Let $\mathbb{B}$ be a commutative unital complex Banach algebra. Let $f_1,\ldots ,f_n$ be elements of $\mathbb{B}$ and let $I$ be the closed ideal generated by $f_1,\ldots ,f_n$. My question is whether we can express each element of $I$ as a finite linear combination of $f_1,\ldots ,f_n$ with coefficients in $\mathbb{B}$.
Either a proof or a counterexample will be appreciated. Thank you.
No, this is not true. For instance, let $\mathbb{B}=C([0,1])$ and let $I$ be the closed ideal generated by the function $f(x)=x^2$. Then $I$ is the ideal of all functions that vanish at $0$ (this follows from Stone-Weierstrass, for instance). In particular, the function $g(x)=x$ is in $I$. However, $g$ is not a multiple of $f$ in $\mathbb{B}$.
What's going on here is that you can find a sequence $(h_n)$ such that $(fh_n)$ converges to $g$, but $(h_n)$ does not converge at all. The functions $h_n$ are unbounded but they are only getting large near $0$, and multiplying them by $f$ makes them bounded since $f$ is small near $0$.