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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.


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 ...


3

Suppose that $T$ is bounded. Then $$ \|x\|^{2}+\|Tx\|^{2}=((I+T^{\star}T)x,x) \le \|(I+T^{\star}T)x\|\|x\| \\ \|x\|^{2} \le \|(I+T^{\star}T)x\|\|x\| \\ \|x\| \le \|(I+T^{\star}T)x\|. $$ The last inequality can be use to show that the range of $I+T^{\star}T$ is closed, and $I+T^{\star}T$ has a bounded inverse on ...


2

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

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


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: ...


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 ...


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

The claim is false in the general infinite dimensional setting. For example, let $X$ be the Banach space $\ell^p$, 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.


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

Since $T^*T$ is Hermitian, the eigenvalues $\lambda_i$ are real. Suppose $v_i$ is an eigenvector corresponding to $\lambda_i$. Then \begin{eqnarray} \langle \lambda_i v_i, v_i\rangle =\langle T^*Tv_i, v_i\rangle= \|Tv_i\|^2\geq 0 \end{eqnarray} The eigenvalues $\lambda_i$ of $T^*T$ are real and nonnegative, and the eigenvalues of $I+T^*T$ are $1+\lambda_i$, ...


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 ...


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

I don't think there is any natural connection. Note that the $V$ in your Spectral Theorem is a unitary, while in Stinespring it can be any operator. Also, in the Spectral Theorem your map goes from a Hilbert space to a Hilbert space, while in Stinespring it maps a C$^*$-algebra into $L(H)$.



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