3
$\begingroup$

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 $|x-x_i| <\frac{\epsilon}{3L}: |f_n(x)-f(x)|<\frac{\epsilon}{3} \tag{3}$

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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – user251257 Jul 12 '17 at 18:05
  • $\begingroup$ @user251257 Not homework, just a problem I found and wanted to solve. What mistake you found? $\endgroup$ – Itay4 Jul 12 '17 at 18:07
  • 1
    $\begingroup$ 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. $\endgroup$ – user251257 Jul 12 '17 at 18:15
  • $\begingroup$ So it should be $\frac{\epsilon}{6}?$ $\endgroup$ – Itay4 Jul 12 '17 at 18:18
  • $\begingroup$ $\epsilon/6/L$. Or just assume without loss generality $f = 0$ and $L = 1$... $\endgroup$ – user251257 Jul 12 '17 at 18:26
3
$\begingroup$

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 \begin{equation} \left|\, f_n(x_j) - f_m(x_j) \right| < \frac{\varepsilon}{3} \text{ whenever } n, m \geq N_j \,. \end{equation} 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 -_-.

$\endgroup$
  • $\begingroup$ 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 ? $\endgroup$ – Itay4 Jul 13 '17 at 4:32
  • 1
    $\begingroup$ No, I understand that part comes from the statement you made above. I'm asking about $|f_m(x_j)-f_m(x)|<\frac{\epsilon}{3}$ $\endgroup$ – Itay4 Jul 13 '17 at 4:59
  • $\begingroup$ 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 $|x-x_j|<\delta$. $\endgroup$ – Matt A Pelto Jul 13 '17 at 5:04
  • $\begingroup$ 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. $\endgroup$ – user251257 Jul 13 '17 at 19:23
  • $\begingroup$ "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. $\endgroup$ – Matt A Pelto Jul 13 '17 at 22:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.