0
$\begingroup$

My question is: Assume $\{f_n(x)\}_{n=1}^\infty$ converge uniform to $f(x)$, e.g., $\max_{x\in R}|f_n(x)-f(x)|=O(n^{-1/2})$, and $f$ be $\alpha$-order Holder continuous. Can we show the Holder continuity of $f_n$ with $n$ large enough?

Thanks a lot.

1 Answers 1

0

I think no, we can't prove that. I guess I have a counterexample for $\alpha=1$ (Lipschitz case) and $f_n$ defined on a compact set. Let $f_n(x):[0,1] \mapsto \mathbb{R}$ be such that $f_n(x)=\sqrt{x}$ if $x \in [0,\frac{1}{2n}]$ and $f_n(x)$ is linear if $x \in (\frac{1}{2n},\frac{1}{n}]$, connecting points $(1/2n,\sqrt{1/2n})$ and $(1/n,0)$. $f_n$ is continuous for every $n$, and $$\max_{x \in [0,1]}|f_n(x)-0|=\max_{x \in [0,1]}|f_n(x)|=\sqrt{\frac{1}{2n}} $$, whence $f_n \rightarrow 0$ uniformly on $[0,1]$ and the zero function is of course Lipschitz. On the other hand, for every $n$, $f_n(x)$ is not Lipschitz since it isn't even differentiable at $x=0^{+}$, because the derivative is unbounded.

Added later note: $\sqrt{x}$ is not holder continuous for every $\alpha \in (1/2,1]$, while it is if $\alpha \in (0,1/2]$, so the same ideas provides counterexamples for $\alpha \in (1/2,1]$.

  • 0
    You gave me a good example! Actually, $f_n$ are still $\alpha$-order holder continuous. I mean whether we can show the holder continuity with a possible low order. – 2017-02-09
  • 0
    I can't understand your first statement: those $f_n$ are not $\alpha$-Holder continuous if $\alpha > 1/2$ (see final note)! I suspect that we should be able to find counterexamples also for $\alpha < 1/2$, i.e the initial proposition is false for every $\alpha$. I'll work on that soon. – 2017-02-09
  • 0
    Take a function which is not $\alpha$-Holder continuous for every $\alpha \in (0,1]$ in the interval $[0,1]$, but continuous and such that $f(0)=0$ (a classic example is $f(x)=1/\log(x)$ defined $0$ at zero, the value that provides continuity). Perform a similar trick as the one I used above and construct a sequence $f_n$ such that near zero coincides with $f(x)$, but converges to zero uniformly as $n$ approaches infinity. It will follow that $f_n$ is not holder continuous for every $\alpha$ and every $n$ (since it coincides with $f$ near zero), but the zero function is. – 2017-02-09
  • 0
    A final insight on the problem: generally speaking, theorems about regularity of sequences of functions and their limit usually work in the opposite direction: under some hypothesis, some "good" properties of the $f_n$ are transferred to the limit function $f$ (think about continuity, measurability, differentiability and so on), but it's harder to find theorems that ensure the regularity of the $f_n$, provided it is given for the limit function. For instance, we can also construct a sequence of discontinuous functions converging to a continuous one. – 2017-02-09
  • 0
    If you think that's a satisfactory answer, could you please accept it? Thank you. – 2017-02-09
  • 0
    Yes, your comments are correct, especially,the last insight about the regularity between sequences and their limit. – 2017-02-10
  • 0
    I mean, formally accept the answer :) there is a bottom you can click on top left of my answer with the classic symbol "V" ! Thank you, you're welcome. – 2017-02-10