2
votes
1answer
51 views

Maximal subspace on which an operator is bounded

Consider the Banach space $X=C[0,1]$ of real continuous function on $[0,1]$ equipped with the supremum norm. Consider the operator $A:D(A)\to X$, $Af=f'$ for each $f\in D(A)=C^1[0,1]$. We can see that ...
0
votes
1answer
11 views

Question about Neighborhood basis

In the Simon Reed text, after defining the strong operator topology it is said: "A neighborhood basis at the origin is given by the sets of the form $\{S \ | \ S \in \mathcal{L}(X,Y), ...
1
vote
1answer
33 views

$\sup_t |T(t)|<+\infty$ implies $\sup_t |T(t)^*|<+\infty$?

Let $X$ be a Banach space. $T(t)$ a family of bounded operators for $t\in\mathbb{R}$. $T(t)^*$ is the adjoint operator of $T(t)$. If $\sup_t |T(t)^*|<+\infty$ , then by Hahn-Banach, there's a ...
4
votes
2answers
68 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 ...
1
vote
1answer
40 views

Compact surjective non injective operator

Let $X$ be an infinite dimensional Banach space. I know that every compact operator $A$ is not bijective or $0\in\sigma(A)$. Fox example the compact operator $A$ defined on $X=C([0,1],\mathbb{R})$ ...
1
vote
1answer
48 views

Projection onto a convex closed set

H, If $K$ is a non-empty convex and closed subset of a uniformly convex Banach space $X$ (Hilbert for example) and $v \notin K$, we know that there exists a unique $k_0\in K$ such that ...
1
vote
2answers
30 views

About locally convex space

Is a Banach space a locally convex space? Why? Recall A locally convex space is a linear topological space in which the topology has a base consisting of convex sets.
0
votes
1answer
53 views

Prove that — the range $R(T)$ of a bounded linear operator $T:X\to Y$ need not be closed in $Y$

Prove that the range $R(T)$ of a bounded linear operator $T:X\to Y$ need not be closed in $Y$.
2
votes
1answer
100 views

Multiplication operator is not jointly continuous in strong topology

How can I show that multiplication operator ($M:\mathcal{L}(X,Y) \times \mathcal{L}(Y,Z) \rightarrow \mathcal{L}(X,Z)$; $M(A,B)=AB)$ is not jointly continuous in strong topology? I have to show that ...
0
votes
1answer
66 views

boundary of a spectrum proof

Let $A$ be a closed unital subalgebra of banach algebra $B$. Prove that ${\delta}{\sigma}_{B}(x)$ is contained in ${\delta}{\sigma}_{A}(x)$ for every $x$ in $B$.
1
vote
1answer
107 views

Convergence in the strong/weak operator topology: nets versus sequences

Let $H$ be a separable infinite dimensional Hilbert space, with orthonormal basis $(e_n)_{n=1}^\infty$. Consider the operators $U_n$ on $H$ such that $U_ne_k = e_1$ if $n=k$ and $U_ne_k=0$ if $n\neq ...
2
votes
0answers
47 views

Examples of extremally disconnected spaces

I am trying to understand the notion of extremally disconnected space (in other words Stonean space), i.e. a space in which any open set has an open closure. Could you help me and give (reasonable) ...
2
votes
1answer
134 views

Relationship between different topologies of bounded operators on a Hilbert space

I am self-studying functional analysis. Given that $B(H)$ are the bounded operators on a Hilbert space, $H$. I would like to ask how to formally prove that the weak topology is weaker than the ...
1
vote
1answer
39 views

Question about strong and norm convergence.

Maybe the answer to this question is so trivial that I can't see it: Why the strong convergence of operators (on an hilbert space) does not imply the norm convergence? Many books make this example: ...
3
votes
2answers
144 views

A problem about Linear Operator

$X$ and $Y$ are Banach Spaces.$ T$ is a linear bounded operator from $X \to Y$. There exists a real number $c$ which is positive, such that for any $y$ belonging to $T(X)$, there exists a $x$ which ...
2
votes
1answer
42 views

A question about compact Hausdorff space

Let $X$ be a compact Hausdorff space and $C(X)$ be the set of continuous functions on $X$. And $F$ is a closed subspace of $X$. If the $f\in C(X)$ such that $f|_{F}=0$ is only zero function( i.e. ...
3
votes
1answer
131 views

On the use of nets when defining operator topologies

Let's consider the strong operator topology and the weak operator topology on bounded operators of a infinite-dimensional Hilbert space $H$. When they define these operator topologies, some authors ...
1
vote
0answers
131 views

Find spectrum of the operator

I need some help to find point spectrum, residual spectrum, continuous spectrum and for this problem. Let $\theta \in\mathbb{C}$. On the complex space $L^2[-1,1]$ consider the operator $Au=v,$ ...
2
votes
1answer
101 views

spectrum of two bounded linear operators

Suppose that L and B are bounded linear operator on H, assume $0\in \rho(L) \cap \rho(L+B)$ and that $L^{-1}$ is compact. Prove that L+B also has a compact inverse.
1
vote
1answer
160 views

Bounded linear operator in weak topology

Let $B$ be a bounded linear operator on $H$. Prove $B\colon (H,w)\to (H,w)$ is continuous. $(H,w)$ is a Hilbert space with its weak topology.
2
votes
2answers
117 views

Are WOT/SOT topologies hereditarily separable?

Just out of curiosity, Are weak and strong operator topologies on $B(H)$ hereditarily separable? In other words, if $S$ is a subset of $B(H)$, where $H$ is a separable Hilbert space, is $S$ ...
2
votes
1answer
146 views

The span of the orthorgonal projections is norm dense in $B(H)$

This is a question in my functional analysis book. How to use the spectral theorem to prove that the span of the orthogonal projections is norm dense in $B(H)$?
10
votes
2answers
939 views

Gelfand-Naimark Theorem

The Gelfand–Naimark Theorem states that an arbitrary C*-algebra $ A $ is isometrically *-isomorphic to a C*-algebra of bounded operators on a Hilbert space. There is another version, which states that ...
3
votes
1answer
384 views

Operators on $C([0,1])$ that is compact or not.

For $f\in C([0,1])$ set $$Hf(x) = \frac{1}{x}\int_0^x f(t)dt.$$ a) Show that $H$ is a bounded operator from $C([0,1])$ into itself which is not compact. b) From a) it follows that $H$ induces a ...
2
votes
0answers
96 views

Form of weakly continuous linear functional

This was originally a problem in Stratila and Zsido's "Lectures on von Neumann algebras" (E.1.2). I've spent so much time working on it, and right now I cannot see how the result can be so simple. ...
2
votes
0answers
39 views

Initial topology of the spectrum mapping $\sigma$

Let $\mathcal{A}$ be a Banach algebra, the map $\sigma$ maps each element $a\in\mathcal{A}$ to its spectrum $\sigma(a)$, which is a compact subset of $\mathbb{C}$. The collection of compact subsets ...
2
votes
0answers
51 views

Topology of $(\mathcal{A},*)$ determined by $\mathcal{A}_{sa}$?

Let $(\mathcal{A},*)$ be a $*$-algebra, we have the following observation: Let $\|\cdot\|_1$ and $\|\cdot\|_2$ be two norms on $\mathcal{A}$ such that the involution is an isometry with respect to ...
4
votes
1answer
144 views

A subset of $\bar{S}\backslash S$ contains an open ball in $\bar{S}$? (operator theory)

E and S are subsets of a metric space. $E$ is a subset of $\bar{S}\backslash S$. Then $\overline{E}\subset(\overline{S}\backslash S^{o})$, but I wonder whether there is some condition that guarantees ...
4
votes
1answer
129 views

Connected components that are relatively open in $\sigma(T)$

Let $T$ be an bounded linear operator on a Banach space $X$. Suppose the spectrum of $T$, $\sigma(T)$ has infinitely many connected components, then $\sigma(T)$ must contain infinitely many ...
2
votes
1answer
86 views

Almost invariant subspaces for WOT closure of an algebra of operators

Let $X$ be a Banach space and $\mathcal{C}\subset\mathcal{L}(X)$ be a collection of bounded linear operators. A subspace $Y$ is said to be almost invariant under $\mathcal{C}$ if for each ...
3
votes
1answer
323 views

Topologies and Continuity in Operator Theory

I am studying Operator Theory right now, but I have not had much exposure to topology. I am trying to pick it up along the way, and I am wondering about a probably simple point: What is the ...