Let $f_n:[0,1] \rightarrow \mathbb{R}$ be a sequence of Lipschitz functions. Each $f_n$ has a Lipschitz constant equal to $M_n>0$.

Suppose that $f_n$ converges uniformly to a function $f$. Then $f$ is Lipschitz.

My attemp:

For all $x,y \in [0,1], x \neq y$:

$|f(x)-f(y)| \leq |f(x)-f_n(x)| + |f_n(x)-f_n(y)| + |f_n(y)-f(y)|$

For $n\geq n_0$, we have $|f(x)-f_n(x)|<|x-y|$ < and $|f(y)-f_n(y)|<|x-y|$ since $f_n$ converges uniformly to $f$.


$|f(x)-f(y)| \leq |f(x)-f_{n_{0}}(x)| + |f_{n_{0}}(x)-f_{n_{0}}(y)| + |f_{n_{0}}(y)-f(y)| \leq (2+M_{n_0})|x-y|$

Then $f$ is Lipschitz with constant equal to $2+M_{n_0}$

Am I right? Is there an easier way to solve this problem?

Thank you.

  • $\begingroup$ (Forget my earlier comment if you read it). you will need an upper bound for the $M_n$. $\endgroup$ – Thomas Feb 27 '16 at 13:56
  • $\begingroup$ Could you explain to me why? $\endgroup$ – Santos Feb 27 '16 at 14:01
  • 1
    $\begingroup$ The claim you have written down is not correct. The answers you received show this by providing counterexamples. These counterexamples all rely on the fact that the Lipshitz constants blow up. As soon as they are bounded, the claim is true. Also your approach to prove the claim is correct then, but incomplete. See my second comment to David's answer to find out how it can be completed. $\endgroup$ – Thomas Feb 27 '16 at 18:29

This is clearly false. For example, you know the theorem of Weierstrass that if $f$ is continuous on $[a,b]$ then there exists a sequence of polynomials $P_n$ so that $P_n\to f$ uniformly on $[a.b]$, right? Those polynomials are Lipschitz on $[a,b]$.

Edit: As mentioned in comments, it is true if the $M_n$ are bounded. In that case there exists $M$ so that $$|f_n(x)-f_n(y)|\le M|x-y|$$for all $x,y$ and $n$. So $$|f(x)-f(y)|=\lim_{n\to\infty}|f_n(x)-f_n(y)|\le M|x-y|.$$(So if the $M_n$ are bounded you don't even need to assume uniform convergence, just pointwise convergence. The point is that if the $M_n$ are bounded then the $f_n$ are equicontinuous; see the Arzela-Ascoli theorem.)

  • $\begingroup$ Now I'm conviced that the statement is false. But I still can't figure out what is wrong with my proof. $\endgroup$ – Santos Feb 27 '16 at 14:40
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    $\begingroup$ You say that for $n>n_0$ we have $|f(x)-f_n(x)|<|x-y|$ because $f_n\to f$ uniformly. You're using the definition of uniform convergence with $\epsilon=|x-y|$. That means $n_0$ depends on $x$ and $y$. The proof would work if you could show that there is one $n_0$ such that for $n>n_0$ you had $|f(x)-f_n(x)|<|x-y|$ for every $x$ and $y$, but that's not true. $\endgroup$ – David C. Ullrich Feb 27 '16 at 14:45
  • $\begingroup$ Thank you very much for your help. $\endgroup$ – Santos Feb 27 '16 at 14:48
  • $\begingroup$ @Santos The proof would also work if there were a bound on the $M_n$, as mentioned earlier. In this case $n_0$ may be chosen depending on $x,y$. $\endgroup$ – Thomas Feb 27 '16 at 16:22
  • $\begingroup$ Yes of course it works if $M_n$ is bounded. I think that your "depending" was a typo for "not depending"? The property he ascribes to $n_0$, namely $|f(x)-f_n(x)|<|x-y|$ for $n>n_0$, certainly cannot hold independent of $x$ and $y$; that would imply that $f=f_n$. But that's presumably not exactly what you had in mind. $\endgroup$ – David C. Ullrich Feb 27 '16 at 16:26

Thiis is not true. Look at $$f_n(x)=\sum_{k=1}^n{\sin(2^kx)\over k^2}.$$ These converge uniformly to a continous function. They are all Lipschitz, but the limit is a nondifferentialble "monster."


Another example: Define $f_n(x) = e^{-1/(nx)}x\sin(1/x)$ for $x \in (0,1],f(0)=0.$ Then each $f_n$ is continuously differentiable on $[0,1],$ hence is Lipschitz there. We have $f_n(x) \to x\sin (1/x)$ uniformly on $[0,1],$ but the limit function is not Lipschitz (proof: its derivative is unbounded on $(0,1].$)


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