For $f \in C(X)$, if $\alpha(f+c)$ belongs to $\overline{\mathcal{A}}$, then $f$ also belongs to $\overline{\mathcal{A}}$ Let $\mathcal{A}$ be an algebra of continuous real-valued functions on a compact space $X$ that contains the constant functions. Let $f \in C(X)$ have the property that for some constant function $c$ and real number $a$, the function $a(f+c)$ belongs to $\overline{\mathcal{A}}$ (the closure of $\mathcal{A}$ w.r.t. the sup norm).  
I need to show that $f$ also belongs to $\overline{\mathcal{A}}$.
I know that by definition of an algebra of functions, if $f$, $g$ $\in \mathcal{A}$, then $fg \in \mathcal{A}$ as well. But, I have no yet proven that if $\mathcal{A}$ is an algebra, then $\overline{\mathcal{A}}$ is an algebra (I'm actually also working on that problem: asked a follow-up to somebody else's unanswered question about that here, if you'd care to help out with that), and thus can't multiply $a(f+c)$ by itself and hope to be able to squeeze an $f$ out of it somewhere.
So, essentially, I'm stuck. Help would be much appreciated! Thanks.

I'll assume $\alpha \neq 0$. As $\mathcal{A}$ is dense in its closure there exists $(g_n)_{n=1}^\infty \subset \mathcal{A}$ such that $g_n \to \alpha(f+c)$ uniformly as $n \to \infty$. Now, fix $\epsilon>0$. We can find $n_0$ large enough such that for all $n \geq n_0$, $$||g_n - \alpha(f+c)||_{\infty} <\epsilon \cdot |\alpha|$$ We then have $$||(\alpha^{-1}g_n -c) - f||_{\infty}<\epsilon$$ for $n$ large. As $\mathcal{A}$ is an algebra, the sequence $f_n := \alpha^{-1}g_n -c$ lies in $\mathcal{A}$. $f_n$ converges to $f$ uniformly, so $f \in \bar{\mathcal{A}}$.
You can also of course prove $\bar{\mathcal{A}}$ is an algebra itself. For that you would need to use compactness, but the proof is not bad.
