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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!

kimtahe6
  • 1,806

2 Answers2

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Proof b)

suppose $a\in G(A)$. We have to show that there is $r>0$ such that $\parallel x-a\parallel <r$, then $x\in G(A)$. Take $r_{0}=\frac{1}{\|a^{-1}\|}$, we show that $S_{r_{0}}(a)\subseteq G(A)$ . If $\parallel x-a\parallel < \frac{1}{\|a^{-1}\|}$, then we prove that $x\in G(A)$. For $\parallel a^{-1}x-e\parallel=\parallel a^{-1}x-a^{-1}a\parallel=\parallel a^{-1}(x-a)\parallel \leq \parallel a^{-1}\parallel \parallel (x-a)\parallel \leq \parallel a^{-1}\parallel \parallel \frac{1}{a^{-1}}\parallel=1$. So $a^{-1}x$ is invertible, thus $x$ is invertible, i.e $x\in G(a)$. So $S_{r_{0}}(a)\subseteq G(A)$.

Hint:Let $A$ is Banach space. If $a\in A$ and $\parallel a\parallel <1$, then $a\in G(A)$.We used from this hint in the proof of question.

Alireza
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Hints for (b):

1) It suffices to prove that the identity element is in the interior.

2) Use a geometric series.

Robert Israel
  • 448,999