New answers tagged banach-algebras
2
We have, as square roots are self-adjoint $\def\abs#1{\left|#1\right|}$
\begin{align*}
\abs v^*\abs v &= [(v^*v)^{1/2}]^*(v^*v)^{1/2}\\
&= (v^*v)^{1/2}(v^*v)^{1/2}\\
&= v^*v
\end{align*}
1
The spectrum of an invertible element $x$ is $\sigma(x^{-1})=\{\lambda^{-1}: \lambda\in \sigma(x)\}$
If $x$ is invertible, $x \cdot v = \lambda v$ is equivalent to $v= x^{-1}x \cdot v= \lambda x^{-1} \cdot v$, that is $x^{-1} \cdot v=\lambda^{-1} v$.
3
Since $\|T^{-k}\| \leq M$ for all $k \geq 1$, we have by definition $\|T^{-k}(x)\| \leq M\|x\|$. $\|x\| = \|id(x)\| = \|T^{-n}(T^{n}x)\| \leq \|T^{-n}\|\|T^n x\| \leq M\|T^n x\|$. It seems we are done.
1
Using nets does not generate any problem. Suppose that $f_j\to f$ pointwise in $M(A)$. Given $x,y\in A$,
$$
f(xy)=\lim f_j(xy)=\lim f_j(x) \,f_j(y)=f(x)f(y).
$$
The non-obvious equality is the last one. The only difference with the case of sequences is that a convergent net need not be bounded; but it is eventually bounded, and so the proof that the limit of ...
1
In fact $A\in\mathcal{A}$ is invertible iff $m(A)\neq 0$ for all $m\in\Omega(\mathcal{A}):=\mathcal{M}(\mathcal{A})\setminus\{0\}$. As the consequence
we have the chain equivalences
$$
\begin{align}
\lambda\in\mathbb{C}\setminus\sigma(A)
&\Longleftrightarrow A-\lambda 1\in\mathrm{Inv}(\mathcal{A})\\
&\Longleftrightarrow \forall m\in ...
2
Let
$$
B=\begin{bmatrix}0&1\\ 1&0\end{bmatrix}, A=\begin{bmatrix}1&0\\0&2\end{bmatrix}.
$$
Then
$$
e^{-zA}Be^{zA}=\begin{bmatrix}e^{-z}&0\\0&e^{-2z}\end{bmatrix}\,\begin{bmatrix}0&1\\ 1&0\end{bmatrix}\,\begin{bmatrix}e^{z}&0\\0&e^{2z}\end{bmatrix}=\begin{bmatrix}0&e^z\\ e^{-z} & 0 \end{bmatrix}.
$$
So
$$
...
1
Hint: Plug $\dfrac{BC + \beta C}{\Vert BC + \beta C\Vert}$ into the norm computation of $(A,\alpha)$ and compare to plugging in $C$ to the norm computation of $(A,\alpha)(B,\beta)$.
1
Otherwise, using power series, $(A-\lambda I)B$ would be invertible for some $B\in\mathcal A$, hence by commutativity, so would be $A-\lambda I$.
3
You are correct; any involution on a semisimple Banach algebra must be continuous for the reason you give.
The question then remains: Does there exists a Banach algebra with a discontinuous involution? The answer can be found in the following article:
Does there Exist More than One Banach *-Algebra with Discontinuous Involution?
R. S. Doran,
The ...
Top 50 recent answers are included
