let $(X, \| \|$) be the normed linear space of bounded uniformly continuous real valued functions defined on $R$. I need to prove that X is complete (under sup norm).
My attempt:
Let {$f_n$} be a Cauchy sequence in X converging to $f$ in the completion of X.
I am checking the uniform continuity of f below.
$|f(x)-f(y)|$ = $ |f(x) - f_n(x) + f_n(x) - f_n(y) + f_n(y) - f(y)|\;$ $\leq\; |f(x) - f_n(x)|+ |f_n(x) - f_n(y)| + |f_n(y) - f(y)|$
By choosing an x in X, I get an $N_1(x,\epsilon/3)$ such that $|f(x) - f_n(x)| \leq \epsilon/3$ $\forall$ n $\geq$ $N_1$. However, for the next two terms I don't know how to deal with them. I can't fix either y nor n because both are interdependent. I want to use the fact that $f_n$-s are bounded somehow into it.
I can prove that f is bounded on X using $|f(x)| \leq |f(x) - f_n(x)| + |f_n(x)|$ once I get uniform convergence of $f_n$.
If I could get help in proving that $f_n$ is uniformly convergent on $f$, then it would be great.
First, $f_n$ Cauchy in $||\cdot||_{\infty}$ norm implies that for every $x\in \mathbb{R}$ $f_n(x)$ is Cauchy, so by completeness of $\mathbb{R}$ $f_n\to f$ pointwise.
Next as $f_n$ is Cauchy given $\varepsilon>0$ there is $N$ such that $m,n\ge N$ implies $||f_m-f_n||_{\infty}<\varepsilon/3$. Taking limit as $m\to\infty$ we get
$$ ||f-f_n||_{\infty}\le\varepsilon/3 $$
so $f_n\to f$ uniformly.
You also need $f$ to be uniformly continuous. For this note that since $f_n$ is uniformly continuous for all $n$, there is $\delta>0$ such that $$|x-y|<\delta \implies |f_N(x)-f_N(y)|<\varepsilon/3 $$
So if $|x-y|<\delta$
$$ |f(x)-f(y)|\le |f(x)-f_N(x)| + |f_N(x)-f_N(y)| + |f(y)-f_N(y)| <\varepsilon/3+\varepsilon/3+\varepsilon/3=\varepsilon $$
Note that you get $|f(x)-f_N(x)|<\varepsilon/3$ and $|f(x)-f_N(y)|<\varepsilon/3$ from uniform convergence (recall how we chose $N$)