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 $[0,1]$.


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 $\leftarrow$


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