0,x_1,x_2,\ldots) $$ and $A$ to be the left shift $$ (x_n) \mapsto (x_2,x_3,\ldots) $$ Then $AB$ is the identity operator, and so invertible, but neither $A$ nor $B$ are invertible. But later on, you ask about $B^{-1} A^{-1}$ implying ...


3

Not necessarily. For example, define on $\ell^p$ $$ L(x_1,x_2,\dots) = (x_2,x_3,\dots)\\ R(x_1,x_2,\dots) = (0,x_1,x_2,\dots) $$ Then $LR = \text{id}_{\ell^p}$ is invertible, but neither $L$ nor $R$ are bijective. You might be able to say something in the case that both $AB$ and $BA$ are invertible.


1

It's not a lucky guess. It's at the heart of Stein's method: you need a characterizing equation for your distribution. There isn't a unique equation, but in fact many and depending on the situation, one might use other characterizing equations. In the case of the normal distribution, this is actually simple integration by parts. Below, all expectations are ...


2

Hint: Use integration by parts to show that $\langle A f, \varphi \rangle = \langle f , A \varphi \rangle$ for all $f, \varphi \in C^\infty_0(\mathbb{R}^2)$. Use this to show that $\langle g, \varphi \rangle = 0$ for all $\varphi \in C^\infty_0(\mathbb{R}^2)$. By the density of $C^\infty_0$, conclude that $g=0$. This argument proves a more general result: ...


0

Is this a simple example? Let $\Omega = [0,1)$ with the Lebesgue measure. Let $\chi_{[0,1/2)}$ be the characteristic function of the interval $[0,1/2)$. Set $$ Pf = f\chi_{[0,1/2)}, \quad f \in L^1(\mu). $$ Clearly, $P : L^1(\mu) \to L^1(\mu)$, $\|Pf\| \leq \|f\|$ for all $f \in L^1(\mu)$ and $Pf \geq 0$ whenever $f \geq 0$. Also, for every $\alpha \gt 0$ ...


1

If $\Omega\subset\mathbb{R}^n$ is bounded, then convolution operators will do: They are compact (if you convolute with, say a standard mollifier), as a consequence of Kolmogorov-Riesz, thus do not fulfill $\alpha \|f\|\leq \|Pf\|$ (note that zero is in the spectrum of compact operators). Furthermore, if you convolute with something positive you'll get ...


0

Answered in the comments: http://en.wikipedia.org/wiki/Adjoint_representation_of_a_Lie_group – Qiaochu Yuan Apr 23 '11 at 18:27 I'm pretty sure that @Qiaochu's right. I'd even go so far as to say that the motivation stems from the finite-dimensional case $A=M_n(\mathbb C)$, where the unitary group $U(n)=\mathcal U(A)$ leaves the subspace of ...


1

The trace is a continuous linear functional on the space of trace-class operators. So you can do what you want provided that $\int_a^b A(x)dx$, $\lim_{x\to a} A(x)$ and $\frac{dA(x)}{dx}$ make sense with respect to the trace-class norm. For the integral, this means that $x\mapsto A(x)$ should be for example Bochner-integrable from $[a,b]$ into the space of ...


0

Your reasoning is correct. In fact, Hahn-Banach is not needed for the isometric part either. On the other hand, it is needed to show that the dual of a subspace $M\subseteq X$ is isometric to the quotient space $X^*/M^\perp$.


2

Observe that $Tr(\rho\ \cdot\ )$ defines a state, say $\omega_\rho$, hence you have the estimate $$|\omega_\rho(U^*[U,O])|\leq\Vert U^*[U,O]\Vert\leq\Vert[U,O]\Vert.$$


3

$$\mathrm{Tr}[\rho U^{\ast }[U,O]]=<\rho ,U^{\ast }[U,O]> $$ defines a bounded linear functional on the trace class on some separable Hilbert space. Since the dual of the trace class (norm $||..||_{1}$) is the set of bounded operators (norm $||..||$), we have the inequality $$ |<\rho ,U^{\ast }[U,O]>|\leqslant ||\rho ||_{1}||U^{\ast ...


1

There's something off here, because the left side is positively homogeneous in $\rho$ but the right side doesn't depend on $\rho$. Are you missing an assumption on the trace norm of $\rho$?


1

Yes, there are such operators. The classical example is the operator $A=\frac{1}{i}\frac{d}{dt}$ on the domain $\mathcal{D}(A)\subseteq L^{2}[0,\infty)$ consisting of all absolutely continuous functions $f \in L^{2}[0,\infty)$ for which $f' \in L^{2}[0,\infty)$. This operator is closed and symmetric, but has no selfadjoint extension. The adjoint $A^{\star}$ ...


1

Here are some thoughts: Let $(a_n), (b_n)\in l^2$. Their inner product is given by: $$ \langle (a_n), (b_n) \rangle = a_1\bar{b_1} + a_2\bar{b_2} + \cdots $$ Hence, $$ \langle T(a_n), (b_n) \rangle = \alpha_1a_1\bar{b_1} + \alpha_2a_2\bar{b_2} + \cdots $$ The defining property of $T^*$ is that it satisfies: $$ \langle T(a_n), (b_n) \rangle = \langle (a_n), ...


1

This is a corollary from Riesz functional calculus and spectral mapping theorem and the proof below is standard( as you can find it in many textbooks on functional analysis ). Setup 1: Let $\sigma(A)$ be the spectrum of $A$. We can write $\sigma(A)=K_1\cup K_2$, where $K_1$ is contained in $Int(C)$ and $K_2$ is outside the closure of $Int(C)$. Setup 2: ...


1

Assuming you're working on a complex Hilbert space, the range condition and the following condition are equivalent to dissipative: $$ \Re(Ax,x) \le 0,\;\;\; x \in \mathcal{D}(A). $$ This last condition persists if you add another bounded operator $C$ satisfying the same condition. In your case, because $D$ is bounded, $$ ...


0

What I am going to say isn't anything new, but sometimes it helps to look more abstractly to see the big picture. Consider a linear map $A : \mathcal{D}(A) \subseteq X \rightarrow Y$ where $X$ and $Y$ are Banach spaces. The graph of $A$ is $\mathscr{G}(A)=\{ (x,Ax) \in X \times Y : x \in \mathcal{D}(A) \}$. The graph is a linear subspace of $X\times Y$. ...


2

Here is one proof of $(\operatorname{aff} C - \operatorname{aff} C) \subset \operatorname{aff} (C - C)$. Note that $S$ is affine iff $S$ can be written as $\{x_0\}+L$ for some linear space $L$. Let $\operatorname{aff} C = \{x_0\} +L$. Then $\operatorname{aff} C - \operatorname{aff} C = \{x_0\} +L + \{-x_0\} +(-L) = L$, hence $\operatorname{aff} C - ...


2

Using the spectral theorem: A selfadjoint operator $T \ne 0$ on a Hilbert space is compact iff $$ T = \lambda_1 E_{1} + \lambda_2 E_{2} + \cdots, $$ where $\{ E_{j} \}$ is a finite or countably infinite set of disjoint orthogonal projections onto finite-dimensional subspaces, and the sequence $\{ \lambda_{j} \}$, if infinite, converges to ...


1

i) if $F(k)$ is the Fourier transform of $f(x)$, then $aF(k)$ is the Fourier transform of $a(-i\nabla)f(x)$. ii) the operator $\gamma$ has kernel $\gamma(x,x')$, in the sense that $(\gamma f)(x)=\int \gamma(x,x')f(x')dx'$; this defines the diagonal of $\gamma$; in particular, the trace of $\gamma$ is the integral of the diagonal, ${\rm tr}\,\gamma=\int ...


0

By definition, the graph of $\overline{A}$ is the closure of the graph of $A$. That means $(x,y)$ is in the graph of $\overline{A}$ iff there is a sequence $x_n \in D(A)$ such that ...


1

We prove that if $A$ and $B$ are linear manifolds then $A+B$ is also a linear manifold. Indeed, let $x,y\in A+B$ and $\lambda\in\mathbb{R}$. Then there exist $x_a,y_a\in A$ and $x_b,y_b\in B$ such that $x_a+x_b=x$ and $y_a+y_b=y$. Then $$ \lambda x+(1-\lambda)y=\lambda(x_a+x_b)+(1-\lambda)(y_a+y_b)=[\lambda x_a+(1-\lambda)y_a]+[\lambda x_b+(1-\lambda)y_b]. ...


0

You have a symmetric operator $P$. That is, $(Pf,g)=(f,Pg)$ for all $f,g \in \mathcal{D}(P)$. A general method for showing that a symmetric operator $P$ is densely-defined and selfadjoint is to show that $(P\pm iI)$ are surjective, which is easy to apply in your case because, for all $g \in L^{2}(\mathbb{R})$, one has $\frac{1}{x\pm i}g \in L^{2}$ and $(P\pm ...


0

The actual question isn't very clear; I am going to assume that question is to show that hypotheses 1 (which I take to mean the set of of common eigenvectors forms a one-dimensional subspace, with the zero vector removed) and 2 imply that there are no other one-dimensional common invariant subspace of these operators. Then the answer is simply that in ...


2

Actually, from your relation, just apply "ri" to get $$\rm{ri\,conv\,rge}A=\rm{ri}(\rm{ri\,conv\,rge}A)\subset \rm{ri\,rge}A\subset\rm{ri\,conv\,rge}A,$$ from which $\rm{ri\,rge}A=\rm{ri\,conv\,rge}A$ is convex.


4

The wedge product as used in e.g. $\mathrm du \wedge \mathrm dv$ is completely separate from logical conjunction $\alpha\land\beta$. There are even two TeX commands: \wedge and \land, where the latter is to be used for conjunction ("logical and"), to keep the two separated in writing your documents.


1

Let $\mathcal{A}$ be unital and $\omega$ be a state. Then $$ 0 \leq \omega(|a - \omega(a){\bf 1}_\mathcal{A}|^2) = \omega(a^*a - a\omega(a^*) - a^* \omega(a) + |\omega(a)|^2{\bf 1}_\mathcal{A}) = \omega(a^*a) - |\omega(a)|^2. $$


1

The following claim is not true in general. Let $X$ and $Y$ be Banach spaces and $X_1 \subset X$ a subspace. If $T: X_1 \to Y$ is a continuous bounded operator, then there exists a continuous linear extension to the whole space $X$, i.e. there exists a bounded linear operator $\hat T: X \to Y$ such that $ \hat{T}\restriction_{X_1} = T$. Take for ...


0

Let $$ S = \{(y_n) \in \ell^{\infty} : \exists C > 0 \text{ such that } |y_n| \leq C/n\quad\forall n\in \mathbb{N}\} $$ If $(x_n) \in \ell^{\infty}$ then $T(x_n) \in S$ with $c = \|(x_n)\|_{\infty}$, and conversely, if $(y_n) \in S$, then the sequence $(x_n)$ defined by $$ x_n := ny_n $$ is in $\ell^{\infty}$ and satisfies $T(x_n) = (y_n)$. Hence, $R(T) = ...


0

Since $H$ is a Hilbert space, I will give another proof using this property. Assume $T$ is compact and let $(x_n)_{n\in\mathbb{N}}$ be a bounded sequence. As it is bounded, we can construct a weakly convergent subsequence, call it $(x_n)_n$ again in abuse of notation, such that $x_n\to x$ weakly as $n\to\infty$. Now, we want to show that $T^\ast x_n \to ...


1

I'm assuming your notation means that, for every $h \in \mathcal{D}(H)$, one has $Ph \in \mathcal{D}(H)$ and $HPh = PHh$. If that is the case, let $R(\lambda)=(H-\lambda I)^{-1}$ for $\lambda\not\in\sigma(H)$ and note that $$ P(H-\lambda I)h = (H-\lambda I)Ph,\;\;\; h \in \mathcal{D}(H) \\ R(\lambda)Pg = PR(\lambda)g,\;\;\; g \in \mathcal{H}. $$ ...


1

You argument is that $$ F(\alpha) = \int_{0}^{\infty}e^{-\alpha t}f(t)dt \\ \int_{0}^{\infty}e^{-\alpha t}\alpha F(\alpha)dt = F(\alpha)=\int_{0}^{\infty}e^{-\alpha t}f(t)dt \\ \implies \alpha F(\alpha)=f(t) $$ Does that help you spot your error?


0

The answer to question 1 is negative. For example, $L(\varphi)=\varphi(0)$ is not of this form. Neither is $L(\varphi)=\varphi'(1/2)$, etc. A continuous functional on $E$ is bounded by some finite collection of seminorms; therefore it is also a continuous functional on $C^m([0,1])$ for some $m$. The space $C^m([0,1])$ is a direct sum $P_m\oplus V_m$ where ...


0

Meanwhile I got it... The derivative translates into: $$\langle\tfrac{1}{h}\{A(t+h)-A(t)\}\varphi,\psi\rangle=\langle\tfrac{1}{h}\{A(h)-A(0)\}e^{itH}\varphi,e^{itH}\psi\rangle$$ Expand the expression: $$|\langle\tfrac{1}{h}\{A(h)-A(0)\}\varphi,\psi\rangle-\langle A\varphi,iH\psi\rangle-\langle iH\varphi,A\psi\rangle|\\ \leq|\langle ...


2

Since the sequence $(A \varphi_n)_n$ is bounded, it suffices to check the weak convergence on a dense subspace, e.g. for $\chi \in\mathcal{D}(A)=\mathcal{D}(A^\ast )$. But for those $\chi$, we have $$ \langle A \varphi_n,\chi \rangle =\langle \varphi_n , A \chi\rangle \to \langle \varphi, A \chi\rangle =\langle A\varphi, \chi\rangle. $$ Here, the last ...


0

Hint: Consider the polynomial $f(X)=X^2.$


2

Let $T\in B(H)$ be positive, with $\|T\|\leq1$. We want to show that, for each basic wot-neighbourhood $V$ of $T$, there exists a projection $P\in V$. Such a neighbourhood $V$ is of the form $$ V=\{X\in B(H):\ |\langle(T-X)y_j,z_j\rangle|<\varepsilon,\ j=1,\ldots,m\} $$ for a fixed $\varepsilon>0$ and $y_1,\ldots,y_m,z_1,\ldots,z_m\in H$. The key idea ...


0

Use the spectral theorem for normal operators. $H = -i \log(U)$ (for any branch of the logarithm).


3

$1.$ $T$ is surjective because of the Fundamental Theorem of Calculus. The reason that $T^{-1}$ does not exist is that $T$ is failing to be injective. $T$ is not injective because any two functions that differ by a constant will get mapped to the same function, e.g., $x$ and $x+c$, $c\in \mathbb{R}$ both get mapped to the constant function $1$. $2.$ Since ...


1

Let $\mathbf{x}=(x_n)$ be any sequence in $l^{\infty}$, then it can be easily seen that give any $\epsilon>0$ you can find a natural number $N$ such that $\frac{\|\mathbf{x}\|}{N} < \epsilon$. So, for all $n\geq N$ you can see that $|x_n|\leq\|\mathbf{x}\|$ and $\frac{1}{n}\leq \frac{1}{N}$, so, \begin{equation}|\frac{x_n}{n}| \leq ...


2

By Fundamental Theorem of Calculus you can see that the range of $T$ is $R(T) =\{f\in C^1[0,1]\mid f(0)=0\}$ where $C^1[0,1]$ is the space of all continuously differentiable functions. $T^{-1}$ is linear is easy to see. But $T^{-1}$ is not bounded. To see this first of all notice that $T^{-1}(x(t)) = x'(t)$, i.e., $T^{-1}$ is the differentiation operator ...


0

Question 1: $T^{-1}$ does not exist because $T$ is not injective. E.g. Let $x(t)=t$ , $y(t)=t+2.$ Then $Tx(t)=Ty(t)=1.$ Observe that $T$ is surjective by the fundamental theorem of calculus (which can also be applied to question 3).


0

That's not true. Consider the operators $T,S \colon L^2[0,1] \to L^2[0,1]$, given by $$ Tx(t) = tx(t), \quad Sx(t) = t^2x(t). $$ Both are self-adjoint, and both have no eigenvalues. But $S \ne T$, as $S1 \ne T1$.


2

The exponential of an idempotent $R$ is very easy to calculate: if $R^2=R$, then $$ e^{i\pi R}=\sum_{k=0}^\infty\frac{i^k\,\pi^k\,R^k}{k!} =I+\sum_{k=1}^\infty\frac{i^k\,\pi^k\,R}{k!} =I-R+R\left(\sum_{k=0}^\infty\frac{i^k\pi^k}{k!}\right)\\ =I-R+e^{i\pi}R=I-2R. $$ Then $$ U(e^{i\theta})=e^{i\pi E(e^{i\theta})}e^{i\pi F(e^{i\theta})} ...


0

Positive semi-definite by definition are symmetric matrices that has positive eigen-values pick for example 2x2 identity matrix and multiply first $a_11$ component by negative one you'll see that this matrix is symmetric however have negative eigen-values in particular it has -1 as an eigen-value.


2

$1.$ If $T$ is bounded then it is not hard to see that $T$ maps bounded sets to bounded sets. Conversely, let us assume that $T$ maps bounded sets to bounded sets. In particular $T(B_X)$ will be bounded, where $B_X$ is the closed unit ball of $X$. So, there exists $c>0$ such that $\|Tx\|\leq c\|x\|$ for all $x\in B_X$. Now, let $y\in X$ be any arbitrary ...


0

What you are saying is absolutely correct. There is another way, more simple and basic, to see this. Let $x$ be a limit point of $M^{\perp}$, i.e,. there exists a sequence $(x_n)$ in $M^{\perp}$ such that $x_n \longrightarrow x$. Since inner product is continuous, it follows that for any fixed $y$; $\langle x_n,y\rangle \longrightarrow \langle x,y\rangle$. ...


2

We can rewrite this equation as $$ u(x) - f(x) =\lambda \int_0^{1/2}u(y)\,dy $$ Note that since the integral does not vary over $x$, any solution $u$ must differ from $f$ by a constant. That is, we may write $u(x) = f(x) + k$ for some constant $k$. The question then becomes whether a suitable constant $k$ exists. Substituting $u(x) = f(x) + k$ into the ...



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