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## New answers tagged operator-theory

0

The equality implies that if $\lvert v\rvert\xi_1 = \lvert v\rvert\xi_2$, then also $v\xi_1 = v\xi_2$, so it is immaterial which element $\xi\in \lvert v\rvert^{-1}(\eta)$ is chosen to define $u(\eta) = v(\xi)$. All choices yield the same result. Thus $u = v\circ \lvert v\rvert^{-1}$ is well-defined although in general $\lvert v\rvert^{-1}$ is not a map ...

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Notice that $\|u(1-u^*u)\xi'\|^2=\|u^*u(1-u^*u)\xi'\|^2=\|(u^*u-u^*u)\xi'\|^2=0$, for every $\xi'$. Thus, $u(1-u^*u)=0$.

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WARNING. This is not a counterexample as it works only in real Hilbert spaces. See comments. The statement is false if $u$ and $v$ are not assumed to be symmetric. Consider the Hilbert space $\mathbb{R}^2$. The operators $$u\mathbf{x}=(-x_2,x_1)$$ and $$v\mathbf{x}=(-2x_2,2x_1)$$ are such that $$(u\mathbf{x}, \mathbf{x})=(v\mathbf{x}, ... 2 This is a corollary of the previous statement in the book: the polarization identity says that for a sesquilinear form \sigma, you can write$$4\sigma(x,y) = \sigma(x+y,x+y)+i\sigma(x+iy,x+iy) - \sigma(x-y,x-y) -i\sigma(x-iy,x-iy).$$In particular, this shows that given two sesquilinear forms \sigma, \sigma', if \sigma(v,v) = \sigma'(v,v) for all v, ... 1 There is a general method for finding sequences which satisfy linear recurrent relations with constant coefficients. In this case, we have the relation$$ x_{n+1}+x_{n-1} = \mu x_n \text{ for } n\geq 1. \text{ (*)} $$Here is how we solve it: consider all the sequences (not nesessarily from l^2) which satisfy (*). They form a linear space (it's easy to ... 1 (x,u^*\alpha y)=(ux,\alpha y)=\bar\alpha(x,u^*y)=(x,\alpha u^*y). 0 All these results are called "spectral theorems". They depend on what hypotheses you adjoin to the problem other than just being self-adjoint. The "nice" case is for compact self-adjoint operators. Here the statement of the spectral theorem is essentially the same as in the finite dimensional case: there is an orthonormal basis of the space made up of ... 0 The correct rigorous version of this is the Spectral Theorem. I don't know which "certain set of conditions" the professor had in mind. In general self-adjoint operators on an infinite-dimensional Hilbert space might have continuous spectrum, in which case there are no eigenfunctions at all (but a physicist might speak of "generalized eigenfunctions" ... -2 Here is a short answer. Integral are sums (Riemann sums) and derivatives are differences (to the limit). Sums and differences are commutative. 7 The zero representation of an algebra on a Hilbert space H is the map that sends every element of the algebra to the zero operator on H. Murphy's book gives the following definitions: If A is a C*-subalgebra of B(H), it is said to be irreducible, or to act irreducibly on H, if the only closed vector subspaces of H that are invariant for ... 0 How do you invert the map (a_1,a_2,\cdots) \rightarrow (a_1,\,a_1/2,\,a_3/3,\,\cdots)? Let b_n=(a_1,\,a_2/2,\,a_3/3,\cdots,\,a_k/k,\,\cdots) , how do you get the original a_n back? Now, to see if T^{-1}:= (a_1,\,2a_2,\,\cdots) is bounded, you need to know what the norm is that is used in l^{\infty} : It is the sup norm \|a_n\|:=\sup_n |a_n|  ... 0 The range of this operator is a subspace of C_{0}, which consisting of elements eventually go to zero. It can be characterized by$$ \{a_{i}\}\in l^{\infty}, \exists N\in \mathbb{N}, |a_{i}*i|\le C, \forall i\ge N $$It is not difficult to see that the inverse map$$ \{a_{i}\}\rightarrow \{ia_{i}\} $$is not bounded on the sequence a_{i}=1,\forall i ... 0 Hahaaa, I got it thanks to my supervisor. =D All formal calculations... Consider the abstract cauchy problem:$$\frac{\mathrm{d}}{\mathrm{d}t}\left(\tau^t\tau_0^{-t}\right)(Z)=\tau^t(\delta-\delta_0)\tau_0^{-t}(Z)=\tau^t\imath\left[V,\tau_0^{-t}(Z)\right]=\left(\tau^t\tau_0^{-t}\right)\imath\left[\tau_0^{t}(V),Z\right]$$so one gets the recursion relation: ... 5 By using the polar decomposition, we can write T=V|T|. So |T|=V^*T\in J, and then J contains a positive non-compact operator. On a side note, this argument also shows that J contains all adjoints of its operators, since now T^*=|T|V^*\in J. So from now on we assume T\geq0, non-compact, T\in J. This means that there is \lambda>0 with ... 1 I think your argument is fine. I fail to see why Murphy feels the need to use approximate units in this argument. 2 Let e_n be the element of \ell_\infty whose m'th coordinate is 1 if m=n and 0 otherwise. The closed linear span of \{e_n\mid n\in \Bbb N\} in \ell_\infty is the space c_0 of sequences that tend to 0. For your operator, we have T(je_j)=e_j; so the range of T contains each e_j. Since the range of a linear operator is a linear space, ... 1 Your argument is not correct. You say that \|\phi (a^*a)\|=\|\phi (a)^*\phi (a)\| implies that \phi (a^*a)=\phi (a)^*\phi (a) , which makes no sense. Also, without the unital condition the statement is trivially false: take \phi (x)=-x . Now, here is an argument using all conditions. Note that  \phi maps selfadjoints to selfadjoints. For a ... 1 Obviously, the norm is at least what you anticipate by setting f = e_n where n almost realizes the supremum of |\lambda_n - \lambda|^{-1}. Conversely, write f as f = \sum_n c_n e_n. Then$$\|(K - \lambda I)^{-1} f\|^2 = \sum_n |c_n|^2 |\lambda_n - \lambda|^{-2} \leq (\sup_n |\lambda_n - \lambda|^{-2}) (\sum_n |c_n|^2).$$Taking the square root ... 2 The space C_0 is a Banach space. This implies that absolutely convergent sequences are convergent, i.e. if$$\sum_n \Vert \frac{p_n}{3^n} \Vert_\infty$$is finite, then \sum_n \frac{p_n}{3^n} \in C_0. But projections have norm at most one, which means$$\sum_n \Vert \frac{p_n}{3^n} \Vert \leq \sum_n 3^{-n} = \frac{1}{1-1/3} < \infty.$$This ... 1 I'm going to build on another problem of yours, Tobias: Why is this operator self-adoint . In the above problem, it is shown that, if A : \mathcal{D}(A)\subseteq H\rightarrow H is symmetric with (A\pm iI) surjective, then A is densely-defined and selfadjoint. As you noted, your operator O is symmetric on its domain. To see that (O\pm iI) are ... 1 You don't need to show x\in \ell^2; that is a part of the set-up. The equality y=Tx can be checked component-wise: you need to check that y_n=(Tx)_n for every n. That is, y_n=nx_n. Here are the two steps leading to the goal:$$x_n =\lim_{m\to\infty} x^{(m)}_n\tag{1}y_n =\lim_{m\to\infty} (Tx^{(m)})_n\tag{2}$$-1 Ok, I think I got it now... Both work perfectly fine as they are always nondegenerate:$$\mathcal{A}_\text{CAR}:\quad a(f\neq0)\neq0\mathcal{W}:\quad W(f)\neq0$$A counterexample is provided by the angular momentum algebra:$$\mathcal{J}:\quad [J_i,J_j]=\imath\varepsilon_{ijk}J_k$$There one has one trivial representation: J_x=J_y=J_z=0 0 Essentially, you need to check that \forall f,g\in dom(G)\cap dom(T) you have$$(f,T[g]/\phi) = (f,T[g/\phi]),$$where (\cdot,\cdot) is a scalar product in L^2 and \phi is the function \frac{1}{1-x^2}. I don't quite see how it could be possible for generic T. 0 I do not think it is self-adjoint in general. If you define the operator \mathcal{O} by \mathcal{O}f(x) = \dfrac{f(x)}{1-x^2}, then G = T\mathcal{O}. \mathcal{O} is a self-adjoint operator which you can see pretty easily. It is a general result that if A,B are self-adjoint, AB is self-adjoint if and only if A and B commute. Hence G = ... 0 Ok, I think I finally got it... (However, if anybody finds bugs, typos, loopholes etc. then please let me know. Thanks!) Framework Given the natural numbers \Omega:=\mathbb{N} and the Hilbert space \mathcal{H}:=\ell^2(\mathbb{N}). Choose the canonical basis by: e_n:=\chi_n Consider the spectral measure E(\{n\}):=P_n. Operator Domain Regard the ... 1 Take the operator e^H, it's defined as$$e^H = \sum_{k=0}^{\infty}\frac{H^k}{k!} = \sum_{k \text{ even}}\frac{H^k}{k!} + \sum_{k \text{ odd}}\frac{H^k}{k!} \; .$$As you pointed out H^3=-a^2H, therefore$$H^{2k+1}=(-a^2)^kH \text{ for } k \in \mathbb{N} \text{ and } H^{2k}=(-a^2)^{k-1} H^2 \text{ for } k \in \mathbb{N}_0 \; . $$Can you figure out ... 1 If H is bounded and selfadjoint with spectrum \sigma, then$$ (\lambda I-H)^{-1}=\int_{\sigma}\frac{1}{\lambda-\mu}dE(\mu),\;\;\; \lambda\notin\sigma. $$Suppose \Gamma is a positively-oriented simple closed rectifiable curve in \mathbb{C} which contains \sigma in its interior, and suppose that f is holomorphic on an open neighborhood ... 0 Define A:H^s_0 \to (H^s_0)^* by \langle Au, v \rangle = (u,v)_{H^s_0} and consider the equation \langle Au, v \rangle = (f,v){L^2}. u exists uniquely, and the solution map T(f) = u, T:L^2 \to L^2 is compact. Then one can apply the Hilbert-Schmidt theorem. 1 Yes. By induction, T^n(f)(x) = \frac{1}{n!x} \int_0^x \log^n\left(\frac{x}{u}\right) f(u)\, du for 0 \le x < 1. The formula can also be written$$T^n(f)(x) = \frac{1}{n!}\int_0^1 \log^n\left(\frac{1}{u}\right) f(ux)\, du.$$If f = 1, then T^n(f) = 1 for all n and thus \|T^n(f) - f(0)\|_\infty = 0 for all n. When f(x) = x^m for some m \ge ... 0 That's a comment but it got a bit too long. As far as I know, classically, the approximation property deals with bounded linear operators. But let us assume your point for a minute. If it is not "bounded linear" then what is you definition of "bounded" operator? E.g., 1) a bounded function, that is a function with bounded range. But a bounded linear ... 1 The graph G(A) of A is a closed linear subspace of \mathcal H ^2. Any closed linear subspace of that is the graph of a closed restriction of A. So you want to take any closed linear subspace S of G(A) such that P(S) is dense in \mathcal H, where P(u,v) = u is the "first coordinate" projection of \mathcal H^2 on \mathcal H. 1 No, it does not. A continuous nonlinear operator can take a bounded set to an unbounded one. I gave an example here, which is also a homeomorphism. Since you don't require A to be a homeomorphism, here's a simpler version: Let \mathcal H be a Hilbert space with orthonormal basis \{e_i : i\in \mathbb N\}. Define a nonlinear map A:\mathcal ... 1 The spectral theorem for an unbounded selfadjoint operator allows you to represent T as$$ Tx = \int_{0}^{\infty}\lambda dE(\lambda)x,\\ \mathcal{D}(T) = \left\{ x \in H : \int_{0}^{\infty}\lambda^{2}d\|E(\lambda)x\|^{2} < \infty \right\}. $$The unique positive square root of T is$$ \sqrt{T}y = ...

1

$\text{dom}(A) = \text{dom}(T)$ is wrong. By the Spectral Theorem, you can essentially assume $T$ is multiplication by the variable $x$ on $L^2(\mu)$ for some positive measure $\mu$ on $[0,\infty)$, with $\text{dom}(T) = \{f \in L^2(\mu): x f \in L^2(\mu)\}$. Then you want $A$ to be multiplication by $\sqrt{x}$, with $\text{dom}(A) = \{f \in L^2(\mu): ... 0 Thinking of$\Bbb{Z}$as a measure space with atomic measure for every integer$\mu(\{n\})=1$,$\forall n\in\Bbb{Z}$, then$l^2(\Bbb{Z})$is just the space of square integrable functions$L^2_\mu(\Bbb{Z})$. The multiplication operator$M_\varphi$, where$\varphi:E\to\Bbb{R}$is a measurable function with a Borel set$E$as domain, is self-adjoint on the ... 1 The unitary shift operator on$l^{2}(\mathbb{Z})$is not unilateral. This operator is the same as multiplication by$e^{i\theta}$on$L^{2}[0,2\pi]$with normalized inner product $$(f,g) = \frac{1}{2\pi}\int_{0}^{2\pi}f(\theta)\overline{g(\theta)}\,d\theta.$$ You can see this because every$f \in L^{2}[0,2\pi]$can be written as the ... 1 If$\text{Ran}(\lambda I - T)$is not dense in$H$, take$v$in its orthogonal complement. Thus for all$x \in H$,$ 0 = \langle (\lambda I - T) x, v \rangle = \langle x, (\overline{\lambda} I - T^*) v \rangle$, and this implies$(\overline{\lambda} - T^*) v = 0$. 1 Yes, this is Goldstine's theorem which asserts that$\iota(V)$is weakly* dense in$V^{**}$. Of course,$\iota(V)$is norm-closed in$V^{**}$as$\iota$is an isometry. 1 Recall for matrices$C,D$that$\sigma(CD)=\sigma(DC)$, where$\sigma(F)$denotes the spectrum of a matrix$F$. Assuming that both$CD$and$DC$are normal, we conclude in particular$\|CD\|=\| DC\|$. Since$A-B$is self-adjoint, we may assume that it is diagonal,$A-B=\mbox{diag }(\lambda_1,\ldots,\lambda_n)$, with$\lambda_1\leq\ldots\leq \lambda_n$. Let ... 1 on p.475 Kolmogorov-Fomin says: "$\dots0$is the only possible accumulation point for the sequence$\{\mu_n\}\dots$" Kolmogorov-Fomin does not exclude the possibility that$\sigma(A)$is finite and there is no accumulation point for$\sigma(A).$1 Here are some references: Reed, Simon, "Methods of modern mathematical physics. II. Fourier analysis, self-adjointness." Schmüdgen, Konrad "Unbounded self-adjoint operators on Hilbert space." What kind of properties do you expect from analytical vectors? If the restriction of a self-adjoint operator$(A,D(A))$to the space of its analytical vectors ... 1 This long answer is based on many little observations; at every step I will use the preceding ones, sometimes implicitly. Tell me if some step is not clear. Call$P_1:=P_{C_1}$and$P_2:=P_{C_2}$. You can assume that$C_1$is compact (if the compact set is$C_2$, with the following argument you will get that$P_2 P^n(x)$converges to some$\ell$fixed by ... 1 One possibility is the following: Decompose$K =K_+ - K_-$and$\phi =\phi_+ -\phi_-$, where$K_\pm$are the positive and negative parts of$K$. Then your integrand is $$(K_+ - K_-)\cdot (\phi_+-\phi_-)= K_+ \phi_+ + K_- \phi_- -(K_+ \phi_- + K_- \phi_+).$$ Now each of the contents of the two brackets is a measurable, nonnegative function. Fubini's ... 2 The situation is the following: you have$U=\lambda I+T$, with$T$compact. And$\{P_M\}$is a sequence of finite-rank projections such that$P_M\nearrow I$. So, given$\varepsilon>0$, by the compactness of$T$you can write$T=P_MTP_M+T_0$, with$\|T_0\|<\varepsilon$. Then $$\|P_MU-UP_M\|=\|P_MT_0-T_0P_M\|<2\varepsilon.$$ For your second ... 1 For a regular Sturm-Liouville eigenvalue problem on a finite interval$[a,b]$, say $$Lf = \left[-\frac{d}{dx}p\frac{d}{dx}+q\right]f = \lambda f,$$ there are two types of standard endpoint conditions: Separated conditions $$\cos\alpha f(a)+\sin\alpha f'(a) = 0 \\ \cos\beta f(b) + \sin\beta f'(b) = 0.$$ The linear ... 1 It means that$P$is the "strong limit" of the$T_n$, i.e., the limit of the$T_n$in the strong topology (Wikipedia link). A more common way of writing it is $$\operatorname{s-lim} T_n$$ You can Google search on the string strong limit s lim for examples. 1 From comments, it sounds like by "Euclidean space" you mean "Hilbert space". Let's call it$H$. It is certainly true that for any bounded linear operator$A : H \to H$, the adjoint map exists. It's really the same as for Banach spaces: define the linear operator$A^*$on$H^*$via$A^* f = f \circ A$, and then use the Riesz representation theorem to ... 1 You should note what norm you are using (the matrix-$2$-norm$\|U\|_2$). Practically this is correct, though because it shows $$\| Ux \|_2 = \|x\|_2 \Rightarrow \frac{\|U x\|_2}{\|x\|_2} = 1 \Rightarrow \|U\|_2 = \sup_{x\ne 0} \frac{\|U x\|_2}{\|x\|} = 1$$ 0 I don't have a concrete example in my head right now. But here is the fact: the strong limits of normal operators are precisely the subnormal operators. 2 This is not true in general, e.g. let$a_1=\begin{pmatrix}\sqrt{2}/2&-\sqrt{2}/2\\\sqrt{2}/2&\sqrt{2}/2\end{pmatrix}$and$a_2=a_1^*$in$M_2(\mathbb{C})$. Since$a_1$is unitary, then$A=C^*(1,a_1,a_2)=C^*(a_1)$. But$a_1$has two eigenvalues (so$a_2$also has two eigenvalues), thus$\Omega(A)=\sigma(a_1)\$ has two elements, but ...

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