# If $f_n \rightarrow f$ pointwise, then $f_n \rightarrow f$ uniformly, given $\exists C$ such that $|f_n(x) - f_n(y)| \le C|x-y|$.

Let $f_n \in C[0,1], \forall n \in \Bbb N$, and let $f:[0,1] \rightarrow \Bbb R$. Suppose, $\exists C$ such that $|f_n(x) - f_n(y)| \le C|x-y|$, $\forall n \in \Bbb N$. Then, if $f_n \rightarrow f$ pointwise, then $f_n \rightarrow f$ uniformly.

This has been my approach so far:

Since $f_n$ is continuous on $[0,1]$, and $[0,1]$ is compact, $f_n$ is uniformly continuous on $[0,1]$.

Fix $\epsilon \gt 0$. Then, $\exists \delta \gt 0$ such that $|f_n(x) - f_n(y)| \lt \epsilon, \forall x,y\in [0,1]$ for which $|x-y| \lt \delta$.

So, choose $x,y \in [0,1]$ such that $|x-y| \lt \delta$. We have: $$|f_n(x) - f(x)| \le |f_n(x) - f_n(y)| + |f_n(y) - f(x)| \lt \epsilon + |f_n(y) - f(x)|$$

Now, this is where I'm stuck as if I apply triangle inequality again to get:

$$|f_n(x) - f(x)| \lt \epsilon + |f_n(y) - f_n(x)| + |f_n(x) - f(x)| \le \epsilon + C|x-y| + |f_n(x) - f(x)|$$ I end up erasing the LHS and getting nowhere.

I have seen an approach where we "temporarily" fix $x$ and set $N$ satisfying the pointwise of convergence of $f_n$ to $f$ at the point $x$. But this doesn't really make sense to me, as then we cannot show uniform convergence since our inequality chain above will depend on the $N$, which itself depends on the chosen $x$.

Any hints are greatly appreciated!!

First show that $$\tag1|f(x)-f(y)|\le C|x-y|.$$ Indeed, if we assume $|f(x)-f(y)|> C|x-y|$ for some $x,y$, then we find $n\gg 0$ with $f_n(x)\approx f(x)$ and $f_n(y)\approx f(y)$ so closely that still $|f_n(x)-f_n(y)|>C|x-y|$. More formally, for sufficiently big $n$, we have both $|f_n(x)-f(x)|<\frac12(|f(x)-f(y)|- C|x-y|)$ and $|f_n(y)-f(y)|<\frac12(|f(x)-f(y)|- C|x-y|)$ and therefore $$C|x-y|\ge |f_n(x)-f_n(y)|\ge |f(x)-f(y)|-|f_n(x)-f(x)|-|f_n(y)-f(y)|>C|x-y|,$$ contradiction.

With the bound $(1)$ we see that $|f_n(a)-f(a)|<\frac12\epsilon$ implies $$|f_n(x)-f(x)|\le |f_n(x)-f_n(a)|+|f_n(a)-f(a)|+|f(a)-f(x)|\le2C|a-x|+\frac12\epsilon,$$ hence $|f_n(x)-f(a)|<\epsilon$ for all $x$ with $|x-a|<\frac\epsilon{4C}$.

Now cover the domain $[0,1]$ with finitely(!) many open intervals of width $\frac\epsilon{2C}$ and make sure $n$ is big enough to have $|f(x)-f_n(x)|<\frac\epsilon2$ in the finitely many interval midpoints. Then $|f(x)-f_n(x)|<\epsilon$ for all $x\in[0,1]$.

Remark: Perhaps unexpectedly, we dis not make use of the fact that the domain $[0,1]$ is comapct, only of the fact that it is bounded (with minor adjustments it would still work for a disconnected bounded domain). In other words, the claim remains true if the functions are instead defined on $(0,1)$, for example.

Hint: (I hope it makes sense)

1. It suffices to show $(f_n)$ is Cauchy.

2. By the dominated-condition, we can plug finitely many points $0=x_0<x_1<x_2<...<x_k=1\in [0,1]$ such that $x_{i}-x_{i-1}$ are small enough so that for any $x,y \in [x_{i-1},x_i]$, $|f_n(x) -f_n(y)|$ is small enough for all $n$.

3. $(f_n(x_i))$ is Cauchy for each $i$.

4. Now use $|f_n(x) -f_m(x)|\leq |f_n(x) -f_n(x_i)| +|f_n(x_i)-f_m(x_i)| +|f_m(x_i) -f_m(x)|$, and notice that $|f_n(x_i)-f_m(x_i)|$ are "uniformly small" for large $n,m$.

Let $\varepsilon>0$ be given, and set $\delta=\min\left[\frac{\varepsilon}{4}, \frac{\varepsilon}{4C}\right]$. Since the collection of open balls $\mathcal{B}: = \{B(\, x, \delta) : x \in [0,1] \}$ is a cover for $[0,1]$, we may refine $\mathcal{B}$ to a finite subcover, say $\{B(\,x_1, \delta), \, \ldots, \, B(\,x_K, \delta)\}$ (Heine-Borel Theorem). Since $f_n$ converges pointwise on $[0,1]$, for each point $x_j \: \left(\,j=1,\ldots, K \right)$ there exists a positive integer $N_j$ so that $$\left|\, f_n(x_j) - f_m(x_j) \right| < \frac{\varepsilon}{4} \text{ whenever } n, m \geq N_j \,.$$ We set $N = \max [N_1, \ldots, N_K]$. Now, given $x \in [0,1]$ we know there exists a positive integer $M_x \geq N$ so that $|\,f_{M_x}(x)-f(x)|\leq \frac{1}{4}\varepsilon$ (since $f_n \to f$ pointwise) and that $|x-x_j| < \delta$ for some $j \in \{i \in \mathbb{N} : i \leq K\}$ (definition of a cover). We combine to notice that if $x \in [0,1]$ and $n \geq N$, then

\begin{aligned} \left|\,f_n(x)- f(x) \right| & \leq \left| \,f_n (x)- f_n(x_j) \right| + \left| \,f_n (x_j)- f_{M_x}(x_j) \right| + \left|\, f_{M_x} (x_j)- f_{M_x}(x) \right| + \left|\, f_{M_x}(x)- f(x) \right| \\ & < \frac{\varepsilon}{4} + \frac{\varepsilon}{4} + \frac{\varepsilon}{4} + \frac{\varepsilon}{4} = \varepsilon \, . \end{aligned} Therefore, $f_n \to f$ uniformly on $[0,1]$ (definition of uniform convergence).