# Prove that {$f_n$} converges uniformly on $[0,1]$ if and only if {$f_n$} is equicontinuous on $[0,1]$ and converges pointwise on $[0,1]$.

Let {$$f_n$$} $$\subseteq C([0,1])$$. Prove that {$$f_n$$} converges uniformly on $$[0,1]$$ if and only if {$$f_n$$} is equicontinuous on $$[0,1]$$ and converges pointwise on $$[0,1]$$.

$$\rightarrow$$

Suppose $$f_n$$ converges uniformly on $$[0,1]$$. Then, since it is uniformly convergent $$\forall\epsilon > 0, \exists N \in \mathbb{N}$$ such that $$\forall n \geq N$$, $$|f_n(x)-f(x)|< \frac{\epsilon}{3}$$ , $$\forall x \in X$$.

Now, for each $$j \leq N$$ the function $$f_j$$ is uniformly continuous so there exist $$\delta_j > 0$$ such that $$d(x, y) \leq \delta_j$$ implies $$|f_j (x) − f_j (y)| < \epsilon$$. The limit $$f$$ is also uniformly continuous, so there exists $$\delta' > 0$$ such that $$|f(x) − f(y)| < \frac{\epsilon}{3}$$ whenever $$d(x, y) < \delta'$$ . Set $$\delta =$$ min$$(\delta',$$ min $$\delta_j)>0$$. If $$d(x,y) < \delta \leq \delta',$$ then for $$n > N$$

$$|f_n(x)-f_n(y)| \leq |f_n(x)-f(x)| + |f(x)-f(y)|+|f_n(y)-f(y)| < \epsilon$$

and thus $$f_n$$ is equicontinuous.

How do we show it converges pointwise? and proving $$\leftarrow$$