Trying to fill in a proof, and I was wondering if the following is true.

Let $f \in C^{\gamma}(\mathbb{[0,1]})$ (ie a $\gamma$-Holder continuous funtion with $\gamma \in (0,1)$). Define the norm $\| \cdot\|_{L}$ and $\| \cdot \|_{\gamma}$ on $C([0,1])$ as follows:

\begin{eqnarray*} \| h\|_L &=& |h(0)| + \sup_{x \neq y} \frac{|h(x)-h(y)|}{|x-y|} \\ \| h \|_{\gamma} &=& |h(0)| + \sup_{x \neq y} \frac{|h(x)-h(y)|}{|x-y|^{\gamma}} \end{eqnarray*}

Suppose $g_n([0,1]),g([0,1]) \subset [0,1]$. If $g_n \rightarrow g$ in $\|\cdot\|_L$, does $f \circ g_n \rightarrow f \circ g$ in $\|\cdot\|_{\gamma}$?


Hint: These inequalities might be useful $$ \begin{align} \frac{|f\circ g(x)-f\circ g(y)|}{|x-y|^\gamma} &=\frac{|f\circ g(x)-f\circ g(y)|}{|g(x)-g(y)|}\frac{|g(x)-g(y)|}{|x-y|^\gamma}\\ &\le\|f\|_L\|g\|_\gamma \end{align} $$ and $$ \begin{align} \frac{|f\circ g(x)-f\circ g(y)|}{|x-y|^\gamma} &=\frac{|f\circ g(x)-f\circ g(y)|}{|g(x)-g(y)|^\gamma}\left(\frac{|g(x)-g(y)|}{|x-y|}\right)^\gamma\\ &\le\|f\|_\gamma\|g\|_L^\gamma \end{align} $$

  • $\begingroup$ I feel like I'm missing something obvious. How do you apply the inequality? Doesn't $ f\circ g - f \circ g_n \neq f \circ(g-g_n)$? $\endgroup$ – user65011 Mar 5 '13 at 1:52
  • $\begingroup$ I don't think this solution is complete either. The difficulty with this problem is you need to control $$|f \circ g_n(x) - f \circ g(x) - f \circ g_n(y) + f \circ g(y)|.$$ As far as I can tell there is no obvious way to split and bound this. $\endgroup$ – ktoi May 4 '18 at 9:43

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