If we define Supnorm on $C(\bar{U})$ is
$$||f||_{C(\bar{U})}:=\sup_{x\in U} |f(x)|.$$ on $C(\bar{U})$ is Banach space
what I know is :
I'm proving this normed space
since $||f||_{C(\bar{U})}=\sup_{x\in U} |f(x)| =0 \iff f=0$
and $||af||_{C(\bar{U})}=\sup_{x\in U} |af(x)| =|a| ||f||$
finally $||f+g||_{C(\bar{U})}=\sup_{x\in U} |f(x)+g(x)|\le \sup _{x\in U} |f(x)|+\sup _{x\in U} |g(x)|$
how to prove completeness ..thank you
If $f_n$ is Cauchy and $\epsilon >0$ there exists $k$ such that $|f_n(x)-f_m(y)| <\epsilon$ for all $x \in \overline {U}$ $\cdots$ (1), for all $n,m \geq k$. In particular $\{f_n(x)\}$ is a Cauchy sequence for each $x$. Hence $f(x)=\lim f_n(x)$ exists. Letting $m \to \infty$ in (1) we get $|f_n(x)-f(y)| \leq \epsilon$ for all $x \in \overline {U}$ for all $n \geq k$. Hence $f_n \to f$ uniformly. Uniform limits of continuous functions are continuous. Hence $f \in C(\overline {U})$ and $\|f_n-f\|_{C(\overline {U})} \leq \epsilon $ for $n \geq k$. Hence $f_n \to f$ in the norm.