I tried to prove the following:
Let $A$ be a commutative non-unital complex Banach algebra and $\chi : A \to \mathbb C$ a character. Then $$ \sigma (a) = \{\chi (a) : \chi \in \Omega (A) \} \cup \{0\}$$
Could somebody please check my proof and tell me if it is correct? Your help is greatly appreciated.
Proof: Note that since $A$ is not unital the spectrum of an element $a$ is defined to be the spectrum of $a$ in the unitization $\widetilde{A}$.
$\supseteq$: Let $\chi $ be a character. Then for $a$ in $A$, $\chi (a-\chi(a)) = 0$ is not invertible hence $a-\chi (a)$ is not invertible in $\widetilde{A}$ hence $\chi (a) \in \sigma (a)$. Also note that since $A$ is not unital none of the elements can be invertible hence $0 \in \sigma (a)$ for all $a$.
$\subseteq$: Let $\lambda \in \sigma (a)$. Then $(a-\lambda)$ is not invertible in $\widetilde{A}$. Since $(a-\lambda)\widetilde{A}$ is a proper ideal of $\widetilde{A}$ it is contained in a maximal ideal $\mathfrak m$ of $\widetilde{A}$. This $\mathfrak m$ corresponds to exactly one non-zero unital homomorphism $\widetilde{\chi}:\widetilde{A} \to \mathbb C$ with $\mathrm{ker}(\widetilde{\chi}) = \mathfrak m$. Note that the restriction of $\widetilde{\chi}|_A$ is a character on $A$. Furthermore, for every $a \in A$, $\widetilde{\chi}|_A (a - \lambda) = 0 = \widetilde{\chi}|_A(a) - \widetilde{\chi}|_A(\lambda)$ so that $\lambda \in \{\chi(a) | \chi \in \Omega (A)\}$.
The claim is that for any $a \in A$, $\sigma(a) = \{\chi(a) \mid \chi \in \Omega(A)\} \cup \{0\}$. Hence, to prove this, you take any element $a \in A$, and show that for that fixed element $a$, $\sigma(a) \subseteq \{\chi(a) \mid \chi \in \Omega(A)\} \cup \{0\}$ and $\sigma(a) \supseteq \{\chi(a) \mid \chi \in \Omega(A)\} \cup \{0\}$. So, you must fix a single $a \in A$ at the start of your proof—the proof works because you could have taken any $a \in A$ at all—and work with this single chosen $a$ throughout. Now:
$\supseteq$: The only change is that you're working with the single element $a \in A$ that you fixed at the beginning of your proof. In particular, if your given $a$ were invertible in $\tilde{A}$, then $aa^{-1} = 1$, so that since $A$ is an ideal in $\tilde{A}$, $1_{\tilde{A}} \in A$, which is a contradiction.
$\subseteq$: Once you've found your character $\tilde{\chi}$, which, a priori, depends on $a$ and $\lambda$, observe that since $a-\lambda \in (a-\lambda)\tilde{A} \subset \mathfrak{m} = \ker\tilde{\chi}$, $0 = \tilde{\chi}(a-\lambda) = \tilde{\chi}(a) - \lambda \tilde{\chi}(1_{\tilde{A}}) = \tilde{\chi}(a) - \lambda$. Hence, if $\chi$ denotes the restriction of $\tilde{\chi}$ to $A$, $\lambda = \chi(a)$ as required. Note that $a$ was fixed at the very beginning of the proof, and that $\lambda$ was fixed throughout this part of the argument.