# $(C(A,\mathbb{R}),\|\cdot\|_\infty)$ is a Banach Space when $A$ is Compact

For a normed vector space $$(V,\|\cdot\|_{V})$$, let $$A\subseteq V$$ be compact, and let $$(C(A,\mathbb{R}),\|\cdot\|_{\infty})$$ be the space of continuous functions from $$A$$ to $$\mathbb{R}$$ with respect to the sup-norm. How can I use specifically the Weierstrass M-Test, and facts about absolutely convergent series in normed vector spaces to show that $$(C(A,\mathbb{R}),\|\cdot\|_{\infty})$$ is a Banach Space?
Can I assume that I have an arbitrary absolutely convergent series $$\Bigl(\sum_{k=0}^n f_{k}\Bigr)_{n\in\mathbb{N}}$$ in $$(C(A,\mathbb{R}),\|\cdot\|_{\infty})$$, and then just let $$M_n=\|f_n\|_{\infty}$$ and then apply the M-Test to show that the sequence converges?

• Yes, that's the idea. Absolute convergence implying convergence is equivalent to completeness in normed linear spaces, see here: math.stackexchange.com/questions/2180369/… – Theo Bendit Apr 4 at 0:21
• If a sequence of continuous functions $(F_n)$ converges uniformly then the limit $F$ is continuous, this is a well-known exercice $|F(x)-F(a)|\le 2\|F-F_n\|_\infty + |F_n(x)-F_n(a)|$. For $n$ large enough $\|F-F_n\|_\infty < \epsilon/3$ and for $x$ close enough to $a$, $|F_n(x)-F_n(a)| < \epsilon/3$ – reuns Apr 4 at 0:24