# Given sequence of $L-$Lipschitz functions which converges pointwise, prove uniform convergence

Let $f_n:[a.b]\rightarrow \mathbb{R}$ be sequence of $L-$Lipschitz functions, that is: $$\forall x,y\in[a,b]: |f_n(x)-f_n(y)|\leq L|x-y|$$ Suppose $f_n \rightarrow f$ pointwise, prove $f_n \rightrightarrows f$

I have all the parts of the puzzle for the proof, and I'm trying to put them all together, I'm using this in my answer.

I would appreciate is you could correct my proof, and if you have an alternative proof, I would be more then happy to see it.

My proof:

Let $\epsilon>0.$

$f_n$ are uniformly continuous on $[a,b]:$

$\tag{1} \exists \delta>0\ \forall x,y\in[a,b]: |f_n(x)-f_n(y)|<\frac{\epsilon}{3}$

$f$ is also $L-$Lipschitz:

$\tag{2} \forall x,y\in[a,b]:|f(x)-f(y)|<L|x-y|=\frac{\epsilon}{3}$

Let us set a partition of $[a,b]$ such as Stephen Montgomery-Smith suggests:

Pick points $x_1,\dots,x_m \in [a,b]$ which are distance $\frac{\epsilon}{3}$ from each other.

For each $1 \le i \le m$, find a number $N_i$ so that for all $n \ge N_i$ we have $|f_n(x_i)-f(x_i)| \le \epsilon/3$.
Let $N = \max_{1 \le i \le m} N_i$

Now given any $x \in [a,b]$, pick $1 \le i \le m$ such that $|x-x_i| <\frac{\epsilon}{3L}: |f_n(x)-f(x)|<\frac{\epsilon}{3} \tag{3}$

\begin{align}|f_n(y)-f(y)| &=|f_n(y)-f_n(x)+f_n(x)-f(x)+f(x)-f(y)| \\ &\leq |f_n(y)-f_n(x)| + |f_n(x)-f(x)|+|f(x)-f(y)| \\ &< \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon \\\end{align}

• If it is homework, you should perhaps elaborate some details. For example, why is (2) is true? Why do the $x_i$’s exist? Why can you pick that particular $i$? In fact there is a minor mistake. However your idea is not bad. – user251257 Jul 12 '17 at 18:05
• @user251257 Not homework, just a problem I found and wanted to solve. What mistake you found? – Itay4 Jul 12 '17 at 18:07
• you have assumed $|x_k - x_j| < \epsilon/ 3$ only. That does not imply $|x-x_i| < \epsilon/3/L$ for a suitable $i$ if $L> 1$. Also, $f_n - f$ is not necessarily $L$-Lipschitz. However, certainly $2L$-Lipschitz. – user251257 Jul 12 '17 at 18:15
• So it should be $\frac{\epsilon}{6}?$ – Itay4 Jul 12 '17 at 18:18
• $\epsilon/6/L$. Or just assume without loss generality $f = 0$ and $L = 1$... – user251257 Jul 12 '17 at 18:26

Let $\varepsilon>0$ be given, and set $\delta=\min\left[\frac{\varepsilon}{3}, \frac{\varepsilon}{3L}\right]$. Since the collection of open balls $\mathcal{B}: = \{B(\, x, \delta) : x \in [a,b] \}$ is a cover for $[a,b]$ we may find a finite subcover, say $\{B(\,x_1, \delta), \, \ldots, \, B(\,x_M, \delta)\}$ (Heine-Borel Theorem). Since $f_n$ converges pointwise on $[a,b]$, for each point $x_j \: \left(\,j=1,\ldots, M \right)$ we may find a positive integer $N_j$ so that $$\left|\, f_n(x_j) - f_m(x_j) \right| < \frac{\varepsilon}{3} \text{ whenever } n, m \geq N_j \,.$$ Setting $N = \max [N_1, \ldots, N_M]$ shows that

\begin{aligned} \left|\,f_n(x)- f_m(x) \right| & \leq \left| \,f_n (x)- f_n(x_j) \right| + \left|\, f_n (x_j)- f_m(x_j) \right| + \left|\, f_m(x_j)- f_m(x) \right| \\ & < \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon \; \: \text{ whenever } \, n,m \geq N \text{ and } x \in [a,b] . \end{aligned}

Since $\mathbb{R}$ is complete, it follows that the sequence of functions $\{\,f_n\}_{n=1}^\infty$ converges uniformly on $[a,b]$ (Cauchy Criterion).

I prefer uniform convergence first, ask $\,f$ questions later -_-.

• Shouldn't the pointwise convergence give you information about the distance between $f_n$ and $f$ and not between elements in the sequence? Also, how do you justify the third part in the inequality ? – Itay4 Jul 13 '17 at 4:32
• No, I understand that part comes from the statement you made above. I'm asking about $|f_m(x_j)-f_m(x)|<\frac{\epsilon}{3}$ – Itay4 Jul 13 '17 at 4:59
• Since the finite collection of open balls $\{B(\,x_1, \delta), \, \ldots, \, B(\,x_M, \delta)\}$ covers $[a,b]$, we know that $x \in B(x_j, \delta)$ for some $1 \leq j \leq M$. Hence $|x-x_j|<\delta$. – Matt A Pelto Jul 13 '17 at 5:04
• Since you rely on the Cauchy criterion, you actually need that $C[a,b]$ is complete. This is true because $[a,b]$ is compact and $\mathbb R$ is complete. This makes thing unnecessarily complicated. – user251257 Jul 13 '17 at 19:23
• "This is true because $[a,b]$ is compact and $\mathbb{R}$ is complete." You note this like it is news to me. Anyways you may write an equally formal rendition that instead of completeness uses the limit function $f$ explicitly, then we may talk about more complicated. For the record, I am already told the limit function $f$ exists for all $x \in [a,b]$. The compactness of $[a,b]$ will also be used in either rendition of the proof. Now obviously there are plenty of proofs on here that use the limit function $f$. So you may always go look at those instead of mine if you don't like it. – Matt A Pelto Jul 13 '17 at 22:09