The space $C(\bar{\Omega})$

This post tries to establish that the above mentioned space is Banach. The proof is straight-forward but it is here for reference sake! Let $\Omega \subset \mathbb{R}^d$ be an open set. Denote by $C(\bar{\Omega})$ the set of all bounded continuous functions $u: \Omega \to \mathbb{R}$ which can be continuously extended to $\bar{\Omega}$ in a bounded manner. (These are exactly the same functions which are bounded and continuous on $\bar{\Omega}$) Now, we put the norm (called the sup norm) $\|f\| = \sup_{x \in \bar{\Omega}} f(x)$ on this space. Clearly, $C(\bar{\Omega})$ is a vector space and $\|f\| = \sup_{x \in \bar{\Omega}} |f(x)|$ defines a norm on $C(\bar{\Omega})$. The only non-trivial thing needed to prove this is that $\sup_x (|f(x)| + |g(x)|) \leq \sup_x |f(x)| + \sup_x |g(x)|$ (note that these two values need not be equal).

$C(\bar{\Omega})$ with the norm defined above is a Banach space

Let $f_n$ be a cauchy sequence in $C(\bar{\Omega})$, i.e. for every $\epsilon > 0$, there exists $N$ such that for all $n,m > N$, we have $\|f_n-f_m\| < \epsilon$, i.e. $\sup_x |f_n(x) - f_m(x)| < \epsilon$. This gives that $f_n(x)$ is a cauchy sequence for each $x\in \bar{\Omega}$. As $\mathbb{R}$ is complete, $f_n(x)$ converges for each $x$. Let the limit be $f(x)$ (i.e. $f(x) = \lim_{n \to \infty} f_n(x)$)

Note that we have convergence $f_n$ to $f$ in sup norm. (Just take the limit $m \to \infty$ in the cauchy criterion). It remains to show that $f$ is a bounded continuous function on $\bar{\Omega}$. $$\sup_x|f(x)| \leq \sup_x|f(x) - f_n(x)| + \sup_x|f_n(x)|$$ For large enough $n$ the first term on the r.h.s can be made small and second term is finite as $f_n$ is bounded. Therefore $f$ is bounded.

As for continuity, we have $$|f(x)-f(y)| \leq |f(x)-f_n(x)| + |f_n(x)-f_m(x)| + |f_m(x)-f_m(y)| + |f_m(y)-f(y)|$$ The first, second and the fourth terms can be made small by choosing large enough $n,m$ independent of $x,y$. The third term can be made arbitrarily small by choosing $y$ sufficiently close to $x$ due to the continuity of $f_m$.