2
votes
1answer
39 views

Why are “not bounded” operators not everywhere defined?

Let $X, Y$ be Banach spaces, $\mathcal{D}(T)$ a subspace of $X$, and $T\colon X\to Y$ a linear map. Such a $T$ is commonly called an unbounded linear operator, where unbounded just means that the ...
1
vote
3answers
51 views

Basis of eigenfunctions in Banach space

I have a question about proving the existence of a basis of eigenfunctions. Assume we have a compact operator $L:\mathcal{B}\rightarrow\mathcal{B}$, where $\mathcal{B}$ is a Banach space of analytic ...
2
votes
0answers
23 views

Invariant subspaces and their complements

The following is Exercise 6.4.2 of Conway's Functional Analysis: Suppose $T\in B(X)$ where $X$ is a Banach space. Prove that $M$ is invariant under $T$ if and only if $M^{\perp}$ is invariant under ...
2
votes
1answer
44 views

Compactness of an operator on $c_0$ in terms of its infinite matrix representation

Let $A\in {\cal B}(c_0)$ (${\cal B}(c_0)$ is linear bounded operators on $c_0$) and for $n\geq 1$, define $e_n \in c_0$ by $e_n(m)=\delta_{nm}$. Put $\alpha_{nm}=(Ae_n)(m)$ for $n,m\geq 1$. we have $M ...
0
votes
0answers
15 views

Composition operator of a non-analytic function

I recently came across a problem that can be reformulated in simple terms involving the composition operator $$ C_g\colon X \to X, \quad f \mapsto f \circ g $$ for functions $$ f,g\colon \mathbb{C} ...
3
votes
1answer
53 views

Give necessary and sufficient conditions for a multiplication on $L^p$ to be compact

Let $(X, \Omega, \mu)$ be a $\sigma-$ finite measure space and for $\phi \in L^\infty(\mu)$ let $M_\phi:L^p(\mu) \to L^p(\mu)$ defined by $M_\phi f = \phi f $ be the multiplication operator. Give ...
2
votes
1answer
20 views

$\sup$ norm of a function

The following is an example of Murphy's C*-algebras and operator theory: I do not know how he concludes $$\int_0^1 |k(s,t) - k(s',t)||f(t)| dt \leq \sup|k(s,t) - k(s',t)|||f||_\infty$$ Please help ...
3
votes
1answer
52 views

Integral operator on $L^p$ is compact

Let $(X,\Omega,\mu)$ be an arbitrary measure space, $1<p<\infty$ , and $\frac{1}{p}+ \frac{1}{q} = 1$. If $k:X. X\to \Bbb C$ is an $\Omega.\Omega-$ measurable function such that $$M = [\int ...
10
votes
1answer
90 views

When is an operator on $\ell_1$ the dual of an operator on $c_0$?

Suppose $T:\ell_1\to\ell_1$ is a continuous linear operator. When can we say that $T$ is a dual, or adjoint, of an operator on $c_0$? In other words, under what conditions can we find a continuous ...
0
votes
2answers
57 views

Existence of a semigroup of bounded operators which is not $C_0$

Let $X$ be any Banach space. Then we can define a $C_0 $ semi group of bounded operators on $X$. But my question is that can we define a semi group of bounded operators which is not $C_0$?
2
votes
1answer
27 views

Kernel of the Extension of a Bounded Linear Operator

Suppose $T\colon E\to F$ is a bounded linear operator between Banach spaces. Moreover let $i\colon E\to E’$ be a dense, compact inclusion of $E$ into some other Banach space $E’$. Finally assume that ...
2
votes
2answers
65 views

How to show that $e^{tA}=\frac{1}{2\pi i}\int_{\{Re \ \lambda =a\}}e^{\lambda t}(\lambda I-A)^{-1}d\lambda$?

Let $X$ be a Banach space and $A:X\to X$ be a bounded operator. We can show that if $|\lambda|>|A|$ then $\lambda I-A$ is invertible and $$(\lambda ...
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 ...
1
vote
1answer
30 views

Let $A: X \to X$ be a Fredholm operator, then $Ax=y$ has a solution iff $Ax=0$ implies $x=0$?

Let $X$ be a Banach space and let $A: X \to X$ be a Fredholm operator, then $Ax=y$ has a solution iff ($Ax=0$ implies $x=0$)? I can't see how this is implied by the common definitions of Fredholm ...
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
2answers
47 views

Strongly continuous semigroup of operators which cannot be extended to a group

Let $X$ be a Banach space. We call a family of bounded operators $(T(t))_{t\in \mathbb{R}}$ a strongly continuous group if it satisfies the properties of the strongly continuous semigroup but for ...
0
votes
1answer
37 views

Restriction of an operator on a Banach space

Let $X$ be a Banach space and $A:D(A)\to X$ be an unbounded linear operator such that for all $\lambda>c$ ($c$ some constant), $(\lambda I-A)^{-1}$ exists and is a bounded operator which satisfies: ...
1
vote
1answer
58 views

Inverse of operator is not continuous in Banach spaces

Let $X$ be a Banach space. If $A:X\to X$ is an invertible bounded operator (injective, surjective and continuous), then $A^{-1}$ is also bounded. Now can I have an example of an unbounded operator ...
6
votes
1answer
199 views

Uniform continuity of the function $x(t)=e^{tA}x$

Let $A$ be a bounded operator on a Banach space $X$. Consider the exponential function $x(t)=e^{tA}x:=\sum_{n=0}^{+\infty}\dfrac{t^nA^n}{n!}x$, for all $t\in \mathbb{R}$, where $x\in X$. If the ...
3
votes
0answers
22 views

Fredholm Integral in Bayesian Appliation

Let $X = x_1, x_2, \ldots, x_n$ be a sequence of Bernoulli random variables with $k$ successes. Suppose that, given $X$, the posterior predictive probability of $x_{n+1} = x$ is known to be $g(x)$ ...
1
vote
2answers
46 views

Is it a compact operator?

Let $$C^{1}_{2\pi}=\{u\in C^{1}((0,2\pi),\mathbf{R}^n):u(0)=u(2\pi)\}$$ $$C_{2\pi}=\{u\in C^{0}((0,2\pi),\mathbf{R}^n):u(0)=u(2\pi)\}.$$ $C_{2\pi}$ is equipped with the norm $$\|u\|_0=max|u(s)|$$ ...
2
votes
2answers
37 views

Bounded, surjective linear operator between Banach spaces

How can I show that for a given surjective linear operator $T: X \to Y$ between Banach spaces, if there exists an $\epsilon > 0$ such that $||Tx|| \geq \epsilon||x||$ for all $x \in X$, then $T$ is ...
3
votes
2answers
48 views

Computing the norm of operator when space is equipped with sup norm and $L^1$ norm

Let $\phi $ be the linear functional $\phi (f)=f(0)-\int^1_{-1}f(t)\:\mathrm{d}t$ a.Compute the norm of $\phi$ as a functional on Banach space $C[-1,1]$ with sup norm. b.Compute the of $\phi$ as a ...
2
votes
1answer
45 views

Riesz Lemma for reflexive spaces

I know the proof of Riesz Lemma: Let $Y$ be a closed (proper) subspace of a normed space $X$. Let $\varepsilon >0$. Then it exists an element $x \in X$ such that $||x||=1$ and $d(x, Y) \geq ...
0
votes
1answer
21 views

Extension of a linear operator

Let $T$ be a linear operator defined on the space of the algebraic polinomials in $[0,1]$ (polinomials with rational coefficients) such that for each $k \in \mathbb{N}, T[x^k]=0$. Is it possibile to ...
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
0answers
31 views

Group of operators such that $|T(t)x|\geq c |x|$

Let $X$ be a Banach space. Can I have an example of a strongly continuous group of operators $T(t)$ such that $$|T(t)x|\geq c |x|, \ t\in\mathbb{R}$$with $c>1$. For $c=1$, I know examples of ...
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 ...
1
vote
1answer
57 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 ...
1
vote
1answer
31 views

Functional defined on the space of functions with compact support

Let $X=C_c(\mathbb{R})$, the space of functions with compact support, normed with the max norm. Define $\Gamma: X \rightarrow \mathbb{R}$ as: $$ \Gamma f= \int_{- \infty}^{\infty} f(t)dt ...
2
votes
2answers
38 views

Fredholm Index: Finite Corank $\Rightarrow$Closed Range [duplicate]

Obviously closed subspaces turn quotient spaces into normed spaces rather than just merely vector spaces. However the dimension involved in Freholm's index are purely algebraic. Why do we thus ...
4
votes
1answer
84 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 ...
0
votes
1answer
64 views

Compact Operator <=> Separable Range

Is it true that a bounded operator is compact iff its range is separable: $$T\text{ bounded}:\quad T\text{ compact}\iff \mathcal{R}(T)\text{ separable}$$
0
votes
1answer
37 views

Existence of nontrivial bounded linear operator?

Are there two normed spaces such that there is no nontrivial bounded linear operator between them: $$\nexists T:X\to Y: T\text{ nontrivial, linear and bounded}$$
2
votes
2answers
119 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: ...
1
vote
1answer
77 views

What is the dual space in the strong operator topology?

Let $X$ be a Banach space, the strong operator topology on the space of bounded linear operators $\mathcal{B}(X)$ is defined by the family of continuous semi-norms $A\to\|Ax\|$, $x\in X$. What is the ...
0
votes
1answer
36 views

Exercise on isometry

Let $X$ be a Banach space and $T$ a linear bounded operator defined on $L(X,Y)$ with $Y$ a normed space. If $T$ is an isometry then $TX$ is a closed subspace of $Y$. I considered a sequence $y_n$ ...
0
votes
1answer
19 views

Bounded operator on continuous functions

Let $X=C([0,1])$ and $T: X \rightarrow X$ defined as $$(Tf)(t)=f(t)+f(0)$$ Prove $T$ is bounded. I was thinking about using the fundamental theorem of calculus in order to get some bounds on $f(0)$ ...
1
vote
1answer
57 views

Creation and Annihilation Operators: Norm Estimate

Given the Fock space: $$\mathcal{F}(\mathcal{h}):=\bigoplus_0^\infty\mathcal{h}^{n}\text{ with } \mathcal{h}^{n}:=\bigotimes_1^n \mathcal{h},\mathcal{h}^0:=\mathbb{C}$$ Define the creation and ...
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 ...
13
votes
1answer
222 views

Injectivity of the operator $(Ax)(t)=\int_0 ^1 k(s,t) x(s)ds$

Let $X=C([0,1],\mathbb{R})$ (equipped with the supremum norm). Let $A$ be the operator defined for each $x\in X$ by $$(Ax)(t)=\int_0 ^1 k(s,t) x(s)ds,$$ where $k:[0,1]\times [0,1]\to \mathbb{R} $ is ...
2
votes
2answers
92 views

Compact operators, injectivity and closed range

Let $X$ be a an infinite dimensional Banach space. $A\in B(X)$ is a compact operator. If its range $Im(A)$ is closed in $X$ then $A$ cannot be injective because $A:X\to Im(A)$ would be a compact ...
0
votes
1answer
54 views

Am I wrong ? (2)

Let $X=C[0,1]$ be the space of real continous functions on $[0,1]$. $X$ is a Banach space with the two norms $$|f|_\infty=\sup_{s\in[0,1]}|f(s)|$$ and ...
1
vote
1answer
40 views

$(X,|.|_A)$ is Banach implies $A$ is closed

Let $(X,|.|)$ be a Banach space. We know that if $A:X\to X$ is a closed operator then $(X,|.|_A)$ is a Banach space, where $|.|_A$ is the norm defined by $$|x|_A=|x|+|Ax|$$ Then using the "continuity ...
1
vote
1answer
49 views

Where am I wrong ??

Let $(X,|.|)$ be a Banach space. $A\in B(X)$ a bounded injective operator. Then we can define another norm on $X$ by $$|x|_A=|Ax|.$$ Since we have $$|x|_A\leq |A||x|$$ Then by the result of continuity ...
1
vote
0answers
40 views

Holomorphic Functional Calculus

Framework: Consider a Banach space: $$(E,\|\cdot\|)$$ Given an unbounded operator: $$T:\mathcal{D}(T)\to E\qquad\mathcal{D}(T)\subseteq E$$ together with its resolvent map: ...
0
votes
1answer
21 views

Injectivity and surjectivity of $\lambda I-A$.

Let us $A$ a square matrix, $\lambda\in \mathbb R^+$, $I$ identity matrix, R a operator, X Banach space. If $$(\lambda I-A) Ru=u \ \ (u\in X)$$ and $$R(\lambda I-A) u=u \ \ (u\in X)$$ then can we ...
12
votes
1answer
201 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
2answers
51 views

Is this operator bounded ??

Let $X$ be the Banach space $X:=\{ f\in C(\mathbb{R},\mathbb{R}),\sup_{t\in \mathbb{R}}|e^{-s^2}f(s)|<+\infty \}$ equipped with the norm $$|f|_X=\sup_{t\in \mathbb{R}}|e^{-s^2}f(s)|$$ I want to ...
3
votes
0answers
114 views

Differentiation of norm in Banach space (explanation of text needed)

Let $Y$ be uniformly smooth Banach space. Consider the convex $C^1$ functional $\Phi:Y \to \mathbb{R}$ defined $$\Phi(y) = \frac{1}{q}\Vert y \Vert^q_{Y}.$$ Its derivative $\varphi:Y \to Y'$ is a ...