Assume sequence of functions $\;f_n(x)\in W^{1,2}_{loc} ((l^n_{-},l^n_{+});\mathbb R^m)\;$ and a function $f:(c,d)\rightarrow \mathbb R^m\;$such that

  • $\;\lim_{n\to \infty} l^n_{-}=c\;\;\;\lim_{n\to \infty} l^n_{+}=d\;$
  • $\;f_n(x)\to f(x)\;$ uniform on bounded intervals $\;\forall x \in (c,d)\;$

Note that $\;-\infty \le c \lt d \le +\infty\;$. Also consider a continuous non-negative function $\;G:\mathbb R^m \rightarrow \mathbb R\;$ which satisfy:

$\;\vert G(f_1)-G(f_2) \vert \le \varepsilon\;$ when $\;\vert f_1-f_2 \vert \le δ\;\; \forall f_1,f_2\;$ for some $\;δ\gt 0\;$ and $\;\varepsilon \gt 0\;$

If $\;f\in C^{0,\frac{1}{2}}\;$ i.e $\;\vert f(x_1)-f(x_2) \vert \le \sqrt {2σ}{\vert x_1-x_2\vert}^{\frac{1}{2}} \;\;\forall x_1,x_2$, show that $\;G(f(x))\ge \varepsilon\;\;\forall x\in (x_n-\frac{δ^2}{2σ},x_n+\frac{δ^2}{2σ})\;$

My attempt:

All I have done is noticing that $\;x\in (x_n-\frac{δ^2}{2σ},x_n+\frac{δ^2}{2σ})\;$ is equivalent to $\;\sqrt {2σ} {\vert x-x_n \vert}^{\frac{1}{2}} \lt δ\;$ but it seems impossible to me to connect it to the Holder estimation for $\;f\;$. In addition I believe that from the condition for $\;G\;$ I could get some information about $\;G(f_n)\;$ and so using the fact that $\;G(f_n) \to G(f)\;$,I might prove the exercise but again I have no clue how!

I've been stuck to this exercise for more than a day, so I came here desperately seeking an answer or hints.I would really appreciate any help.

Thanks in advance!

  • $\begingroup$ Are you missing a term $G(f(x_n))$ somewhere in the claim? $\endgroup$ – Thomas Jul 7 '17 at 15:41
  • $\begingroup$ (or some statement about $G(f_n(.))$?) $\endgroup$ – Thomas Jul 7 '17 at 15:44
  • $\begingroup$ @Thomas No, I mentioned $\;G(f_n(x))\;$ because I believe that by condition for $\;G\;$ i.e. $\;\vert G(f_1)-G(f_2) \vert \le \varepsilon\;$, I could get some information about $\;G(f_n(x))\;$ and then using the fact that $\;G(f_n(x)) \to G(f)\;$ , I might prove the ex. But again all this could be all wrong $\endgroup$ – kaithkolesidou Jul 7 '17 at 15:50
  • $\begingroup$ With what you have written down (in the yellow box) you could just choose $G\equiv 0$ and every condition is fullfilled, but certainly not the claim. Also there is nothing known about $G(f_n)$, so you cannot conclude anything about $G(f)$ from $f_n$. Finally there is no relation of $G$ to any given data. Some prerequisite is missing or the statement is plain false (see $G\equiv 0$) $\endgroup$ – Thomas Jul 7 '17 at 16:03

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