Tag Info

New answers tagged

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 ...


2

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, $u(1-u^*u)=0$.


1

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 ...



Top 50 recent answers are included