Operator theory is the branch of functional analysis that focuses on bounded linear operators, but it includes closed operators and nonlinear operators. Operator theory is also concerned with the study of algebras of operators.

learn more… | top users | synonyms

0
votes
0answers
47 views

Semigroups: Entire Elements (II)

Problem Given a Banach space $E$. Consider a contraction C0-group: $$T:\mathbb{R}\to\mathcal{B}(E):\quad\|T(t)x\|\leq\|x\|$$ Define its generator by: $$Ax:=\lim_{h\to0}\frac{1}{h}(T(h)x-x)\in E$$ ...
0
votes
0answers
21 views

Self adjoint operator with countable eigenvalues

It has been explained here that any self adjoint operator on a seperable Hilbert space inhabits at most countable many eigenvalues. Now I wonder, can one construct an operator $$ T : L^2([0,1]) \to ...
0
votes
0answers
14 views

Sums of two closed and closed / continuous operators

Let $X$ be a normed space and $A_j:D(A_j)\rightarrow X$ (j=1,2) linear. (i) If $D(A_1)=X$, $A_1$ continuous and A_2 closed. Do we have $A_1+A_2:D(A_1)\cap D(A_2) \rightarrow X$, $x\mapsto A_1x+A_2x$ ...
1
vote
0answers
32 views

Questions about the Kapalansky density theorem

I'm studying Takesaki's Theory of operator algebras book by myself. The following is a theorem from that book: I have several questions about this proof: 1- He claims, in the first line of ...
1
vote
0answers
27 views

A property of exponential of operators 2

Let $X$ be a Banach space. The other day I asked if all bounded operators $A:X\to X$ satisfy the following property: (P): All bounded nonzero trajectories $t\mapsto e^{tA}x$ satisfy $$\inf_{t\in ...
14
votes
1answer
243 views

A property of exponential of operators

Let $X$ be a Banach space. $A\in B(X)$ is a bounded operator. we can define $e^{tA}$ by $$e^{tA}=\sum_{k=0}^{+\infty}\frac{t^kA^k}{k!}$$ I am interested in this property: If $x\in X$, such that the ...
2
votes
1answer
64 views

Operator classification

Imagine that we have a second-order Sturm-Liouville problem on an interval $(a,b)$. What is the relationship between the structure of solutions and the dimension of $\ker(T^* \pm i)$, does anybody ...
0
votes
0answers
31 views

Is adjoint operator of generator of an analytic semigroup be a generator of analytic semi-group?

Let $X$ be a Banach space. The adjoint semigroup {$T(t)^\prime:t≥0$} consisting of all adjoint operators $T(t)^\prime$ on the dual space $X^\prime$ is, in general, not strongly continuous where ...
0
votes
1answer
20 views

Isolated eigenvalues of a self adjoint operator

If $X$ is a separable Hilbert Space and $T : X \to X$ selfadjoint and bounded, then the point spectrum $$ \sigma_p(T) $$ is only countable as explained here. I have the following three questions: ...
1
vote
2answers
51 views

Definition of operator norm

I want to show $T=d/dx$ is unbounded on $C^1[a,b]$ with $b>1$. Take a sequence $f(x)=x^n$, and $\|T\|=\sup_{x\in[a,b]}\frac{\|Tx\|}{\|x\|}=\frac{\|n\cdot b^{n-1}\|}{\|b\|}$. I want to claim as $n$ ...
0
votes
2answers
25 views

Is the Norm of the Square Root of an Operator equal to the Square root of the Norm of the Operator

Suppose we have a positive operator $A \in \mathcal{B}(\mathcal{H})$, does it follow that $$\|A\|^{1/2} = \|A^{1/2}\|?$$ If not, is there some relation between these quantities?
1
vote
2answers
17 views

Why $\|f-g\| \leq \sup_{h\in H}\frac{\|h\|}{\|Kh\|}\|K(f-g)\|$?

Let $f,g\in L^2$ with Lebesgue measure. and $K:L^2\to L^2$ be some linear and continuous operator. Show that $$\|f-g\| \leq \sup_{h\in H}\frac{\|h\|}{\|Kh\|}\|K(f-g)\|$$ where $h\in H\subset L^2$.
0
votes
0answers
36 views

Tensor product of $C^*$- algebras

We know from the paper of Douglas and Howe (enter link description here) that the commutator ideal $\mathcal{I}$ of $\mathcal{A}(C(T^2))$, the $C^*$-algebra generated by Toeplitz operators with ...
1
vote
1answer
38 views

Showing an Operator is Well-Defined and Bounded

Let $\{e_n\}_{n \in \mathbb{N}}$ be an orthonormal system within $\ell^2$. Fix a sequence $\lambda = (\lambda_1, \ldots , \lambda_n , \ldots) \in \ell^{\infty}$ and define $ \displaystyle Tf = ...
6
votes
3answers
84 views

Why is $\partial_z\partial_{\bar z}=\frac14\left(\partial_r^2+\frac1r\partial_r+\frac1{r^2}\partial_{\theta}^2\right)$?

I have to show the identity I wrote in the title: it should be $\partial_z\partial_{\bar z}=\frac14\left(\partial_r^2+\frac1r\partial_r+\frac1{r^2}\partial_{\theta}^2\right)$ but some computation ...
-1
votes
1answer
64 views

Norm of the product of an isometry and a bounded operator

Let $A$ be a bounded operator and $V$ a linear isometry, both defined on a complex Hilbert space $H$ (infinite dimensional). I could easily prove that $\|VA\|=\|A\|$. But, I just couldn't prove that ...
-1
votes
1answer
49 views

Semigroups: Product Rule [closed]

Given a Banach space $E$. Consider C0-semigroups: $$S,T:\mathbb{R}_+\to\mathcal{B}(E)$$ Then the product rule holds: $$(TS)'(t)x=T'(t)S(t)x+T(t)S'(t)x$$ How to prove this from scratch?
1
vote
2answers
41 views

Operator norm with $\inf$

Let $T: V \to W$ be a linear operator. The operator norm is defined as $$ \|T\| = \sup_{v\in V: \|v\|_V = 1} \|Tv\|_W$$ Does $$ \|T\|' = \inf_{v\in V: \|v\|_V = 1} \|Tv\|_W$$ define a norm? I ...
4
votes
1answer
38 views

Justifying an equality involving a closed operator $A$

Justify the equality $$A \int_0^\infty e^{-\lambda t} S(t) u \, dt = \int_0^\infty e^{-\lambda t} AS(t) u \, dt$$ used in (16) of §7.4.1. (Hint: Approximate the integral by a Riemann sum and recall ...
2
votes
1answer
31 views

Order zero maps in matrix algebra

Let $a$ and $b$ are two elements in a $C^*$algebra $A$. We say $a\perp b$ if $ab=ba=a^*b=ab^*=0$. We say a completely positive map $\phi: A \rightarrow B$ is of order zero if for any positive elements ...
0
votes
0answers
58 views

Semigroups: Nonexample

Given a C*-algebra $\mathcal{A}$. Consider a *-derivation $\delta$. Does it always generate a group: $$\tau(t)=e^{it\delta}$$ But a group of automorphisms is a contraction group: ...
0
votes
0answers
16 views

A question involving normed spaces and strictly convex spaces

Let $(X, \| \cdot \|_X)$ be a normed space and let $\| \cdot \|$ be a norm on $X$ such that $(X, \| \cdot \|)$ is strictly convex. How can I find a strictly convex space $(Y, \| \cdot \|_Y)$ and a ...
1
vote
1answer
26 views

A question involving norms

Let $(X, \| \cdot \|_X), (Y, \| \cdot \|_Y)$ be normed spaces and $T : X \rightarrow Y$ a bounded operator. Let $x, y \in X$ and let the norm on $X$ $$ \|x\| = \|x\|_X + \|Tx\|_Y. $$ I can't show that ...
2
votes
1answer
128 views

Classification of operators

I have a collection of questions about the limit point/circle concept and self-adjointness that are kind of connected, so I would like to ask them in a row. Apparently, an operator that is limit ...
3
votes
1answer
25 views

If $\sup_{T\in \tau}|y^*(Tx)|<\infty$ then $\tau$ is bounded in $L(X,Y)$

Let $X$ be a Banach space, $Y$ a normed vector space and $\tau\subset L(X,Y)$. Show that if $\sup_{T\in \tau}|y^*(Tx)|<\infty$ for all $x\in X,y^*\in Y^*$ then $\tau$ is bounded in $L(X,Y)$. ...
0
votes
1answer
19 views

computation with polar decomposition of bounded operator on hilbert space

I am trying to prove the following homework problem: Let $T \in B(H)$ (so $T$ is a bounded operator on a Hilbert space $H$), and let $T = U|T|$ be the polar decomposition of $T$. Prove that if $T$ is ...
0
votes
0answers
26 views

Properties of functional calculus

Suppose we have a self-adjoint bounded operator $S$ on a Hilbert space $\mathscr{H}$ with the property that $||Sx||<||x||$ for each $x\in\mathscr{H}\setminus\{0\}$. Now assume that ...
1
vote
1answer
27 views

Trying to show that $(c_0, \| \cdot \|_s)$ is strictly convex, where $\| x \|_s = \underset{i = 1}{\overset{\infty}{\sum}} \frac{1}{2^i} | x_i |$

I'm trying to show that $ (c_0, \| \cdot \|_s) $ is a strictly convex space, where $$ \| x \|_s = \underset{i = 1}{\overset{\infty}{\sum}} \frac{1}{2^i} | x_i |,$$ $ x = (x_1, x_2, ..., x_i, ...) \in ...
2
votes
1answer
84 views

Integration by parts to find the adjoint operator

On the interval $(0,1)$ consider the differential operator $Lu=u''''+u'$ with boundary conditions $u(0)+u'(1)=u(1)+u'(0)=0$ $2u(0)+u''(1)=2u(1)+u''(0)=0$ $(1)$ I want to find the adjoint ...
2
votes
2answers
33 views

Properties of resolvent operators

I am asked to prove the identities of $(12)$ and $(13)$, which are given on page 438 of the textbook PDE Evans, 2nd edition as follows: THEOREM 3 (Properties of resolvent operators). (i) If ...
0
votes
1answer
73 views

Spectrum of Laplacian on Half line. $\left [0, \infty \right)$

I would like to calculate the spectrum of Dirichlet and Neumann Laplacian of the domain $\left [0,\infty \right)$. To be precise, Define the Operator $T$ on $L^2\left[0,\infty\right)$ as $Tf=-f''$ ...
0
votes
1answer
15 views

Unclear passage of a theorem concerning compact operators (Schauder fixed point theorem)

I'm looking at this proof of Schauder theorem and I am struggling with a passage. This is my problem: Let $X$ be a Banach space, $K \subset X$ a convex, close and bounded set and $F:K \rightarrow ...
1
vote
1answer
31 views

How to solve this operator equation?

May be $\alpha(t)$ a non-invertible Operator on a Hilbert space for a real-valued Parameter $t$. Also, let be $A(t),B(t)$ arbitrary Operators also dependent on $t$ and $f(t)$ an $L^\infty$-integrable ...
0
votes
0answers
21 views

Does a pseudodifferential operator $A$ commute with the Resolvent of $A^*A$?

Suppose we have a pseudodifferential operator $A$ (not necessarily bounded) for which the heat operator $$ e^{-tA^*A} := \frac{i}{2 \pi} \int_\Gamma e^{-t\lambda} (A^*A - \lambda)^{-1} \;d\lambda ...
0
votes
1answer
24 views

GNS Construction on non-unital algebra

STATEMENT: If A has a multiplicative identity 1, then it is immediate that the equivalence class $ξ$ in the GNS Hilbert space H containing 1 is a cyclic vector for the above representation. If $A$ is ...
0
votes
0answers
30 views

Finite dimensional C*-algebras and spectrum of each its elements

Let $A$ be a finite dimensional C*-algebra. when I say finite dimensional I mean there is $x_1,...,x_n\in A$ such that $A=span\{x_1,...,x_n\}$(C*- algebra generated by $\{x_1,...,x_n\}$)(is it ...
1
vote
1answer
288 views

Prove that the Set of Bounded Linear Operators is Banach

Let $B(V,V')$ be the vector space formed by set of linear operators $T:V\rightarrow V'$. where $V,V'$ are normed vector spaces. Equip $B(V,V')$ with the norm $$ \|T\|=\sup\frac{\|T(x)\|}{\|x\|} $$ ...
2
votes
1answer
16 views

Compactness of translation operator in weighted spaces

Let $x,v\in\Bbb R^d$, $t\in \Bbb R$ and $m(x,v)$ be a smooth strictly positive function rapidly decaying on infinity - think $m(x,v) = \exp(-|x|^2-|v|^2)$. Define Banach spaces $X$ and $Y$ by ...
2
votes
1answer
88 views

Show self-adjointness with eigenvalue expansion.

I was wondering if anybody here knows how to show that the negative Laplacian is self-adjoint on the 2 nd order Sobolev space of the two-sphere? I read that it is a rather cumbersome calculation, but ...
0
votes
1answer
33 views

Finite dimensional C*-algebra

Let $A$ be a simple and finite dimensional C*-algebra. We first note that $aAb\neq 0$ for every nonzero $a,b\in A$. Let $B$ be a maximal abelian self-adjoint subalgebra of $A$. Being finite ...
1
vote
1answer
39 views

Spectral Measures: Stone's Formula

Given a Hilbert space. Consider a Hamiltonian: $$H:\mathcal{D}(H)\to\mathcal{H}:\quad H=H^*$$ Then Stone's theorem says: ...
0
votes
0answers
35 views

Positive element of a C*-algebra

Let $A$ be an abelian C*-algebra and $p$ be a projection in $A$. To show $p$ is an extreme point of $A^+_{\|.\|\leq 1}$ suppose there is $b,c\in (A^+)_{\|.\|\leq 1}$ such that $p= \frac{1}{2}(b+c)$ ...
0
votes
2answers
148 views

Eigenvalues of an operator induced in a quotient space

Give an example of a vector space $V$, an operator $T \in \mathcal L(V)$ and a $T$-$\space$invariant subspace $U$ of $V$ such that $T/U$ has an eigenvalue that is not an eigenvalue of $T$. Attempt: I ...
2
votes
3answers
70 views

Why if $T$ is not continuous, then for each $n\in N$, there exists $x_n\in X$ such that $||Tx_n||\ge n||x_n||$

If $X$ $Y$ are Banach space, $T$ is a linear operator between them. I don't understand the following statement: If $T$ is not continuous, then for each $n\in N$, there exists $x_n\in X$ such that ...
1
vote
1answer
37 views

Is the minimiser of the quadratic form of a semi-bounded self-adjoint operator an eigenstate?

I am wondering whether the following fact, for which I know well the proof when $H$ is a Schroedinger operator (see Lieb-Loss, Analysis, Chapter 11), is also true in the general setting used below, ...
0
votes
1answer
31 views

Domain of operator via spectral theorem

Assumptions: Suppose $T$ is a self-adjoint operator (possibly unbounded) from the (dense) domain $D(T)$ on a Hilbert space $H$, hence $T:D(T)\rightarrow H$. Assume that $f$ is a continuous function on ...
2
votes
2answers
30 views

Product unbounded operators

Let $A : D(A) \subset H \rightarrow H$ be unbounded and $B$ be a bounded operator, both of them are self-adjoint, then $(AB)^* = B^*A^*$ and $(BA)^* = A^*B^*$, right? I just wanted to be sure that ...
0
votes
1answer
57 views

Spectral theorem for unbounded self-adjoint operators, questions about the proof

I want to understand the proof of the Spectral theorem for unbounded self-adjoint operators. First the theorem: Let H be a separable complex Hilbert space, $A:D(A)\subseteq H\to H$ a densily defined ...
0
votes
1answer
22 views

If $A$ is an abelian C*-algebra, and $\tau$ is pure then it is a character on $A$

If $A$ is an abelian C*-algebra,and positive linear functional $\tau$ is pure then it is a character on $A$. Murphy in his book(C*-algebras and operator theory) has below proof: While I think we can ...
0
votes
1answer
45 views

Why if $T$ is not a bounded operator then exists $ (x_n) $ that converges to $ 0_{X} $ for which $ \| T(x_n) \| \geq n^2 $ for all $ n $?

Let $X$ and $Y$ be normed spaces. Suppose that $ T: X \to Y $ is a linear operator and assume that $T$ is not bounded. Why with these assumptions can I say that exists a sequence $ (x_{n})_{n \in ...