# Tag Info

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 $c \neq 0$)

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

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

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.

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 ...

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

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 $A^{-1}=A_0^{-1}(I-(A_0-A)A_0^{-1})^{-1}$.

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

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 $$w^*w=(v^*+q-p)(v+q-p)=v^*v+q-p=r+q-p=s,$$ $$ww^*=vv^*+q-p=p+q-p=q. ... 1 Yes. The functional calculus preserves approximation by polynomials. So, from$$ 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} $$for any polynomial p. Now using a sequence of polynomials that converges uniformly to f, you get that$$ ...

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 $T = \begin{bmatrix} a & 0 ... 1 According to wikipedia, an operator is a function whose domain and codomain are both vector spaces or modules. Since$\mathbb{R}, \mathbb{Q}, \mathbb{C}$are all (one-dimensional) vector spaces, many familiar functions are also operators. However, a general function might be from a domain that is not a vector space, and hence not be an operator, e.g. ... 1 Unilateral shifts (including weighted shifts) are never normal. 1 First for the eigenvalues: By the spectral theorem (and since$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 ...

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