# Pointwise Convergence of Lipschitz Functions is Uniform

Say we have $f_n:[a,b]\to \mathbb{R}$ , such that for all $n$ we have $|f_n(x)-f_n(y)|\leq L|x-y|$ and $f_n \to f$ pointwise.

Is the convergence uniform?

I started with an attempt to prove it by showing Cauchy Criterion for uniform convergence : for any $c\in[a,b]$ $$|f_n(x)-f_m(x)|=|f_n(x)-f_n(c)+f_n(c)-f_m(c)+f_m(c)-f_m(x)|\\\leq|f_n(x)-f_n(c)|+|f_n(c)-f_m(c)|+|f_m(c)-f_m(x)|\leq 2L|x-c|+|f_m(c)-f_n(c)|.$$

Now let $\epsilon>0$, there exists $n,m$ such that $|f_m(c)-f_n(c)|<\frac \epsilon 2$ from pointwise convergence. Also, if we take $x\in(c-\frac \epsilon {4L},c+\frac \epsilon {4L})$, we get $|f_n(x)-f_m(x)|<2L\frac \epsilon {4L}+\frac \epsilon 2=\epsilon$. So we have uniform convergence in $(c-\frac \epsilon {4L},c+\frac \epsilon {4L})$ for all $c\in[a,b]$. We can get finite cover of these covers, where we have uniform convergence, and take the maximum $N$ from all of those intervals to get uniform convergence in $[a,b]$.

However, My friend presented me with a possible counter-example : $f_n(x)=nxe^{-nx}$ in [0,1], where the derivative is bounded, and the convergence is only pointwise to 0, and not uniform.

I couldn't find where my proof fails, and Id be glad if someone can point it out for me.

The counter-example is wrong: the sequence $f_n(x)=nxe^{-nx}$ is not uniformly Lipschitz in $[0,1]$ since, as $n\to +\infty$, $$\frac{f_n(1/n)-f_n(0)}{1/n-0}=ne^{-1}\to +\infty.$$

Moreover your proof is correct. See also Given sequence of $L-$Lipschitz functions which converges pointwise, prove uniform convergence

• This means that $|f_n(x)-f_n(y)|\leq L_n|x-y|$? That is, that the Lipschitz constant is dependent on $n$?
– Sar
Commented Aug 3, 2018 at 17:58
• @Sar Exactly, and we have that $(L_n)_n$ is not bounded. Commented Aug 3, 2018 at 17:59
• And another question about the correctness of my proof, I wasnt sure that taking intervals dependent on $\epsilon$ is okay. On the one hand, we fix an $\epsilon$ at the start of the proof, however, as $\epsilon$ gets smaller, the interval does too, and at the limit it is just a point, Where uniform convergence makes no sense. Is my use of Heine-Borel nessecary to adress this issue? Or maybe there's another way to solve this?
– Sar
Commented Aug 3, 2018 at 18:01
• For an alternative proof see math.stackexchange.com/questions/2356399/… Commented Aug 3, 2018 at 18:05
• Thank you, much appriciated.
– Sar
Commented Aug 3, 2018 at 18:06

+1 for Robert's response...

Let $\varepsilon>0$ be given, and set $\delta=\min\left[\frac{\varepsilon}{4}, \frac{\varepsilon}{4L}\right]$. Since the collection of open balls $\mathcal{B}: = \{B(\, x, \delta) : x \in [a,b] \}$ is a cover for $[a,b]$, 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 $[a,b]$, 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 $x \in [a,b]$ we know there exists a positive integer $M_x \geq N$ so that $|\,f_{M_x}(x)-f(x)| < \frac{1}{4}\varepsilon$ (since $f_n \to f$ pointwise) and that $|x-x_j| < \delta$ for some $j \in \{1, \ldots, K\}$ (definition of a cover). We combine to notice that if $x \in [a,b]$ 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 $[a,b]$.