Let $\alpha \in (0,1]$. A function $f: [0,1]\rightarrow \mathbb{R}$ is defined to be $\alpha$-Hölder continuous if

$$ N_{\alpha}(f)=\sup\{ \frac{|f(x)-f(y)|}{|x-y|^\alpha} : x,y\in[0,1] \ \ \ \ x\neq y \} < \infty $$

(a) Suppose $\{f_n\}$ is a sequence of functions from $[0,1]$ to $\mathbb{R}$ such that for all $n=1, 2, 3, \cdots $ we have $N_{\alpha}(f_n)\leq 1$ and $|f_n(x)|\leq 1$ for all $x\in [0,1]$. Show that $f_n$ has a uniformly convergent subsequence.

(b) Show that (a) is false if the condition "$N_{\alpha}(f_n)\leq 1$" is replaced by "$N_{\alpha}(f_n) < \infty$".

End of question.

My problem is part(b) confused me. See, for part (a) $f_n$ is uniformly bounded on compact set $[0,1]$ and equicontinuous, hence Arzela Ascoli theorem can be applied and the conclusion followed.

The way I showed equiconinuity is this,

Given $\epsilon >0$, take $\delta < \epsilon^{\frac{1}{\alpha}}$, hence if $|x-y|<\delta$, $|f_n(x)-f_n(y)|<\epsilon$ for all $n\in \mathbb{N}$ and $x\in[0,1]$.

Now suppose that in Part (b), we let $N_{\alpha}(f_n)=M<\infty$, isn't it that if we take $\delta< (\frac{\epsilon}{M})^{\frac{1}{\alpha}}$ then equicontinuity still holds and Arzela Ascoli theorem still apply? Why not? What is the example in (b)? I am really confused. Hope someone can help me with this. Thanks.

  • 2
    $\begingroup$ The way the question is formulated you cannot conclude $N_\alpha(f_n)\le M$. You can only assume $N_\alpha(f_n)\le M_n$, but $M_n$ may be unbounded. $\endgroup$ – user20266 Aug 5 '12 at 12:13
  • $\begingroup$ wow, yes, I completely miss that. Thanks. $\endgroup$ – KWO Aug 5 '12 at 12:15
  • $\begingroup$ Dear Thomas, do you have an example for (b), I am thinking about $x^{1/n}$ but having some difficulties showing $N_{\alpha}(f_n)$ is finite. $\endgroup$ – KWO Aug 5 '12 at 12:17

I think you misinterpreted in b., since it's a case contained, after scaling in the first one.

Consider a sequence of functions $f_n$ such that $f_n(k2^{-n})=0$ for each $k$ and $f_n((2k+1)2^{-n-1})=1$, with linear interpolation. Then $\lVert f_n\rVert_{\infty}=1$ and $N_{\alpha}(f_n)=2^{(n+1)\alpha}$. A subsequence cannot be uniformly equicontinuous. Indeed, let $\{n_k\}$ an increasing sequence of integers. We will see that the condition of equi-continuity fails for $\varepsilon=1/2$. For $f_{n_k}$ we can take at most $\delta=2^{-(n_k+1)}$, but it won't work for $f_{n_{k+1}}$.

$f_n$ for $n=2$.

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  • $\begingroup$ sorry still try to visualize your function. $\endgroup$ – KWO Aug 5 '12 at 12:24
  • $\begingroup$ Got it, thanks a lot. $\endgroup$ – KWO Aug 5 '12 at 12:28
  • $\begingroup$ @KWO I've included a picture. $\endgroup$ – Davide Giraudo Aug 5 '12 at 12:34
  • $\begingroup$ Thanks again, it is very nice of you. $\endgroup$ – KWO Aug 5 '12 at 12:38

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