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.

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$(X,|.|_A)$ 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 ...
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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 ...
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28 views

Application of operator theory in ODE and PDE

I am looking for references of applications of operator theory (especially spectral theory) in ODE, PDE and possibly SDE. I have learnt operator theory in the general set up, but only know little ...
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19 views

Normal bounded operator

Let $T$ be a bounded normal operator on a Hilbert space. Now I have to show that $T$ is self-adjoint if and only if $\sigma(T) \subset \mathbb{R}$. I already know that for an Abelian unital ...
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1answer
48 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 $X=C([0,1],\mathbb{R})$ ...
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38 views

Spectrum of an operator

Let $X=C([0,1],\mathbb{R})$ the Banach space of continuous real functions in $[0,1]$ equipped with the supremum norm. We define the operator $A$ for each $x\in X$ by $$(Ax)(t)=\int_0 ^t x(s)ds, \ \ \ ...
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382 views

is $T$ compact operator?

is $T$ compact operator? $T:C[0,1]\rightarrow C[0,1]$: $x(t)\mapsto x(t^2)$ where $t\in[0,1]$ with supremum norm Could you please help.
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72 views

Frechet/Gateaux differentiability of an integral operator L^2 --> R

Let $f: R \rightarrow R$ be a continuously differentiable function on the real numbers (if needed also infinitely many often differentiable). Define the Operator $F : L^2([0,1]) \rightarrow R$ for $x ...
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112 views

Fredholm operator norm

I have seen here, that the operator norm of a Fredholm operator $T_k(f)(s):=\int_0^1 k(s,t) f(t) dt $, where $k \in L^2([0,1]^2)$ and $f \in L^2([0,1])$ is not equal to the $L^2$ norm of the Kernel. ...
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43 views

I have to decide whether an operator is closed

So here is my problem, I have to decide whether the following operator is closed, $$\frac{\mathrm{d}}{\mathrm{d}x}:C^2([0,1])\subset C^0([0,1])\rightarrow C^0([0,1])$$ with the $||\cdot||_{\infty}$ ...
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Prove that $A\int_0^\infty S(t) u dt=\int_0^\infty S(t) A u dt$ if A is a closed operator

From Wikipedia: Closed linear operators are a class of linear operators on Banach spaces. They are more general than bounded operators, and therefore not necessarily continuous, but they still ...
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20 views

Proof of Strong Operator Convergence Theorem

Recall the theorem : $T_n \in B(X,Y)$ where $X,\ Y$ are Banachs, is strongly convergent iff (a) $ \parallel T_n \parallel $ is bounded (b) $T_nx$ is Cauchy where $x$ is in total subset ...
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1answer
68 views

Proving that $AB-BA=cI$ for nontrivial $c \in \mathbb{C}$

I have a homework question I can`t solve: Let $X$ be a normed linear space, $A,B \in B(X)$. Show that there exists no nontrivial $c \in \mathbb{C} $ such that $AB-BA=cI$. Thanks alot already guys! I ...
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21 views

Doubt on eigenvalues of normal operators

I'm trying to understand the solution of the following problem: $T$ is a normal operator. If $T( v)=\lambda v$, then $T^*(v)=\bar\lambda v$: The solution is: I didn't understand why we ...
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41 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: ...
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67 views

How to find if it is a compact operator

How to find if it is a compact operator: $F\colon C[0,1]\rightarrow C[0,1]$ : $x(t)\mapsto \int^1_0 \cos(t^2+s^2)x(s)ds$ Could you please help with this question.
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57 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 ...
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24 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 ...
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63 views

Exponentiation of imaginary operator

It is very easy to prove that if $D=\dfrac{d}{dx}$, then $(e^{nD}f)(x)=f(x+n)$ about $x=m$ in the real numbers. Proof: $$(e^{mD}f)=\sum^\infty_{n=0}\dfrac{D^nf}{n!}m^n\\ \implies ...
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71 views

Understanding the definition of a Integral

Definition: Let $X$ be a Banach space and $I$ the identity operator on $X$. A family $\{T(t)\}_{t\geq 0}$ of bounded linear operators from $X$ into $X$ is a semigroup of bounded linear operator on ...
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44 views

derivation of divergence from nabla operator

For a two dimensional orthogonal curvilinear coordinate system $(t_1, t_2)$, we have the position vector $r$, where $h_i = | \frac{\partial r}{\partial t_i} |$ are the scale factors and $a_i$ are the ...
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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 ...
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25 views

Cauchy Schwarz inequality with an operator

The standard Cauchy-Schwarz inequality is given by, $|\langle\Phi|\Psi\rangle|^2\le\langle\Phi|\Phi\rangle\langle\Psi|\Psi\rangle$ But now I'm intressted in what happens to ...
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54 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: ...
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1answer
32 views

Hermitian and Diagonal Matrix Norm inequality

I have a matrix inequality that I think is true, but I can't prove. $D_1$ and $D_2$ are diagonal matrices with non-negative entries. $M_1$ and $M_2$ are positive definite matrices. I want to show ...
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49 views

Question about finite rank operators

Let $X$ be a normed space, $\mathcal{F}(X)$ the algebra of all operators on $X$ with finite fank, then $\mathcal{F}(X)$ is the unique minimal ideal of $\mathcal{K}(X)$ the algebra of all compact ...
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202 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 ...
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2answers
52 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 ...
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an invariant of $C^{*}$ algebras

consider the following property (invariant) for complex $C^{*}$ algebras: "$T(x)=x^{*}$ is the only non zero $\mathbb{R}$-linear map on $A$ which satisfies $T(x)T(y)=T(yx)$." Questions: 1)Some ...
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58 views

Norm of the multiplication operator

Let $f \in L^\infty[0,1].$ It is clear that the norm of the multiplication operator $M_f : g \mapsto fg$ on $L^p[0,1]$ is $\|f\|_\infty.$ What happens in the noncommutative situation? Let us ...
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Verification of a contraction

Let $A\colon \text{dom}(A) \to \mathcal{H}$ be a densely defined symmetric operator on a Hilbert space $\mathcal H$. The symmetry implies that $$ \|(A + i)f\|^2 = \|Af\|^2 + \|f\|^2 \quad \text{for ...
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24 views

Covolution (space) over compact Lie groups

Let $G$ be a compact Lie group. Is there any way one can characterize the functions $\phi$ of the form $\phi=\psi\ast \psi^\ast$ in $C^\infty(G)$ where $\psi\in C^\infty(G)$? Here as usual ...
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40 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. ...
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35 views

Which operators commute with integration?

Leibniz proved that $D_y\cdot D^{-1}_x=D_{x}^{-1}\cdot D_y$, where $D_x=\frac{\partial}{\partial x}$. It follows that $D^n_y\cdot D^{-1}_x=D_{x}^{-1}\cdot D^n_y$ where $n \in \mathbb{N}.$ I've not ...
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30 views

Question about the operator norm on $\mathbb R^2$

So here is my question, I have to decide whether the following statement is true Let $T$ be an isomorphism on $\mathbb R^2$. Then $$\|T\|=\frac{1}{\|T^{-1}\|}$$ I am pretty sure that the statement ...
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1answer
57 views

Linear-Bounded Functional Proof

How can I prove that the following functional defined in $C[a,b]$ is both linear and bounded? $$ f: C[a, b] \to \mathbb R, $$ $$ f(x)=\int_{a}^{b}x(t)y_{0}(t)dt $$ for all $x = x(t) \in C[a,b]$, ...
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about analytic operator valued function

What's the definition of an analytic operator valued function ? Can we consider an analytic operator valued function as analytic function?
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275 views

The difference between hermitian, symmetric and self adjoint operators.

I am struggling with the concept of hermitian operators, symmetric operators and self adjoint operators. All of the relevant material seems quite self contradictory, and the only notes I have to go ...
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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 ...
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28 views

How would you define the square of the linear operator

If you define the linear operator norm of $A:X\to Y$ to be $$\|A\|_{op} = \inf\{C>0: \|Ax\|_Y \leq C\|x\|_X \text{ for all } x \in X \}$$ Then how would you define $\|A\|_{op}^2$? My guess is you ...
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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 ...
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Closed Graph Theorem

Let $(x_n)$ be a Schauder basis of $X$ and $(y_n)$ an equivalent one to $Y$. They are supposed to be equivalent, hence for every sequence $(a_n)$ the series $\sum_{n \in \mathbb{N}} a_nx_n$ converges ...
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Existence of invariant states in a $C^*$-algebra

Let $\mathcal{A}$ be a C*-algebra and $\{\tau_t\}_{t\in\mathbb R}$ a weakly-continuous group of *-automorphisms. I've read the claim (without proof) that for any state $\eta$ (that is $\eta$ is a ...
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25 views

Decreasing Function Projected onto Simplex

Consider $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$, defined as $f(x) := a x + b$, where $a<0$ and $b \in \mathbb{R}_{\leq 0}^{n}$. Note that $f$ is decreasing: $$ x \geq y \Longrightarrow f(x) ...
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40 views

can we say that $P_{X\oplus Y}=P_X+P_Y$?

if $X$ and $Y$ are two closed subspaces in H then can we say that $P_{X\oplus Y}=P_X+P_Y$? Here $P_X$ is orthogonal projection on X.
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about proof of linear operators

$T\colon H^1[0,1]\to L^2[0,1]$ $$ T_{x}(t)=x(t) $$ $H^1[a,b]$ is the space of all continuous differentiable functions with the norm $$ \|x\|_{H^1}=\left(\int_{a}^{b}x^2(t)dt+\int_{a}^{b}((x′(t))^2dt ...
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1answer
35 views

Zhou operator theory book, Kaplanskys formula

In Zhou's operator theory book, Kaplanskys formula has stated that if $P$ and $Q$ are projection in a von neumann algebra $A$ acting on $H$, then $P\vee Q-Q\sim P-P\wedge Q$. In the proof, it says ...
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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 ...
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44 views

Projection of smallest norm

This is related to the question: Compactness of set of projections If $X$ is a infinite dimensional Banach space, then any finite dimensional subspace $E$ is complemented, that's an immediate ...
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25 views

Saturated Monotone and Increasing Mappings

Let $A : \mathbb{R}^n \rightarrow \mathbb{R}^n$ be a monotone mapping, i.e., $$ \left( A(x) - A(y) \right)^\top \left( x-y\right) \geq 0 $$ for all $x,y \in \mathbb{R}^n$. Let $B : \mathbb{R}^n ...