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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

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  • $\begingroup$ 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
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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 $f_n$.

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  • $\begingroup$ 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

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