UPDATE
I have a problem:
Let $\mathcal A$ be a commutative Banach algebra. Denote $G(\mathcal A)$ is the set of all invertible elements in $\mathcal A$.
Prove the following assertions:
a) $G(\mathcal A)$ is a group under multiplication in $\mathcal A$;
b) $G(\mathcal A)$ be an open set in $\mathcal A$;
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For a): $G(\mathcal A)=\{x \in \mathcal A \ \mid \exists y:=x^{-1} \in \mathcal A \ \text{such that} \ x • x^{-1} = x^{-1}• x = e\}$
I tried to use the definition of the group:
$G(\mathcal A) \ne \emptyset$: Because $e \in G(\mathcal A)$;
Associativity: Because $\mathcal A$ be a Banach algebra, so
i/ $x(yz)=(xy)z,\ \forall x,y,z \in \mathcal A$;
ii/ $x(y+z)=xy+xz,\ \ (x+y)z=xz+yz,\ \forall x,y,z \in \mathcal A$;
iii/ $(\alpha x)y=x(\alpha y)=\alpha(xy),\ \forall x,y,z \in \mathcal A, \ \forall \alpha \in \Bbb C$
Hence, for all $x, y$ and $z$ in $G(\mathcal A)$ then $(x • y) • z = x • (y • z)$.
- Closure
For all $x, y \in G(\mathcal A)$, the result of the operation, $x • y=y • x$, is also $\in G(\mathcal A)$.
- Inverse element:
For each $x \in G(\mathcal A)$, there exists an element $y:=x^{-1} \in G(\mathcal A)$ such that $x • x^{-1} = x^{-1}• x = e$, where $e$ is the identity element.
Therefore, $G(\mathcal A)$ is a group under multiplication in $\mathcal A \blacksquare$.
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For b): $G(\mathcal A)$ be an open set in $\mathcal A$;
I think that We need to prove that
$\exists U \subset G(\mathcal A) $, where $U$ is a neighborhood of $x \in G(\mathcal A)$
And use the implicit function theorem.
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We apply the theorem:
If $A$ is a Banach algebra, $x \in G(\mathcal A)$ and $h \in \mathcal A$ such that $\|h\| \le \dfrac{1}{2}\|x^{-1}\|^{-1}$ then $x+h \in G(\mathcal A)$.
Proof
- Since $\|h\| < \dfrac{1}{2}\|x^{-1}\|^{-1}$ we have: $$\left \| x^{-1}h \right \| \le \left \| x^{-1} \right \|\cdot \left \| h \right \|<\dfrac{1}{2}<1$$ Whence, $e+x^{-1}h \in G(\mathcal A)$.
On the other hand, we have performed: $$x(e+x^{-1}h)=x+h, x \in G(\mathcal A)$$ Hence, $x +h \in G(\mathcal A)$.
- Now, we'll show that, there exists an open ball with center $x \in G(\mathcal A)$ and radius $r=\dfrac{1}{2}\|x^{-1}\|^{-1}$ such that: $$B\left ( x;\dfrac{1}{2}\|x^{-1}\|^{-1} \right )\subset G\left ( \mathcal A \right )$$
Indeed, We take $y \in B\left ( x;\dfrac{1}{2}\|x^{-1}\|^{-1} \right )$ then $$\left \|y-x \right \| <\dfrac{1}{2}\|x^{-1}\|^{-1}\overset{\text{theorem (*)}}{\rightarrow} x+(y-x)=y\in G\left ( \mathcal A \right )\square $$
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I have forgotten anything in my solution above?
Any help will be appreciated! Thanks!
Hints for (b):
1) It suffices to prove that the identity element is in the interior.
2) Use a geometric series.