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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 $A^*A$.


0

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 ...


0

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) & ...


0

$\exists n_0$, s.t. $\|T_1^{n_0}\| \leq (r(T_1)+\epsilon )^{n_0}$ and $\|T_2^{n_0}\| \leq (r(T_2)+\epsilon )^{n_0}$. $$ \begin{split} \|(T_1+T_2)^n\| &\leq \sum_{k=1}^n \begin{pmatrix}n \\ k \end{pmatrix} \|T_1^k T_2^{n-k}\|\\ &\leq \sum_{k=1}^n \begin{pmatrix}n \\ k \end{pmatrix} \|T_1^{n_0}\|^{\lfloor k/n_0 \rfloor} \|T_1\|^{r_k} ...


1

I'll try to put something together to help you a bit here: If you are working with manifolds, you usually only have local descriptions of a manifold $M$, given by coordinate systems $x:M\cap U\rightarrow \mathbb{R}^n$ (here $U$ is some neighbourhood on $M$ diffeomorphis to a ball in Euclidean space). On a manifold you can do differential calculus by doing ...


1

Let's just try to find an eigenfunction for a given $|\lambda|< 1$: we want an $f$ that solves $$ f\left( \frac{x+1}{2}\right) = 2\lambda f(x)- f(x/2) . \quad\quad\quad\quad (1) $$ Start out with an arbitrary (integrable) function on $(0,1/2)$, then use (1) for $0<x<1/2$ to define $f(t)$ for $1/2<t<3/4$, then reenter (1), with ...


4

Essentially, the problem is $u''+\lambda u=0$ on $[-\pi,\pi]$ with periodic boundary conditions. (Note that this interval has the same length as the circumference of the circle; this choice of parametrization ensures that the Laplace-Beltrami operator on the circle directly corresponds to the Laplacian on this interval.) The solutions to the DE itself are ...


2

Your proposed approach is fine: plug in $g$ into both sides of your equation and you will get $$g'' = \sum_{n\geq 1} -a_n n^2 \cos(nx) - b_n n^2 \sin(nx)$$ $$\lambda g = \frac{\lambda a_0}{2} + \sum_{n\geq 1} a_n \lambda \cos(nx) + b_n \lambda \sin(nx).$$ There is something special and fortuitous about the basis you have chosen for $g$: compute ...


3

Not a solution, but a sketch. You need to find a manageable expression for $\Delta f$ if $f:S^1 \rightarrow \mathbb{R}$. One option is to work with a (almost onto) chart of $S^1$, e.g. by using an arclength parametrized curve $c:(-\pi,\pi)\rightarrow S^1$ and by looking at the resulting equation for $f\circ c$ instead. The fact that $f$ is assumed to be ...


1

You have $$ (\lambda I-T)(x_1,x_2,\ldots)=((\lambda-1)x_1, (\lambda-\frac12)x_2,\ldots). $$ Define an operator $S $ by $$S (x_1,x_2,\ldots)=(\frac{x_1} {\lambda -1},\frac {x_2}{\lambda -\frac12},\ldots). $$ By the choice of $\lambda $ the linear operator $S $ is well-defined and bounded: by construction, $S (\lambda I-T)=(\lambda I-T)S=I $. So $S=(\lambda ...


0

Yes. Suppose $A $ is $m\times n $. Let $e_1,\ldots,e_n $ and $f_1,\ldots,f_m $ be the canonical bases of $\mathbb R^n $ and $\mathbb R^m $ respectively. Write $A=USV $ the singular value decomposition. Let $v=Ve_j $; as $V $ is a unitary, $\|Ve_j\|=1$. Then, since $S_{pq}=0 $ if $p\ne q $, \begin{align} |A_{ij}| & =|\langle Ae_j,f_i\rangle| =|\langle ...


0

Let $H$ be the subspace spanned by the orthonormal set $\{1,e^{ix},e^{2ix},\cdots\}$ in $L^2[0,2\pi]$ with the inner product $$ (f,g)=\frac{1}{2\pi}\int_{0}^{2\pi}f(t)\overline{g(t)}dt $$ Consider the operator $$ T = -i\frac{d}{dx}(e^{ix}f(x)). $$ Then $T(e^{inx})=(n+1)e^{i(n+1)x}$ for $n=0,1,2,3,\cdots$. So this is the same as ...


1

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2

You are making this much to difficult because you are ignoring what you were told. You were told that the differential operator was $d^2/dx^2$, the "Laplacian", but only in one dimension. The general solution to $d^2y/dx^2= 0$ is the linear function $y= ax+ b$. The solution to $d^2G/dx^2= \delta(x- x')$, then, is a "broken line: $G(x, x')= px+ q$ for ...


1

The proof of the converse is simple: if $\sigma(T)=\sigma(S)$, then $C(\sigma(T))=C(\sigma(S))$. Then $G_2\circ G_1^{-1}$ is the $*$-isomorphism you are looking for. Now, it is not true that the statement you gave implies unitary equivalence; you cannot ignore multiplicity. Consider $$ ...


1

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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