# Tag Info

1

A counter example is the Volterra operator, it is a bounded linear operator between Hilbert spaces, with no eigenvalues while the spectrum is $\{0\}$. It is defined as $$V:L^2(0,1)\to L^2(0,1), f\mapsto \left(t\mapsto \int_0^t f(x)\mathrm{d}x\right).$$

1

Normally the singular values of a matrix $A$ are defined as the (positive) square roots of the eigenvalues of $A^*A$.

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You need a small variation of your idea. First, we may assume without loss of generality that $\|T_1\|\leq1$, $\|T_2\|\leq1$. Now you take $n_0$ such that $\|T_1^{n}\| \leq (r(T_1)+\epsilon )^{n}$ and $\|T_2^n\|\leq(r(T_2)+\epsilon)^n$ for all $n\geq n_0$. Also, since $\|T_1^k\|^{1/k}\to r(T_1)$ and $\|T_2^k\|^{1/k}\to r(T_2)$, there exist $c_1,c_2>0$ ...

2

Any eigenvalue of any operator has an entire vector space of eigenfunctions. So $au_k$ is also a eigenfunction of $\lambda_k$ for any $a\ne0$. This gives him the freedom to choose his eigenfunctions to be normalized. If $v_k$ is an arbitrary eigenfunction of $\lambda_k$, then he can define $$a = \left(\int_{-\infty}^{\infty}v_k^2(x)dx\right)^{-1/2}$$ and ...

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The book by Gerald B. Folland, "a course in abstract harmonic analysis" is the best one. For spectral theory you can read part of " a course in functional analysis" written by J.B. Conway. Also you can read books on C*-algebras and operator theory by Murphy or Kadison.

2

I don't fully understand the question: if you agree that $\left\{\frac{1}{\sqrt{2\pi}}, \frac{1}{\sqrt{\pi}}\cos kx, \frac{1}{\sqrt{\pi}}\sin kx\right\}$ are a basis for the function space of $2\pi$-periodic functions, it follows immediately that any function in this space can be expressed as a linear combination of these basis functions (and the $a_i$ and ...

1

Any solution of $-g''=\lambda g$ is a linear combination of $\cos\sqrt\lambda x$ and $\sin\sqrt\lambda x$. Because you want your eigenfunctions to be functions on the circle, you need the period to be an integer: $$\cos\sqrt\lambda (x+2\pi)=\cos\sqrt\lambda x, \ \ \ \ \ \ \sin\sqrt\lambda (x+2\pi)=\sin\sqrt\lambda x.$$ For this you need ...

1

First, some comments about notation and normalization: If the wave kernel is given by $$W(t,x,y)= \sum_{k \geq 1} e^{it \sqrt{- \lambda_k}} \mu_k(x) \mu_k(y)$$ and the wave trace satisfies $$w(t)=\sum_{k \geq 1} e^{it \sqrt{- \lambda_k}}=\int_{-\infty}^{\infty}W(t,x,x)dx$$ then we must have $\int_{-\infty}^\infty \mu_k(x)^2\,dx=1$. This is natural for the ...

2

I think the kernel is \begin{align*} W\left(t,x,y\right) & =\sum_{n\geq1}e^{-tn}e^{in\left(x-y\right)}\\ & =\frac{1}{e^{\left(t-i\left(x-y\right)\right)}-1},\quad t>0. \end{align*} Looking at pg 25 of the linked pdf, I think the following makes more sense: \begin{align*} W\left(t,x,y\right) & ...

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It seems as if you confused what to assume and what to prove in the second part. If you want to show the existence of $\pi$, you cannot include it in the definition of $H$. But if you get your arguments sorted out, it's quite obvious. Let $\pi=G_2^{-1}\circ G_1^{-1}$ and $f\colon \sigma(S)=\sigma(T)\to \mathbb{C},\,z\mapsto z$. By definition, $G_1(f)=S$ and ...

1

Suppose you have a basis $V_W$ for $W$ and a basis $V_{W^{\perp}}$ for $W^{\perp}$. Using Gram-Schmidt's method, you find a orthonormal basis $V^{'}_W$ for $W$ and a orthonormal basis $V^{'}_{W^{\perp}}$ for $W^{\perp}$. Now, you know that $W \cap W^{\perp} = \emptyset$, so given $u_1 \in V^{'}_{W}$ and $u_2 \in V^{'}_{W^{\perp}}$, you have that they are ...

1

The case where $\mathcal{N}(T-\lambda I)\ne \{0\}$ is covered. So assume $\mathcal{N}(T-\lambda I)=\{0\}$ and $\lambda\in\sigma(T)$. Because $T-\lambda I$ is normal, then $$\|(T-\lambda I)x\|=\|(T^*-\overline{\lambda}I)x\|,\;\;x\in H,$$ which also implies that $\mathcal{N}(T^*-\overline{\lambda}I)=\{0\}$. Therefore, $$... 2 Consider the map f defined by x \mapsto \frac{Ax}{\sum_i (Ax)_i} defined on the (topological) disk D that consists of vectors x satifying x_1 \ge 0, x_2 \ge 0, \ldots, x_n \ge 0, x_1 + x_2 + \ldots + x_n = 0 (i.e., D is the standard simplex in the positive octant). Then$$ f : D \to D $$is a continuous map of a closed disk to itself (this ... 0 Yes. Steiner symmetrization decreases the first Dirichlet eigenvalue (also known as the fundamental frequency), unless the domain is already symmetric. And a triangle that is not equilateral can be Steiner-symmetrized in a nontrivial way. More generally, Pólya and Szegő conjectured that among all n-gons of fixed area the regular n-gon has the lowest ... 2 The Spectral Theorem for A is given in terms of a Borel Spectral measure E$$ Ax = \int_{-\infty}^{\infty}\lambda dE(\lambda)x, $$and x \in \mathcal{D}(A) iff$$ \int_{-\infty}^{\infty}\lambda^2 d\|E(\lambda)x\|^2 < \infty. $$The operator e^{iA^2} is defined through the functional calculus as$$ e^{iA^2}x = ...

1

What you are missing from the statement in the book is that $H$ is separable. Consider first $B(H)$. Then $K(H)$ is weakly dense in $B(H)$, and it is not hard to show that $K(H)$ is separable. Now consider a von Neumann algebra $M\subset B(H)$. You have an inclusion of unit balls $M_1\subset B(H)_1$. The previous exercise in the book proves that, since ...

3

Given $f$ and $\epsilon$, choose a polynomial $p$ with $\Vert f-p\Vert_{\infty,X}<\epsilon$ (where $\Vert\cdot\Vert_{\infty,X}$ is the supremum norm oin $X$). Now see the corresponding polynomial function in $\mathcal{A}$, $p:\mathcal{A}\to\mathcal{A}$. (Remember: the functional calculus respects this notation, i.e., $p(a)$, in the functional calculus, is ...

0

Note: There is nothing about completeness of $\mathcal{H}$ needed to carry out of the following steps. Because $P_n$ is monotone, then $(P_nx,x)$ is monotone in $n$ for each fixed $x$, and is bounded above by $(x,x)$, which forces convergence of $\lim_n(P_n x,x)$ for all $x$. Then, using polarization, the following expression must also have a limit in $n$ ...

0

Consider the two cases of $V_n$ being monotone increasing and monotone decreasing in $n$. In the first case you have $P_n P_{n+1}=P_n$ and in the second you have $P_n P_{n+1}=P_{n+1}$. At any rate you have either $P_n(P_{n+1}- P_{n})=0$ or $P_{n+1}(P_{n+1}-P_{n})=0$. First consider $V_n$ is increasing. So $(P_{n+1}-P_n)(z)$ is in $V_n^\perp$. This means ...

1

Assume $V_n \subseteq V_{n+1}$ for all $n$. Then $P_{n+1}P_n=P_n$. Using adjoint, $P_nP_{n+1}=P_n$ must also hold because $P_k^*=P_k$ for all $k$. Therefore, for all $x$, $$(P_nx,x) = (P_nx,P_nx)=\|P_nx\|^2=\|P_nP_{n+1}x\|^2 \le \|P_{n+1}x\|^2=(P_{n+1}x,x).$$

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