3
$\begingroup$

I have a general question about sequence of functions $(f_n)$ from $[0,1]$ to reals which are non-uniformly Lipschitz (i.e., each has a Lipschitz constant $D_n$ but we don't know if it is bounded) and converge uniformly to a function with Lipschitz constant $D$. I have been thinking whether if this was sufficient to say that the set $\{D_n, n \in \mathbb{N}\}$ is bounded. I have seen here examples which give examples of non-Lipschitz functions converging to Lipschitz ones so it seems this idea may not be correct. And indeed I was able to build a counter example which is

$f_n(x) = nxe^{-n^nx}$

The derivative of these functions at $0$ is $n$, so there are elements of this sequence with arbitrarily large Lipschitz constant (and each $f_n$ is indeed Lipschitz since their derivative is continuous in $[0,1]$). Moreover for these functions the maximum is achieved at $x=\frac{1}{n^n}$ with value $\frac{1}{en^{n-1}}$ so they uniformly converge to $0$.

Thus this builds a counterexample. Now my question is could one add some more assumptions to such a sequence so that the Lipschitz constants would be bounded (of course no easy assumptions like the functions being differentiable and their derivatives converging etc, it should be a topological property).

Thanks

|cite|improve this question
$\endgroup$
  • $\begingroup$ Do you mean that the set $\{D_{n}:n\in\mathbb{N}\}$ is bounded? Each $D_{n}$ is bounded in your counter-example. $\endgroup$ – T. Eskin Feb 24 '13 at 14:43
  • $\begingroup$ Yes I meant the set $\{D_n : n \in \mathbb{N}\}$ to be bounded. For the unbounded case there is already the example $\frac{1}{n}\sqrt(x)$ $\endgroup$ – Sina Feb 24 '13 at 14:45
  • $\begingroup$ This counter-example seems to do the job. And as you noted, it is easy to make a sequence of non-Lipschitz functions converging uniformly to a Lipschitz function. Take e.g. $f_{n}(x)=\frac{1}{n}$ for $x\in\mathbb{Q}$ and $f_{n}(x)=0$ for $x\in\mathbb{R}\setminus\mathbb{Q}$. Then $f_{n}\to 0$ uniformly, but each $f_{n}$ is not even continuous. $\endgroup$ – T. Eskin Feb 24 '13 at 14:51
0
$\begingroup$

If I understand correctly, the question is: given that $f_n\to f$ and $f$ is Lipschitz, what additional assumptions do we need to conclude that $(f_n)$ is a uniformly Lipschitz sequence?

Concerning

it should be a topological property

I remark that Lipschitz-ness is not a topological property, it is a metric property. If we are on a topological space without a metric, the concept of being Lipschitz does not exist.

Here is a metric assumption: the sequence $(f_n)$ converges in the Lipschitz norm $$ \|f\|_{\rm Lip} = |f(x_0)|+\sup_{x\ne y}\frac{|f(x)-f(y)|}{d(x,y)} \tag1$$ The definition of norm (1) involves choosing a base point $x_0$ in our metric space; the term $|f(x_0)|$ is necessary so that the constant functions get nonzero norm.

Let's check. Suppose $\|f_n-f\|_{\rm Lip}\to 0$ and $f$ is Lipschitz. Then there is $N$ such that for all $n\ge N$ we have $\|f_n-f\|_{\rm Lip}\le 1$. Hence $\|f_n\|\le 1+\|f\|_{\rm Lip}$, which is a uniform upper bound on the Lipschitz constants of $f_n$.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Thanks it does not easily solve my problem (like some other easier property to check but it gives some new insight). And I guess this is the one of the most you can expect without having derivatives. $\endgroup$ – Sina Feb 28 '13 at 9:31

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.