Questions tagged [banach-algebras]

A Banach algebra is an algebra over the real or complex numbers which is equipped with a complete norm such that |xy| ≤ |x||y|. The study of Banach algebras is a major topic in functional analysis. If you are about to ask a question on C*-algebras or von Neumann algebras please use (c-star-algebras) or (von-neumann-algebras) instead (or in addition). Further related tags: (operator-algebras), (operator-theory), (banach-spaces), (hilbert-spaces).

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Continuity of adjoint of algebra homomorphism

Consider unital (complex) abelian Banach algebras $A$ and $B$ with corresponding ideal spaces $\Sigma_A$ and $\Sigma_B$. Suppose $\rho:A\to B$ is a continuous algebra homomorphism that maps $1_A$ to $...
Oskar Vavtar's user avatar
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1 answer
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Uniqueness of non-zero homomorphism in abelian Banach algebra

Consider $\mathbb{C}^3$ equipped with product $\bullet$ given as $$(x_1,x_2,x_3)\bullet(y_1,y_2,y_3)~=~ (x_1y_1,x_1y_2+x_2y_1,x_1y_3+x_2y_2+x_3y_1)$$ and norm $\|(x_1,x_2,x_3)\|=|x_1|+|x_2|+|x_3|$. $A:...
Oskar Vavtar's user avatar
1 vote
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Define multiplication on $L^1(\mathbb{R})$

I want to define multiplication on $L^1(\mathbb{R})$, of course, convolution is an allowed multiplication which makes $(L^1(\mathbb{R}),*)$ is a Banach algebra, I want to know if there are other ...
InnocentFive's user avatar
1 vote
0 answers
63 views

For a self-adjoint element $a$ in a unital Banach $*$-algebra, is it true that $\|a\| \mathbb{1} -a \geq 0$?

In a Banach $*$-algebra positive elements can be defined as finite sums of the sort $\sum\limits_k b_k^*b_k$, is it true then that, for a self-adjoint $a$, the expression $\|a\|\mathbb{1} -a$ is a ...
Felipe Dilho's user avatar
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What are the maximal ideals in the unitization of Banach algebras? [closed]

Under what conditions the complex Banach algebra A is a maximal ideal in its unitization(where unitization of A is $‎‎A^\sharp\simeq A\oplus\mathbb{C}$)? Indeed, we know that Banach algebra A is a ...
E Nasrabadi's user avatar
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Character space of abelian Banach algebra is weak* closed in the unit ball of dual space.

Say $A$ is an abelian Banach algebra and $\Omega(A)\subset A^* := B(A, \mathbf{C})$ is the set of non-zero algebra homomorphisms. Consider now $\Omega(A)$ with the weak$^*$ topology. I am trying to ...
Anton Odina's user avatar
3 votes
1 answer
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$\mathcal{C}^2[0,1]$ is a Banach Algebra

The following is problem 13 here. Consider functions in $\mathcal{C}^2[0,1]$ and $a,b>0$. In this case, if we define: $$\lVert f \rVert:=\lVert f \rVert_\infty+ a \lVert f' \rVert_\infty +b \lVert ...
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2 answers
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Existence of a non-zero continuous function vanishing at infinity.

I am currently reading the book '$C^*$-Algebras and Operator Theory' by Gerard J. Murphy and I have trouble understanding two statements on page 4. Let $\Omega$ be a locally compact Hausdorff space ...
Anton Odina's user avatar
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1 answer
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Property of the Gelfand transform

I'm reading this script and I got stuck on the last Let $X$ be the set of all non-zero multiplicative linear functionals on the unital commutative Banach algebra $\mathcal{A}$ then $\mathrm{sp}(\hat{A}...
Davide Modesto's user avatar
2 votes
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Exponential boundedness of a strongly continuous semigroup $(T_t)_{t>0}$.

Let $T = (T_t)_{t>0} $ be a strongly continuous semigroup (of bounded operators) on a Banach space $E$, i.e, $ \lim_{t \rightarrow z} \|T_tx- T_{z}x \|, \forall z >0, \forall x \in E $. Note ...
Jeffrey Jao's user avatar
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An isolated point in the spectrum must be a point in the approximate point spectrum??

Is the following claim and proof correct?? I can't seem to find an error. Let $X$ be a complex Banach space, and $A\in \mathcal{B}(X)$ a bounded operator. Then, if $\lambda\in \sigma(A)$ is isolated ...
user760's user avatar
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Definition of Poles in spectral theory

In usual complex analysis, a pole $z_0\in \mathbb{C}$ of a function $f$ that is holomorphic on a punctured disk $0<|z-z_o|<R$ is defined as a zero of function $\frac{1}{f}$. But in spectral ...
user760's user avatar
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$L^1(G)$ is a Banach *-Algebra [duplicate]

Let G be a locally compact group with left Haar measure $\mu$. In Principles of Harmonic Analysis, it's affirmed that $L^1(G)$ is a Banach *-Algebra, where the multiplication operation is convolution ...
Pedro Lourenço's user avatar
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Trying to understand the functional calculus of a unitary operator.

I am trying to understand the theorem on this Wikipedia article. It is stated as follows. Theorem. Let $x$ be a normal element of a $C^*$-algebra A with an identity element e. Let $C$ be the $C^*$-...
caffeinemachine's user avatar
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Need help understanding proof that $\hat{x} (\hat{A}) = \sigma(x)$

From Functional Analysis: Spectral Theory, by V. S. Sunder, page 97. Let $A$ denote a unital commutative Banach algebra, then $\hat{x} (\hat{A}) = \sigma(x)$, where $\hat{A}$ is the collection of all ...
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Inequality between exponential in unital Banach Algebra

So, I am trying to solve the following statement: Suppose that $\sigma(a) \subseteq \{\lambda \in \mathbb{C}: \text{Re} \lambda < 0\}$. Show that there exist $M, \omega \in \mathbb{R}$ with $\omega ...
Tipeg's user avatar
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Is the given space is a banach algebra and find it’s maximal ideal space?

Let $ \mathcal{A} =\left\{f \in C(\mathbb{T}): \forall n \in \mathbb{N}^{+}, \int_{0}^{2 \pi} f\left(e^{i \theta}\right) e^{i n \theta} \mathrm{d} \theta=0\right\}$.prove $ \mathcal{A}$ is a banach ...
hicozhao's user avatar
6 votes
2 answers
459 views

Example of operator that commutes with a given multiplication that is not itself a multiplication.

I am aware that given a diagonal matrix $D$ whose diagonal entries are all distinct, any matrix $A$ that commutes with $D$ must be itself diagonal. I am also aware that this result does not generally ...
SecretlyAnEconomist's user avatar
3 votes
1 answer
207 views

Decomposition of the spectrum of a closed Banach subalgebra

In the book An Introduction to Operator Algebras by Kehe Zhu the following theorem is stated: Theorem. Suppose $A$ is a closed subalgebra of a unital Banach algebra $B$ with $\hat{1}\in A$. If $x \in ...
ferolimen's user avatar
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Maximal ideal space of an $L^\infty$ space is extremely disconnected

I'm reading the book An Introduction to Operator Algebras by Kehe Zhu, in a such book all the algebras are assumed to be unital and associative. Once said that, the autor proves that if $A$ is a ...
ferolimen's user avatar
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Let $I$ be a semiprime ideal of $\mathbf{A}$. Suppose $a\mathbf{A}a \subseteq I$ for some $a$. Show that $a \in I$.

Let $\mathbf{A}$ be an algebra. An ideal $I$ of $\mathbf{A}$ is called semiprime if $\mathbf{A}/I$ has no nonzero nilpotent ideal. Can someone supply a proof of following fact? Let $I$ be a ...
Math Lover's user avatar
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3 votes
1 answer
197 views

Maximal ideal space of the $n$-dimensional ball algebra

Let $B_n \subset \mathbb{C}^n$ be the $n$-dimensional open ball of radius $1$ centered at the origin. Let $A$ be the set consisting of holomorphic functions in $B_n$ which are continuous in $\overline{...
SCarlsen's user avatar
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How to salvage this quick proof of the exponential sum law in Banach algebras

From the power series definition of the exponential function in the real or complex numbers you can quickly prove the sum rule by considering the function $$ f(x) = \exp(a + b - x) \exp(x), $$ proving ...
sudgy's user avatar
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1 answer
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Resolvent inequality follows it's boundness

Following is from my lecture notes. I have resolvent map $R(\cdot,x): \mathbb{C} \rightarrow A$, where $x\in A$ Banach algebra. Domain is whole complex numbers because for the sake of contradiction it ...
user3342072's user avatar
1 vote
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Banach subalgebra generated by commutative set equals the centralizer of its centralizer

Let $A$ be a unital Banach algebra and $S\subset A$ a subset consisting of commuting elements. Let $C(S)$ be the centralizer of $S$, that is, $y \in C(S)$ iff $ys=sy$ for all $s \in S$. I have already ...
ferolimen's user avatar
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The set of homomorphism on $C_\infty(X)$ for a locally compact $X$

Let $X$ be a locally compact Hausdorff space, which is not compact. $C_b(X)$ is a Banach algebra of all bounded continuous functions with the sup norm. Let $C_\infty(X)$ be the Banach algebra of ...
Anna  Vakarova's user avatar
1 vote
1 answer
57 views

The set of unital homomorphisms

Let $X$ be a locally compact Hausdorff space, which is not compact. $C_b(X)$ is a Banach algebra of all bounded continuous functions with the sup norm. Let $C_\infty(X)$ be the Banach algebra of ...
Anna  Vakarova's user avatar
2 votes
0 answers
38 views

Prove that dual of a unital Banach algebra is nonempty using resolvent and Liouville's theorem

Let $\mathcal{A}$ be a unital Banach algebra over complex numbers $\mathbb{C}$. For every $a \in \mathcal{A}$, let $\sigma(a)$ be the spectrum of $a$. Define the resolvent of $a$ to be $ R(a,z) = (...
Anna  Vakarova's user avatar
0 votes
1 answer
87 views

The algebra of all bounded linear operators acting on a complex Banach space is a prime algebra.

Let $X$ be a Banach space over the complex field $\mathbb{C}$ and $\mathcal{B}(X)$ the algebra of all bounded linear operators on $X.$ I want to show that $\mathcal{B}(X)$ is a prime algebra. My ...
Akhter's user avatar
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2 votes
1 answer
66 views

Connected component of the constant function $1$ in $C(S^1,\mathbb{C})$

Let $C(S^1,\mathbb{C})$ be the space of continuous functions $f:S^{1} \subset \mathbb{C} \longrightarrow \mathbb{C}$ endowed with the supremum norm, that is, $ \lVert f \rVert= sup_{z \in S^{1}} |f(z)|...
ferolimen's user avatar
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0 answers
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Abstract index is an invariant of Banach algebras

I just started to read the book An introduction to Operator Algebras by Kehe Zhu. In this book the author defines the index group of a Banach algebra $A$ (which is assumed to be a unital algebra) as ...
ferolimen's user avatar
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3 votes
1 answer
213 views

Let $\mathcal{B}_0(X, Y)$ is a Banach space. Does this imply that $Y$ is a Banach space?

Consider $$\mathcal{B}_0(X, Y) =\\{T\in\mathcal{B}(X,Y): \overline{T(B_X[0,1])}\subset Y \text{compact}\\}$$ where $B_X[0, 1]=\{x\in X: \|x\|\le 1\}$ Claim: $\mathcal{B}_0(X, Y)$ is a Banach space ( ...
Sourav Ghosh's user avatar
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1 vote
1 answer
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If all unital Banach algebras contain maximal ideals, then why doesn't $M_n(\mathbb{C})$ have proper ideals?

In Murphy's textbook on C*algebra, he writes: So all unital Banach algebra should contain maximal ideals, but on a StackExchange post, there is an explicit example $M_n(\mathbb{C})$ of an unital (non-...
Bill's user avatar
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2 votes
1 answer
33 views

What are the most general structures that admit a singular value decomposition for each element?

I was looking at *-algebra and C*-algebra and as far as I understood, the latter guarantees each element to have a singular value decomposition, or at least, that every element $a$ is such that $u a v$...
NYG's user avatar
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0 answers
43 views

Multiplicative linear functional on B* algebra

Let $\mathcal{B}$ be a $B^*$ algebra and $\phi$ is a multiplicative linear functional on $B$. Is it true that $\phi(a^*)=\overline{\phi(a)}$? Do we need any additional assumption here? Thank you in ...
user490489's user avatar
7 votes
1 answer
93 views

Wiener Lemma for matrix valued functions

The ordinary Wiener lemma states that if $f(x):=\sum_{n \in \mathbb{Z}} a_n \exp(inx)$ ($\sum |a_n|<\infty$) and if $f(x)\neq 0$ everywhere, then $g:=1/f$ can also be written as $$g(x)= \sum b_n \...
P.Jo's user avatar
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1 answer
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show that $A=\{f \in C^1 [0,1]\}$ is semi simple

Let $A=C^1 ([0,1])$ with the norm $$ \|f\|=\|f\|_{\infty}+\|f'\|_{\infty}, \quad f\in C^1([0,1]) $$ I want to show that $A$ is semi simple. Recall that A is semisimple if $Rad(A)=0$ where $Rad=\...
gigggadag's user avatar
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0 answers
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Structure of the solutions of the matricial ODE: $\dot U(t) = -U^2 - BU$.

Let $A$ and $B$ be a fixed $n \times n$ positive semi-definite matrices. Given an $n \times n$ matrix $U$, define another $n \times n$ matrix by $F(U) := U^2 + U B$. Consider the ODE $$ \dot U(t) = -F(...
dohmatob's user avatar
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1 vote
1 answer
59 views

Spectrum of a bounded function

Let $S$ be a non empty set and $B=\ell^{\infty}(S)$ the Banach algebra of bounded functions on $S$. Show that $$\sigma_B(f)=\overline {f(S)}.$$ I just know that $\lambda\in\mathbb{C}\setminus\sigma_B(...
gigggadag's user avatar
1 vote
0 answers
40 views

Completion of a Banach algebra in the strict topology

Recall that a Banach algebra is faithful when $aA=(0)$ implies $a=0$ and $Aa=(0)$ implies $a=0$. A faithful (not necessarily unital) Banach algebra $A$ embeds in its multiplier algebra $\mathit{Mult}(...
Amaru's user avatar
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0 answers
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Computation of a derivative in a Banach algebra

Let A be a complex unital Banach algebra with the unit e and φ : A → A a continuous antilinear mapping such that φ(exp(x)) is invertible, for every x in A. Let f be the following map: f(λ) = [φ(exp(λx)...
Chadi Benchakroun's user avatar
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0 answers
60 views

Extending a Banach algebras morphism to the multipliers

(The question below is inspired from the question Extending a $C^*$-algebras morphism to the multipliers and its answer.) A Banach algebra A is said to be faithful when for each $a\in A$, $aA=(0) \...
Amaru's user avatar
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1 vote
1 answer
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Examples of free noncommutative Banach algebras

In abstract algebra, we have the notion of free algebra. But does the concept of free Banach algebras also make sense? Suppose $\mathcal{A}:=k\langle X_1, \cdots, X_n \rangle$, where $n>1$, is a ...
user760's user avatar
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0 votes
1 answer
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Surjective $*$-homomorphism is automatically bijective?

I had to prove some statements for two unital $C^*$ algebras $A, B$ and $\Phi: A \to B$ a surjective $*$-homomorphism. I started out the proof by stating By the rank-nullity theorem, $A$ and $B$ are ...
BrakkoFTW's user avatar
1 vote
0 answers
33 views

How do $C^*$-algebras results behave in the real case?

I have been studying $C^*$-algebras and operator theory in general and I have some questions. A common example of a $C^*$-algebra is $M_n(\mathbb C)$ the $n\times n$ complex matrices. Since $M_n(\...
Luis Dias's user avatar
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0 votes
1 answer
87 views

Show the Banach algebra is not a $C^*$-algebra

Working with the Banach algebra $\ell^1(\mathbb{Z})$ and the involution $f^* (n)=\bar{f(-n)}$, I want to show this is not a $C^*$ algebra. I know I need to find some function $f$ such that $||f^* f||...
user avatar
0 votes
1 answer
69 views

Prove $C(X)$ is a Banach algebra by first showing for all $x$, $\vert fg(x)\vert \le \Vert f \Vert \cdot \Vert g \Vert$

I have to prove that $C(X)$ is a Banach algebra. I know that $C(X)$ is a Banach space and that $\langle C(X),+,*\rangle$ is a ring. I know I need to show that $\Vert fg\Vert \le \Vert f \Vert \cdot \...
Tacosi's user avatar
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1 vote
1 answer
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Where is the mistake in computing $\sigma(T)$?

Suppose that $T:\ell_{2}\to \ell_{2}$ is an operator, such that $$(Tx)_{k} = \dfrac{k}{k+1}x_{k+1}$$ Find $\sigma(T)$. Suppose that $|\lambda|>1$. Let $x_{k}\in \ell_{2}$. Then $$\dfrac{kx_{k+1}}{...
matheg's user avatar
  • 498
5 votes
2 answers
73 views

Source and/or detailed proof for $\exp(\lambda (A + B)) = \exp(\lambda A)\exp(\lambda B)$ for commuting elements in a Banach algebra

Every time I need to revisit the proof that $\exp(\lambda (A + B)) = \exp(\lambda A)\exp(\lambda B), \lambda \in\mathbb{K}$ for commuting elements in a Banach algebra, I find myself struggling to ...
Cartesian Bear's user avatar
1 vote
1 answer
133 views

Spectrum of normal operators on a Hilbert space

I would like to understand the following characterization of normal operators on a Hilbert space, for which I could not find a proof. Let $T$ be a normal operator. Let $\lambda \in \mathbb{C}$, then $\...
kiyopi's user avatar
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