1
vote
1answer
49 views

Riesz Functional Calculus vs. Holomorphic Functional Calculus

"Functional calculus" is a word used to describe the practice of taking some functions or formulas defined on complex numbers, and apply them in some way to certain kinds of operators, despite that ...
0
votes
0answers
27 views

Question about the Averson's proof of the bicommutant theorem.

In the Averson's proof of the bicommutant theorem is proved that, if $A$ is a self-adjoint algebra of operators with trivial null space and $T \in A''$, for every $\epsilon>0$, $n=1,2..$ and every ...
1
vote
1answer
15 views

Commutant of a set of operators and norm topology.

In the references I have it's remarked that the commutant $S'$ of a set $S$ in $B(H)$, where $H$ is a Hilbert space, is closed in the weak operator topology. And this is true because if ...
2
votes
0answers
83 views

Spectrum of $Tu=\int^1_{-1} (1-|x-y|)u(y)dy $

Consider the operator $$ Tu(x)=\int^1_{-1} (1-|x-y|)u(y)dy $$ We want to find the spectrum of $T$. The kernel is certainly bounded and so this operator is Hilbert-Schmidt, so $T$ is compact. We ...
0
votes
1answer
36 views

Prove that for a bounded self adjoint operator, $\langle Tx,x\rangle \geq 0$ is equivalent to $\sigma(T)\subset [0,\infty)$

Prove that for a bounded self adjoint operator, the following are equivalent: A: $\langle Tx,x\rangle \geq 0$ B: $\sigma(T)\subset [0,\infty)$ What I have said so far: Since $T$ is self adjoint, ...
1
vote
1answer
26 views

Skew-adjoint differential operator $B$ with spectrum $\sigma(B)=i(-\infty,-1]$

Consider the Hilbert space $X=L^{2}\left(\mathbb{R}^n\right)$ and the Schrödinger operator $A=i\Delta$ defined on the domain $D(A)=H^2(\mathbb{R}^n)$. It is known that the spectrum of $A$ is ...
3
votes
2answers
43 views

Spectra of operators on different spaces

Can the same operator when defined on two different spaces have different spectra? For example and operator defined on $C_0$ and on $\ell_2$?
1
vote
1answer
48 views

Spectrum of a finite rank operator

If $ T\in B(H)$ is a finite rank operator, then there are orthonormal vectors $e_1,...,e_n$ and vectors $g_1,...,g_n$ such that $Th=\sum_{i=1}^n (h,e_i )g_i$, then we can easily see that $T$ is ...
2
votes
1answer
46 views

Definition of exponential for operators

if I have a self-adjoint operator $T:D(T) \rightarrow L^2$, then I define its unitary exponential operator by $$e^{iT}(f) := \lim_{k \rightarrow \infty} e^{iT_{k}}(f),$$ where $T_k(f):=\frac{1}{2} ...
1
vote
1answer
40 views

Relation between $\epsilon$-pseudospectrum of operators

If $H$ is a Hilbert space and $\sigma_{\epsilon}(T)$ denotes the space of all $\epsilon$-pseudospectrum of the operator $T$ and $S, T\in B(H)$ be such that $TS=ST=0$, why ...
1
vote
3answers
62 views

Prove that if $T=T^*$ and $\sigma(T)=\{\lambda\}$, then $T=\lambda I$

Show that if $T$ is a self adjoint linear operator on a Hilbert space such that the spectrum contains a single point $\lambda$, then $T=\lambda I$. Then, show this is false if $T$ is not self ...
8
votes
2answers
92 views

Ways to calculate the spectrum of an operator

Friends, I am learning some very basic stuff of spectral theory and kind of lost, in some sense. I am trying to find ways to compute the spectra of different operators, when they work and don't work. ...
3
votes
1answer
57 views

Is $AA^*$ and $A^*A$ self-adjoint?

if I have a densely defined closed linear operator $A$ and $A^* = -A$(same domain also closed). Is this sufficient that $AA^*$ and $A^*A$ are proper self-adjoint operators, assuming that we can also ...
1
vote
1answer
54 views

Domain of densely-defined second derivative operator, and its factorization

Let $$-d_x^2: \{f \in L^2[0,1];f \in AC^1[0,1] , f(0)=f(1)\} \rightarrow L^2[0,1]$$ be the second derivative operator. Here $AC^1[0,1]$ is the space of functions whose first derivative is absolutely ...
2
votes
2answers
53 views

Decomposition of a positive semidefinite self-adjoint operator?

If I have a positive semi-definite self-adjoint operator $H:D(H) \rightarrow L^2$, is it true that there is always a decomposition $H=A^* A$ available? If this is true, what can we say about the ...
4
votes
2answers
107 views

Spectral theory - continuous spectrum

imagine that I have some differential operator $D$ that is defined on an interval $[a,b]$. Now, assume that we take the boundary conditions in such a way that this operator is self-adjoint. Then, I ...
1
vote
1answer
43 views

Fredholm index for 1-d Schroedinger operator

if I look at a Schroedinger-operator $-\frac{d^2}{dx^2} +V$ on a compact intervall $[a,b] \subset \mathbb{R}$ and I take boundary conditions that this operator is self-adjoint (for example periodic ...
3
votes
2answers
103 views

Two questions in spectral theory: the spectrum of the Fourier transform and the Hamiltonian of the hydrogen atom.

I have the following two questions: The Fourier transform defines a unitary (provided that it is normalized properly) map $\hat{\cdot}:L^2(\mathbf{R})\rightarrow L^2(\mathbf{R})$. I figured out its ...
1
vote
1answer
101 views

Spectral theory

I have absolutely no idea about Spectral theory and want to ask some fundamental questions. 1.) What does it mean that the resolvent of an operator is Hilbert-Schmidt? (Cause I saw a theorem that was ...
3
votes
1answer
68 views

Spectrum of normal elements in C*-algebras

Let $\mathcal{A}$ be a C*-algebra and $x \in \mathcal{A}$ a normal element. Can you show that $\left\{ \phi(x) : \phi \text{ is a state on } \mathcal{A} \right\}$ is the closed convex hull of the ...
0
votes
1answer
61 views

C*-Algebra: Positive Operator

Let $\mathcal{A}$ be a C*-algebra. If $A\in\mathcal{A}$ is a selfadjoint element then $A^*A=A^2$ has positive spectrum since: ...
1
vote
1answer
88 views

Mathieu differential equation

Given the operator $T (\psi)(x):= \psi''(x)-2q \cos(2x)\psi(x)$ with $T : D(T) \subset L^2[0,2\pi] $ I was wondering: What is the right domain $D(T)$ for this operator if we want to solve the ...
4
votes
1answer
101 views

Proving the spectral theorem for unbounded self-adjoint operators

Let $A$ be (densely-defined) self-adjoint operator on a (complex) Hilbert space $H$. Then, the Cayley transform $U=(A-i)(A+i)^{-1}$ is a unitary operator, and $(A\pm i)^{-1} \in B(H)$. Using the ...
0
votes
2answers
51 views

Spectrum of a bounded operator $T$ satisfying $T^n=I$

Let $\mathcal{H}$ be an infinite dimensional Hilbert space, suppose $T\in \mathcal{B}(\mathcal{H})$ is a bounded operator and suppose that $n$ is the smallest natural number so that $T^n=I$. Let ...
1
vote
1answer
43 views

Spectral radius of an operator equals its norm

Let $X$ be a Banach space and $A:X\to X$ a bounded operator. We know that the spectrum of $A$ is always included in the ball $B(0,|A|)$ and the spectral radious $r(A)$ is the smalest radius such that ...
0
votes
0answers
46 views

Spectrum of a bounded operator and Liouville's theorem

Let $X$ be a Banach space and $A:X\to X$ be a bounded operator. We know that the spectrum of $A$ is not empty, because otherwise we find a contradiction by using the holomorphy of the function ...
0
votes
1answer
54 views

Spectrum of the operator $A(f,g)=(g,\Delta f-f)$

Let $\Omega$ be an open set in $\mathbb{R^n}$. We consider the product Hilbert space $H=H^1_0(\Omega)\times L^2(\Omega)$ with the norm $$|(f,g)|^2=\int_\Omega (|\nabla f|^2+|f|^2+|g|^2 ) dx$$ We ...
1
vote
0answers
27 views

Skew adjoint operator with uncountable spectrum

Let $H$ be a Hilbert space. I just want an example of a skew adjoint operator $(A^*=-A)$ with uncountable spectrum. I also want an example for unbounded differential operators. The only example I ...
1
vote
1answer
66 views

“right shift” il $L^1$

Let $X=L^1(\mathbb{R})$ be the space of Lebesgue integrable functions $f:\mathbb{R}\rightarrow \mathbb{C}$ with the usual norm. Let $T\in B(X)$ be defined by $$(Tf)(t)= f(t+1)$$ I need to find the ...
0
votes
2answers
53 views

Normal Operators: Spectrum vs. Numerical Range

Disclaimer: As I realized in the comments that this works for normal operators I decided to modify this question. Besides, I got the proof now - thanks to T.A.E.! Prove that for normal operators the ...
2
votes
2answers
62 views

Adjoint of sum of two operators

Let $A$ be self-adjoint and $B$ symmetric (which means densely defined for me as well) with $A$-bound less than $1$. Does this imply that $(A+iB)^*=A-iB$ ?
1
vote
1answer
56 views

Operator: not closable!

Is there an operator between Banach spaces with the following properties: $$T:\mathcal{D}(T)\subseteq X\to Y:\text{ injective, dense range, continuously invertible, not closable!}$$ (Note that the ...
2
votes
1answer
42 views

Cayley Transform: well defined?

Why is the Cayley backtransformation well-defined: $$A_U:=\imath(1+U)(1-U)^{-1}$$ In general $1-U$ is not invertible for example for $U=1$.
4
votes
1answer
83 views

What makes compact operators special?

I would like to understand why compact operators are considered so special to consider them as an extra class of operators. Over Hilbert spaces these (as far as I know) these are the ones with ...
2
votes
2answers
118 views

Resolvent Set: Definition

Given Banach spaces: $X,Y$ Consider a linear operator: $T:\mathcal{D}(T)\to Y$ (not necessarily bounded nor closed nor closable nor densely defined) Define for the shorthand the shifted operator: ...
4
votes
2answers
70 views

Operators $A$ such that $e^A$ is norm preserving

Let $X$ be a Banach space. $A$ a bounded operator. We can define the exponential of $A$ by $$e^{A}=\sum_{n=0}^{+\infty}\frac{A^n}{n!},$$ which is also a bounded operator. Is there any sufficient ...
0
votes
0answers
18 views

When can we get discrete spectrum?

Suppose that $T$ is a densely defined closed operator on a separable Hilbert space $H$. Form $N = T^*T$. Assume further that $T$ has a finite dimensional kernel and satisfies the commutation relation ...
0
votes
1answer
47 views

Positive Operator: Norm Estimate

In class we encountered the statement: $$H\geq C\mathrm{Id},C>0\Rightarrow\|\mathrm{e}^{-\beta H}\|<1,\beta>0$$ How does one prove this? Moreover what about the weakened version: $$H\geq ...
0
votes
1answer
45 views

How to find the spectrum $\sigma_p(P)$

How to find the spectrum $\sigma_p(P)$: Let $P:H\rightarrow H$ be an orthoprojection, $P\neq 0, P\neq I$. could you please help
2
votes
2answers
72 views

How to show: $A_y$ has no eigenvectors if $y$ is not constant on any subinterval of $[0,1]$

Let $y\in C[0,1]$ and $A_y : C[0,1]\rightarrow C[0,1]: x\mapsto xy$ How to show: $A_y$ has no eigenvectors if $y$ is not constant on any subinterval of $[0,1]$. Could you please help.
3
votes
1answer
55 views

Want to show that an operator is not surjective

So here is my problem, Let $$M_1:L^1\rightarrow L^1$$ $$f(x)\mapsto \arctan(x)f(x)$$ In order to compute the spectrum of $M_1$ I am investigating for which $\lambda\in\mathbb C$ the following map is ...
-1
votes
1answer
42 views

Polynomial Calculus on Spectrum: well defined?

Consider a bounded operator over a Banach space: $T\in\mathcal{B}(E)$ Apply polynomial calculus on the the chosen operator: $p(T),p\in\mathbb{C}[X]$ Why do we need to prove that when two polynomials ...
0
votes
1answer
53 views

Root of polynomial implies vanishing remainder. Application to spectral theory!

Framework: Consider a unital ring: $e\in R$ and a given polynomial: $p\in R[X]$ (Note that I do not require the ring to be an integral domain.) Problem: If it has a root then it factorizes: ...
1
vote
1answer
39 views

Is it possible to consider an approximation to a (non-self adjoint) operator with a self adjoint one?

In operator theory it's wonderful if we have a self-adjoint operator (non necessarily bounded) due to all the work that has been done using their symmetry,... etc. I.e there are many powerful tools. ...
0
votes
0answers
8 views

Comparison of subsets of spectrum

Let $X$ be a Banach space. $A$ is a linear closed and densely defined operator and $S$ is a bounded invertible operator. I want compare $\sigma_{e,S}(\lambda S - A)$ to $\sigma_{e,S}(A)$. Here ...
1
vote
3answers
48 views

Examples of spectrum of compact operators on the sequence space $l_2$

Suppose $T$ is a compact operator on the sequence space $l_2$, and let $\sigma(T)$ be its spectrum. Is it possible to find a $T \ne 0$ such that $\sigma(T) = \{0\}$? Also, is it possible to find $T$ ...
1
vote
1answer
42 views

Multiplication operator has no no eigenvalues

Statement: Let $M_x$ denote the multiplicative operator acting on $L^2([0,1], \, dx)$ by $M_x(f) = xf$. Show that $M_x$ has no eigenvalues Attempt: Let $M_x(f) = xf$ then we should have $M_x(f) = ...
0
votes
0answers
17 views

about spectrum of the sum of two operators

It is well known that if $A$ and $B$ are commuting bounded operators on a Banach space then it is familiar that $$ \sigma(A+B)\subseteq\sigma(A)+\sigma(B)$$ I would to ask if I can find suffisant ...
0
votes
1answer
21 views

stability of essential spectra

Let $X$ be a Banach space. $A$ and $B$ are linear closed and densely defined operators and $\lambda\in\rho(A)\cap\rho(B)$ such that $(\lambda - A)^{-1}-(\lambda - B)^{-1}$ is a Frehholm perturbation ...
0
votes
1answer
36 views

How do I show a left inverse of a bounded linear operator on Banach space?

If $A$ is a bounded linear operator on a Banach space X, with a left inverse $A_l^{-1}$, and P is a projection (also on X), how do I show that $A_l^{-1}P$ is also a left inverse of A (i.e. ...