Spectrum of a bounded function

Question

Let $S$ be a non empty set and $B=\ell^{\infty}(S)$ the Banach algebra of bounded functions on $S$. Show that $$\sigma_B(f)=\overline {f(S)}.$$ I just know that $\lambda\in\mathbb{C}\setminus\sigma_B(f)\iff f-\lambda e$ is a unit in $B.$

Answer 1

If $y\in f(S)$, say $y=f(x)$ for $x\in S$, then the function $f-1y$ vanishes at $x$, so is not invertible (wrt pointwise multiplication), so $y\in\sigma_B(f)$. So $f(S) \subset \sigma_B(f)$, and in fact since $\sigma_B(f)$ is closed, $$\overline{f(S)} \subset \sigma_B(f).$$ Conversely, if $y\not\in \overline{f(S)}$, $f-1y$ vanishes nowhere, so is invertible (wrt pointwise multiplication). Morevoer the inverse is bounded because $y$ is bounded away from $f(S)$ (I'll let you check the details). This shows $$\sigma_B(f) = \overline{f(S)}.$$