# Why do uniformly continuous functions form a Banach space (with the sup norm)?

Let $A$ be a metric space, and $U_{b}(A)$ the set of all bounded and uniformly continuous functions $g:A\rightarrow \mathbb{F}$.
Let $||\cdot||$ be the sup norm with $||g||=\sup_{x\in A}|g(x)|$.

How do I prove that $U_b(A)$ is a Banach space?

I know that being a Banach space means that $U_b(A)$ given the sup norm must be complet, i.e. each Cauchy sequence of functions $(g_n)_n$ must converge.
Further 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. $d(x,y)<\delta$.
I was thinking (assuming that $\mathbb{F}$ is complete), that we can fix a $a\in A$ and say that $(g_n(a))_n$ converges to a limit $g(a)$ in $\mathbb{F}$, if $(g_n)_n$ is Cauchy in $U_b(A)$. Then we can take $g$ to be the limit of $(g_n)_n$.

Is this a correct way of thinking? How do I finish my proof and where do I use the uniform continuity?

EDIT:
Uniform limit:
Let $\epsilon > 0$. Let $N$ s.t. for $n,m \geq N$ we have $\|g_n-g_m\| < \epsilon$. Then for all $n \geq N$

$$|g_n(x) - g(x)| = \lim_{m \to \infty} |g_n(x) - g_m(x)| \leq \epsilon$$

for all $x$ and so $\|g_n- g\| \leq \epsilon$.

EDIT 2:
I am having difficulties showing the uniform continuity of $g$. Any and all help would be greatly appreciated!

• This is the correct way. Remains to show that the pointwise limit $g\colon A\to \Bbb F$ is $\in U_b(A)$ – Hagen von Eitzen Sep 17 '16 at 13:00
• @HagenvonEitzen But I also have to show that g is the pointwise limit, how would I do that? – user368886 Sep 17 '16 at 13:02
• You defined $g$ as the pointwise limit, didn't you? – Hagen von Eitzen Sep 17 '16 at 13:08
• You know that $\lim_{n\to\infty}g_n(a)=g(a)$ for all $a\in A$. You need to show that $g$ is bounded. You need to show that $g$ is uniformly continuous. You need to show that $\lim_{n\to \infty}\|g_n-g\|=0$. – Hagen von Eitzen Sep 17 '16 at 13:22
• Let $\varepsilon > 0$ be given. Pick an $n$ such that $\lVert g - g_n\rVert < \varepsilon/3$. Use the uniform continuity of $g_n$. – Daniel Fischer Oct 3 '16 at 11:57