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Let ‎‎$‎A‎$ ‎be a‎n ‎abelian ‎Banach ‎algebra ‎with ‎identity.‎ ‎

We know that

a) If ‎$‎I‎$ ‎is a‎ ‎maximal ‎ideal of ‎‎$‎A‎$‎, ‎then ‎there ‎is a‎ ‎non-zero ‎homeomorphism ‎$‎\varphi\colon A ‎\to ‎\mathbb{C}$ with $I=‎\ker‎\varphi.$‎‎ ‎‎‎

b) If‎‎‎ $‎\varphi\colon A ‎\to ‎\mathbb{C}‎$is a‎ ‎non-zero ‎homeomorphism, ‎then $\ker\varphi$ ‎is a‎ ‎maximal ‎ideal of $A.$

Now, ‎can ‎we ‎use ‎these ‎conditions ‎to ‎prove ‎the following two assertions: ‎

  1. ‎‎$‎ a ‎\in ‎A‎$ ‎is ‎invertible ‎if and ‎only ‎if there is $‎\varphi\in\Omega(A)‎‎$ ‎so that $\varphi(A)\neq0.$

  2. ‎‎$‎ a ‎\in ‎A‎$ ‎is ‎invertible ‎if ‎and only ‎if ‎‎$ ‎a‎$ ‎is not ‎contained ‎in a ‎proper ‎ideal of ‎‎$‎A‎$‎?

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The stated facts about the character space of $A$ are not necessary. In fact,

1: This statement is not true. Consider $A=C([0,1])$, define $f\in C([0,1])$ by$f(x)=x$ for $x\in[0,1]$, and let $\varphi\in\Omega(A)$ be evaluation at $1$: $$ \varphi(g)=g(1) \tag{$g\in C([0,1])$} $$ Then $\varphi(f)=1\neq0$, but $f$ is not invertible.

2: If $a$ is invertible, $I\subset A$ is a proper ideal, and $a\in I$, then $1=a^{-1}a\in I$, so $A=I$, a contradiction. Conversely, if $a\in A$ is not invertible, then the set $\{ca:c\in A\}$ is a proper ideal in $A$ (it does not contain the identity) that contains $a$.

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${}$2. Suppose that $a$ is invertible. Then $1=aa^{-1}$, so $A$ is the only ideal of $A$ containing $A$. Conversely, if $a\in A$ is contained in a proper ideal of $A$, then $ab\neq 1$ for all $b\in A$, so $a$ is not invertible.

${}$1. In unital Banach algebras, maximal ideals are closed. Use the fact that an ideal of a complex, abelian Banach algebra $A$ is maximal if and only if it is the kernel of some non-zero algebra homomorphism $A\to \mathbb{C}$. This gives you one implication. The other one is of course false.

See also this thread for discussion of necessity of your hypotheses.

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