Sequence of Lipschitz continuous functions approaching Holder continuous function

Question

Let $\{f_n\}$ be a sequence of uniformly bounded Lipschitz functions with Lipschitz constant $L_n$, i.e. for each $n\in\mathbb{N}$ $$ |f_n(x) - f_n(y)| \leq L_n |x-y| $$ It is easy to see if the Lipschitz constants are bounded, then a subsequence will converge uniformly to a Lipschitz function. If the Lipschitz constants are unbounded do we know any conditions that will allow a subsequence to converge to an $\alpha$-Holder continuous function?

Answer 1

Here is a statement that may answer your problem. If not, the method can be adapted (for example, by replacing geometric sequences with powers of $n$).

Let $(f_n)_{n \ge 0}$ be a sequence of functions from $\mathbb{R}$ to $\mathbb{R}$, which converges uniformly to some function $f$ Assume that the functions $(f_n)_{n \ge 0}$ are Lipschitz, with Lipschitz constants $(L_n)_{n \ge 0}$. Assume that we have constants $\alpha>1$, $\beta>1$, and $A>0$, $B>0$ such that for every $n \ge 0$, $$||f_n-f||_\infty \le A\alpha^{-n} \textrm{ and } L_n \le B\beta^n.$$ Then $f$ is locally Holder with exponent $\gamma = \ln\alpha/\ln(\alpha\beta)$. More precisely, for every $x$ and $y$ in $\mathbb{R}$ $$|y-x| \le 2A/B \implies |f(y)-f(x)| \le 2\alpha(2A)^{1-\gamma}B^\gamma|y-x|^\gamma.$$

Proof. Let $x$ and $y$ be two distinct real numbers. For every $n \ge 0$ \begin{eqnarray*} |f(y)-f(x)| &\le& 2||f-f_n||_\infty + |f_n(y)-f_n(x)|\\ &\le& 2A\alpha^{-n} + B\beta^n|y-x|. \end{eqnarray*} Set $g(t) := 2A\alpha^{-t} + B\beta^t|y-x|$. We want to choose $n \ge 0$ such that $g(n)$ is as small as possible.

The unique solution $t_0$ of the equation $2A\alpha^{-t} = B\beta^t|y-x|$ is given by $(\alpha\beta)^{t_0} = (2A/B)|y-x|^{-1}$ so $\beta^{t_0} = (2A/B)^{1-\gamma}|y-x|^{\gamma-1}$ and $g(t_0) = 2(2A)^{1-\gamma}B^\gamma|y-x|^\gamma$.

If $|y-x| \le 2A/B$, then $t_0 \ge 0$, so $|f(y)-f(x)| \le g(\lfloor t_0 \rfloor) \le \alpha g(t_0)$.
We are done.

Remark: I did not compute the exact minimum of $g$, but this is almost optimal. Indeed, the functions $t \mapsto 2Aa^{-t}$ and $t \mapsto Bb^t|y-x|$ are positive, the first is decreasing from $\infty$ to $0$ whereas the second one increasing from $0$ to $\infty$, hence the value $g(t_0)$ is at most twice the minimum of $g$.