# Tag Info

1

I'll assume that you meant for $\{ x_{n}\}$ to be a sequence of unit vectors. Otherwise, you could let $x_{n}=0$ for all $n$, and there would exist such a sequence regardless of whether or not $i\lambda \in\sigma(T)$. Note that there is nothing gained by introducing $i$ into the discussion, especially because you made no assumption about $\lambda$ being ...

0

To paraphrase a famous slogan, "all mathematics is operator theory". This is even more true of applicable mathematics. I would guess that eigenvalue and spectral calculations account for a significant fraction of the world's CPU time spent on computation.

0

If you describe the behaviour of an electron in the hydrogren atom with the help of a Hamiltoninan Operator, then you end up with a differential operator (first in $L^2(\mathbb{R^3})$, but you can use some symmetry properties to end up with a Sturm-Liouville operator in $L^2(\mathbb{R^+})$). This hamiltonian operator has a spectrum which is bounded from ...

1

Your reference is talking about matrices. Which spectral mapping theorem are you interested in? For example, here's one version (see e.g. Rudin, "Functional Analysis", theorems 10.28 and 10.33): Suppose $T$ is a bounded linear operator on the complex Banach space $X$, $\Omega \subseteq \mathbb C$ open with $\sigma(T) \subset \Omega$, and $f$ analytic in ...

0

Let $T$ and $S$ be as stated, and let $\{ \lambda_{j}\}$ be the distinct eigenvalues of $T$. Let $P_{j}$ be the orthogonal projection of $X$ on $\mathcal{N}(T-\lambda_{j}I)$. Then $Tx=\sum_{j=1}^{\infty}\lambda_{j}P_{j}x$ and $Sx=\sum_{j=1}^{\infty}\sqrt{\lambda_{j}}P_{j}x$. Suppose $L\in\mathscr{L}(X)$ satisfies $0 \le L=L^{\star}$ and $L^{2}=T$. As you ...

2

If you do not need any control over the norm of the vector $\alpha$, then, yes, such a vector exists. Take any vector with norm less than $\min (\frac{\varepsilon}{2}, \frac{ \varepsilon}{ 2 \Vert T \Vert})$. If you add the assumption that $\Vert \alpha \Vert=1$, such a vector $\alpha$ need not exist. Consider the operator $T(x)=-x$. Its restriction to any ...

1

This result is true for any bounded operator with $\|T\|\leq1$. Fix $x\in X$ with $\|x\|\leq1$. Put $$C=\tfrac12\,\overline{\text{conv}\,\{T^nx:\ n\in\mathbb N\}}\subsetneq B_X$$ (every element in $C$ has norm at most $1/2$). This is the most general possible result: let $T=(1+\delta)I$ for some $\delta>0$. A subset $C$ as desired satisfies ...

2

If your underlying space $X$ is a Banach space, the answer is "Yes". To prove this, assume that $X$ is infinite-dimensional, and let us show that there exists a compact operator on $X$ whose range is infinite-dimensional. It is a standard fact that since $\dim(X)=\infty$, one can find a biorthogonal sequence $(e_n,e_n^*)_{n\in\mathbb N}$ in $X\times X^*$, ...

0

You could think of the following sequence of operators(it converges strongly to zero but the sequences of adjoints does not) take Sn to be the n-th power of the left shift.

0

You can write $S = \mathcal F^{-1} \circ \mathcal M \circ \mathcal F$, where $\mathcal F$ is the Fourier transform, and $\mathcal M$ is multiplication by $m_1$. Go from there.

2

1) $\|e^{itB}\|=1$ if $t\in\mathbb R$ and $B=B^*$. This is simply because $e^{itB}$ is a unitary. 2) The expression $bU-aV$ is the real part of $\lambda T$: that is, $2(bU-aV)=\lambda T+(\lambda T)^*$. So it is selfadjoint, and $e^{2i(bU-aV)}$ is a unitary as in part 1. 3) Note that up to here you haven't used that $AT=TA$. This forces $AT^n=T^nA$ for all ...

0

Let us show that if $Y$ is infinite-dimensional, then multiplication is not jointly continuous on $\mathcal L(X,Y)\times \mathcal L(Y,Z)$ with respect to the strong operator topology. Choose any non-zero $x_0\in X$. It is enough to show that the set $\mathcal M:=\{ (A,B);\; \Vert BAx_0\Vert <1\}$ is not an $SOT\times SOT$-neighbourhood of $(0,0)$ in ...

0

A compact operator has a closed range iff it has a finite dimensional range. Without loss of generality,we can assume that the range of A is closed,otherwise we can consider the restriction of A to E where E=T−1(F),In all cases the theorem assures that F is finite dimensional. to prove the theorem,consider the canonical map associated with A and the fact ...

0

In the following paper, an example is given of a bounded operator $T$ on a separable Hilbert space $H$, such that the range of $T$ is closed but the range of $T^2$ is not closed. Barnes, Bruce A. Restrictions of bounded linear operators: Closed range. Proc. Amer. Math. Soc. 135 (2007), 1735-1740. Open access.

0

You have shown that $P=T^{\star}T$ satisfies $P^{2}=I$. Let $x \in X$ be given, and define $y=x-Px$. Notice that $Py=Px-P^{2}x=Px-x=-y$. Therefore, $$0 \ge -\|y\|^{2}=(Py,y)=(T^{\star}Ty,y)=(Ty,Ty)=\|Ty\|^{2} \ge 0.$$ So $y=0$, which implies $x=Px$. This is true for all $x$. Thus $P=I$, which means $T^{\star}T=I$. So $T^{2}=I$ becomes ...

1

So since $f$ is non-zero, $S$ is not zero That's correct, but and hence we must have $E(w)$ is the identity. isn't. All that follows from $S = E(w)S$ and $S\neq 0$ is that $E(w)$ is the identity on the range of $S$, but $\mathcal{R}(S)$ need not be the full space (and isn't, generally). $E(w)$ is a projection, so $S = E(w)S$ is equivalent to ...

1

We can view the random operator as a map from $H$ to the space of random variables, $$A \colon H\rightarrow (\Omega\rightarrow H)$$ but also as an operator-valued random variable, $$X_A \colon \Omega\rightarrow (H\rightarrow H)$$ by letting $$X_A(\omega)(x)=A(x)(\omega).$$ Thus, given $\omega\in\Omega$, $X_A(\omega)$ is an operator on $H$ and hence, I ...

2

I see no indication that $M$ and $M'$ are supposed nontrivial invariant subspaces. But by part $(d)$ if theorem 12.22, If $\omega \subset \Delta$ is open and nonempty, then $E(\omega) \neq 0$ and the monotonicity, whenever the spectrum contains more than one point, there are partitions of $\Delta$ into disjoint Borel sets $\omega$ and $\omega'$ such ...

1

By the Lebesgue differentiation theorem, $$(Df)^\prime(t)=\frac{d}{dt} \int_0^t f(s)ds=f(t)$$ for a.e. $t\in[0,1]$. Therefore, $$\langle Df, Dg\rangle_{C^\prime}=\int_0^1 (Df)^\prime(t) (Dg)^\prime(t) dt = \int_0^1 f(t) g(t) dt = \langle f,g\rangle_{L^2[0,1]}$$ so $D$ preserves the inner product. By the characterization of absolutely continuous functions by ...

1

This is a typo. The second $x$ should be a different letter. Since $x\in M=\mathrm{im}\,E(\omega)$, there is a $y\in H$ such that $E(\omega)y=x$. Therefore $$Tx=TE(\omega)y=E(\omega)Ty$$ so $Tx\in M$. It is possible that $E(\omega)=0$, but the point is that there exists $\omega$ such that $E(\omega)\not=0$, because otherwise $E(\omega)=0$ for all Borel sets ...

4

Let $F\subset T(X)$ be a closed (in $Y$) subspace. Then $E = T^{-1}(F)$ is a closed subspace of $X$, and $T\lvert_E \colon E \to F$ is a compact surjective operator. Since $F$ is closed in $Y$, the open mapping theorem implies that $F$ is finite-dimensional. So: the image of a compact operator cannot contain an infinite-dimensional Banach space.

1

I don't know if I correctly understood your question because I don't own Rudin's book. However, if you understood why $S$ commutes with every spectral projection $E(W)$, then the fact that $S$ commutes with $f(T)$ follows from the following reason. Every $f\in L^\infty$ is the norm limit of finite linear combination of simple function (i.e. \sum_{i=1}^n a_i ... 0 First write $$Mv=\lambda v \qquad \text{and} \qquad Mv^*=\lambda' v^*$$ now complex conjugating the second equation and noting M is real $$(Mv^*=\lambda' v^*)^* \rightarrow M^*v=\lambda'^*v \rightarrow Mv=\lambda'^*v$$ and finally since the eigenvalue of a given eigenvector is unique you have that \lambda = \lambda'^* \quad \text{or}\quad \lambda^* ... 0 The matrix \pmatrix{1&1\\-1&1} should be one example of what you're looking for. 2 It suffices to show that \|(T-T^*)x\|=0, for all x\in H. We have \begin{align} \|(T-T^*)x\|^2&= \langle (T-T^*)x,(T-T^*)x\rangle \\& =\langle Tx,(T-T^*)x\rangle-\langle T^*x,(T-T^*)x\rangle\\&= \langle x,T^*(T-T^*)x\rangle-\langle x,T(T-T^*)x\rangle \\&=\langle x,T^*Tx-x\rangle- \langle x,x-TT^*x\rangle=2\|Tx\|^2-2\|x\|^2. \end{align} ... 1 Note that V^*V is a positive element with \|V^*V\|=1. This means that \sigma(V^*V)\subset[0,1]. If you consider the function f(x)=x(1+h(x)). If h is chosen appropriately, then |f(x)|\leq1 for all x (because the term 1+h(x) is positive only on the small neighbourhood of t). Now functional calculus gives you ... 1 The first problem you get into is the issue of the domain on which A and B are defined. Normally there is some large common domain for A and B, but AB or BA may not be defined on the same domain, which makes comparisons of AB and BA hard. A simple one-dimensional problem illustrates the complexity. For example, consider the one-dimensional ... 0 What are the hypothesis on K? Observe that a strongly exposed point is a denting point and a denting point is an extreme point. So, for instance take the set K as the unit ball of c_0. This set is convex and closed but it has no extreme point. Hence, K can not be the closed convex hull of any kind of extreme points of K. Items 1 and 2 can be ... 0 It's hard to know what you can assume at this point. Do you know at this point that \sigma(x^{\star}x)\subseteq [0,\infty)? Then you should be able to show what you want by looking at r_{A}=\inf\sigma_{A}(x^{\star}x) and comparing this to r_{B}=\inf\sigma_{B}(x^{\star}x) in order to conclude that x^{\star}x is invertible in B. That gives you a left ... 0 The spectral theorem is a general argument. For a specific normal operators, there may be a concrete "diagonalizing" unitary. Take the shift S on l^2(\mathbb{Z}). In this concrete example, the diagonalizing unitary is the Fourier transform \mathcal{F} : l^2(\mathbb{Z}) \rightarrow L^2(\mathbb{T}). $$The operator$$ \mathcal{F} S\mathcal{F}^{-1} ... 3 LetH=L^{2}[0,\pi]$. The subspace$D=\mathcal{C}^{\infty}_{c}(0,2\pi)$consisting of infinitely-differentiable functions on$(0,2\pi)$which are compactly supported in$(0,2\pi)$is a dense subspace of$H$. Let$\mathcal{D}(A$) be the domain of twice absolutely continuous functions on$[0,2\pi]$for which $$f'\in H, f'' \in H, f(0)=f(\pi)=0. ... 3 Consider H=L^2(\Omega,\mu) where \Omega\subseteq\mathbb C is compact, and multiplication operator T:H\to H,\ (Tf)(z)=zf(z). Then T is bounded and normal. By the spectral theorem T=\int_{\mathbb C}\lambda dE(\lambda). In this case you can give an explicit formula for the spectral measure E:$$(E(X)f)(z)=\chi_X(z)f(z),$$where \chi_X is the ... 1 Indeed we also have T_F \to T in the strong operator topology. The strong operator topology is defined by the seminorms$$p_F(S) = \sup \{\lVert S(x)\rVert : x \in F\},$$where F traverses the finite subsets of X. The construction immediately yields$$p_F(T_F - T) = 0$$for any linearly independent finite F\subset X, and it is straightforward to ... 0 Let V=T+S. To show that V is closed, we must verify that$$x_n\to x \ \text{ and } \ Vx_n \to y \implies y=Vx \tag{1}$$Since S is bounded, Sx_n\to Sx. Therefore, Tx_n = Vx_n-Sx_n\to y-Sx. Since T is closed, y-Sx = Tx. Thus y=Vx as required. 1 I wouldn't really call the idea of using rank-one projections a "trick". Here is what I consider an even more straightforward approach, which uses the same idea in a more explicit way: Let \theta:B(H)\to B(H) be a *-automorphism. Fix an orthonormal basis \{\xi_j\} of H, and write E_{jj} for the corresponding rank-one projections, i.e. ... 1 A linear operator is injective if and only if it has a left-inverse, and it is surjective if and only if it has a right-inverse. To show that the uniqueness of the one-sided inverse implies the invertibility of A, show that if A is not invertible, but has either a left- or a right-inverse, the one-sided inverse is not unique. Say A has a ... 2 Your proof is fine, though it uses more machinery than necessary (in particular, you need not appeal to sequences): The norm-closed unit ball of a reflexive normed space is weakly compact. This follows from Alaoglu's theorem and the fact that the natural embedding from such a space onto its second dual is a weak-weak* homeomorphism. Now you need only use ... 1 Let H^{2}(D) be the Hardy space of holomorphic functions on the unit disk D. That is, f \in H^{2}(D) iff f = \sum_{n=0}^{\infty}f_{n}z^{n} has a power series expansion in D with \|f\|^{2}=\sum_{n=0}^{\infty}|f_{n}|^{2} < \infty. The shift operator S defined by (Sf)(z)=zf(z) is an isometry, i.e., \|Sf\|=\|f\|; and the spectrum \sigma(S) ... 1 There's a classic trick to produce the unitary. I was told once who this is due to, but I've unfortunately forgotten. Recall that, if \xi,\eta \in H, then \eta \otimes \overline \xi denotes the rank-1 operator which sends \zeta \mapsto \langle \xi,\zeta \rangle \eta. Here, the inner-product is conjugate linear in the 1st slot. It's easy to check that ... 1 Your question raises the point of what is meant by an "isomorphism of von Neumann algebras". Any C^*-isomorphism will of course preserve order, and then it will be normal: if \varphi:A\to B is a *-isomorphism and \{a_j\}\subset A^{\rm sa} is a bounded increasing net with least upper bound a, then \varphi(a) is the least upper bound for ... 4 Yes, we could also define the strong resp. weak operator topologies as the initial topology with respect to the evaluation maps p_x, where Y is endowed with the strong resp. weak topology. For the weak operator topology that is obvious from the transitivity of initial topologies, since the weak topology is just the initial topology with respect to the ... 0 If T_gh=0, then g=0 on [1/2,1]; any polynomial that is zero on an interval is equally zero, so g=0. That is, h is separating. The commutant of A agrees with the commutant of the weak closure of A. This is the set of multiplication operators by all essentially bounded functions (i.e., L^\infty[0,1]). In particular T_k, with k the ... 1 This has nothing to do with topology nor von Neumann algebras. Any invertible idempotent in a ring is equal to the identity: Let P be an invertible idempotent. Then there exists X with XP=PX=I. So I=XP=XPP=IP=P. 3 All separable Hilbert spaces are isomorphic, so it doesn't really matter which Hilbert space you use. That said, you might as well use one that is convenient: in this case, one that is connected in some way to the closed subset. That's one hint. Another hint: multiplication operators are normal. 3 Hint Multiply this equality on the right by A and on the left by B. 1 It is true if you assume that T is self-adjoint (i.e. symmetric), meaning that$$ (Tx,y)=(x,Ty), \quad \text{for all}\,\, x,y\in H, \tag{1} $$and assuming that$$ |(Tx,x)|\le \|x\|^2, \quad \text{for all}\,\, x\in H.\tag{2} $$Note that your inequality holds even for T=-2I, and thus we NEED to assume these two additional things: (1) and (2). So ... 0 We know that$$ 0=T^k-I=\prod_{j=0}^{k-1} (T-\omega^j I), $$where \omega=\exp(2\pi i/k). But as T is positive semidefinite, all its eigenvalues are real and non-negative, and this implies that T-\omega I is non-singular, where \omega\ne 1 is a kth root of 1. In particular$$ S=\prod_{j=1}^{k-1} (T-\omega^j I) $$is non-singular, and hence$$ ... 2 Solution: The problem appears to me to be false. Let$B$be the Banach algebra of continuous functions on the annulus$\mathscr{A}=\{ z \in \mathbb{C} : 1/2 \le |z| \le 1\}$with$\|b\|=\sup_{z \in \mathscr{A}}|b(z)|$. Let$A$be the subalgebra of$\mathscr{A}$generated as the closure in$B$of polynomials in$z$. Let$b(z)=z$. Then$\sigma_{B}(b) = ...

0

No. A homomorphism should map $0$ to $0.$ But $p(T)=0$ does not imply $p=0.$ Take e.g. $T=T^*$ with finite spectrum $\{\lambda_1,\dots,\lambda_n\}$ and $p(x)=(x-\lambda_1)\dots(x-\lambda_n).$ Note, that $p(x)\mapsto p(T)$ is a homomorphism $\mathbb C[x]\to B(H).$

4

Via a faithful state, we can think of $A$ as represented in some $B(H)$. We have $$H=\ker T\oplus \overline{\text{ran} T^*}.$$ For $x\in\ker T$, we have $Tx=0$ and then $$\|T^*x\|^2=\langle T^*x,T^*x\rangle=\langle TT^*x,x\rangle=\langle T^2x,x\rangle=0;$$ so $T^*x=0$ and $T=T^*$ on $\ker T$. Taking adjoints on $TT^*=T^2$ we have $TT^*=T^*T^*$, that ...

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