# Tagged Questions

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) ...

26 views

### Riemann Lebesgue Lemma for locally compact ableian groups

I'm looking for a reference (or proof here) of the generalized RL Lemma for LCAGs. One result is that if $G$ is a LCAG then $$\{ \hat{f} : f \in L^1(G)\} \subset C_0(\Gamma)$$ where $\Gamma$ is the ...
31 views

### How to generate weak topology by the family of semi-norm?

How to generate a weak topology by the family of semi-norm $$\{P_i \colon A \rightarrow B, i \in I\}$$ in which $$P_i(a)=\lVert ia\rVert+\lVert ai\rVert$$ when $A$ is a Banach algebra and $I$ is ...
28 views

### Functoriality in $K$-theory for $C^*$-algebras or Banach algebras

I'm trying to clear up some confusion I'm having over how one establishes functoriality in $K$-theory for $C^*$-algebras or Banach algebras. Let me stick to $K_0$. Given a *-homomorphism (or bounded ...
29 views

### Norms on unitization of a Banach algebra

Let $A$ be a non-unital Banach algebra, and let $A^+$ be its unitization. Then $||(a,z)||_1=||a||+|z|$ is a Banach algebra norm on $A^+$. Can we also make $A^+$ a Banach algebra by giving it the norm ...
39 views

32 views

17 views

### Definition of $K_1(A)$ for a Banach algebra $A$

When defining $K_1(A)$ for a Banach algebra $A$, one may consider $\bigcup_{n\in\mathbb{N}}\{x\in GL_n(A^+):x\equiv I_n\mod M_n(A)\}$ and take the quotient by the component containing the identity, or ...
46 views

### Clarification on Wolfram Mathworld's explanation of the connection between Gelfand Transform and Fourier Transform

http://mathworld.wolfram.com/GelfandTransform.html In the definition, what does $x$, $\hat x(\phi)$, and $\phi$ represent exactly if we were to consider definition of the Fourier transform? Can ...
32 views

### Derivation into dense ideal of Banach algebras

Let $A$ be a Banach algebra and $I$ be an ideal of $A$. A derivation $D:A\to I$ is a linear bounded map, with the following property: $$D(ab)=aD(b)+D(a)b,\qquad a,b\in A.$$ Suppose that $I$ is dense ...
20 views

### What are the appropriate morphisms for forming inductive limits of Banach algebras?

For Banach algebras, if we take the morphisms to be bounded homomorphisms, the inductive limit construction may not result in a Banach algebra. (I remember seeing this fact but I don't know a specific ...
31 views

### How does the product of sets of complex numbers give a character?

I'm working through this "Introduction to Banach Algebras" and just after proposition 8.2 they say: If $A$ is a commutative Banach algebra, $a\in A$ and $\phi\in M(A)$, then $\phi(a)\in sp(a)$. ...
56 views

### Distributivity of projective tensor product over direct sum

Let $I$ is a non-empty set and $\{A_i\}_{i\in I}$ is a family of Banach algebras and $B$ is a Banach algebra. Define \ell^1-\oplus_{i\in I}A_i=\{a=\{a_i\}_{i\in I}: \|a\|_1=\sum_{i\in I}\|a_i\|<\...
39 views

### How are $M_n(A)$ and $M_n(\mathbb{C})\otimes A$ identified as Banach algebras?

When $A$ is a Banach algebra, one can make $M_n(A)$ into a Banach algebra by equipping it with various norms. One can also make the algebraic tensor product $M_n(\mathbb{C})\otimes A$ into a Banach ...
92 views

### Which Hilbert space is isometrically isomorphism with $B(E)$ for some Banach space $E$.

Consider $H$ as a Hilbert space. How can I find a Banach space $E$, for that, $H=B(E)$ where $B(E)$ is the set of bounded linear operator on $E$? (At least under some conditions on $H$) Also if the ...
148 views

### Is every Hilbert space a Banach algebra?

Let $H$ be a Hilbert space. Could we say that, always there is a multiplication on $H$, that makes it into a Banach algebra? If not, under which conditions does it exist?
Let $L^1 (\mathbb{Z})$ be the space of all functions $f:\mathbb{Z}\rightarrow \mathbb{C}$ such that $\left\{\|f\|=\sum_{k\in \mathbb{Z}}|f(k)|<\infty\right\}$. Clearly, $L^1 (\mathbb{Z})$ is a ...