Limit of cauchy sequence of uniformly continuous functions is uniformly continuous

I have posted the following question recently
Why do uniformly continuous functions form a Banach space (with the sup norm)?
I know now that space of uniformly continuous bounded functions with the sup norm is a Banach space (since it's complete).

But how can I prove that the limit of a Cauchy sequence in this space is indeed uniformly continuous.

The pointwise limit has to be the limit of a Cauchy sequence of uniformly continuous bounded functions. However I am not able to the fact that it is uniformly continuous. Please help me with this, (I am desperate).

The definition of uniform continuity is:
for every $\epsilon$ there exists a $\delta>0$ such that $|g(x)-g(y)|<\epsilon$ for all $x,y\in A$ s.t. $d(x,y)<\delta$.

• Obvious fact : you can't approximate (in the $\sup$ norm) a discontinuous function by a continuous function. Hence, if $\|f-f_n\|_\infty \to 0$ where $f_n$ is a sequence of continuous functions, then $f$ has to be continuous too. Finally, if you restrict to a compact set (say $[a,b]$), continuity implies uniform continuity. – reuns Sep 18 '16 at 20:17
• Now when the set isn't compact, you can show the uniform continuity by considering the translate operator $T_\epsilon g(x) = g(x+\epsilon)$. For any $\epsilon > 0$, choose $n$ such that $\|f_n - f\| < \epsilon$, and since $f_n$ is uniformly continuous, we can choose $\delta$ such that $\|T_{\delta}f_n - f_n\| < \epsilon$. Hence $\|T_{\delta} f - f\| = \|T_{\delta} f -T_{\delta} f_{n} + T_\delta f_n - f_n + f_n - f\|\le \|T_{\delta}f - T_{\delta}f_{n} \|+ \|T_{\delta}f_{n}-f_n \| + \|f_{n} - f\| < 3\epsilon$ – reuns Sep 18 '16 at 20:40