Let $I$ be a finite open interval, $d\in\mathbb N$, $\Omega\subseteq\mathbb R^d$ be bounded and open and $$|f|_{\alpha,\:\beta}:=\sup_{\substack{(s,\:x),\:(t,\:y)\:\in\:\overline I\times\overline\Omega\\(s,\:x)\:\ne\:(t,\:y)}}\frac{|f(s,x)-f(t,y)|}{|s-t|^\alpha+\left\|x-y\right\|^\beta}$$ and $$\left\|f\right\|_{\alpha,\:\beta}:=\sup_{(t,\:x)\:\in\:\overline I\times\overline\Omega}|f(t,x)|+|f|_{\alpha,\:\beta}$$ for $f:\overline I\times\overline\Omega\to\mathbb R$ and $\alpha,\beta\in(0,1]$.
Let $g_n:\overline I\times\overline\Omega\to\mathbb R$ for $n\in\mathbb N$ and assume $$c:=\sup_{n\in\mathbb N}\left\|g_n\right\|_{\frac\alpha2,\:\alpha}<\infty\tag1$$ for some $\alpha\in(0,1]$. Moreover, let $f\in C^1(\overline I\times\overline\Omega\times\mathbb R)$.
Why can we conclude that $$\sup_{n\in\mathbb N}\left\|f(\;\cdot\;,\;\cdot\;,u_n)\right\|_{\frac\beta2,\:\beta}<\infty\tag2$$ for some $\beta\in(0,1]$?
We might first note that, since $f$ is differentiable, it is Lipschitz continuous on each compact subset of its domain. Hence, $f$ is Lipschitz continuous on $\overline I\times\overline\Omega\times[-c,c]$. Denote the Lipschitz constant by $L$.
This should yield \begin{equation}\begin{split}&|f(s,x,g_n(s,x))-f(t,y,g_n(t,y))|\\&\;\;\;\;\;\;\;\;\;\;\;\;\le L\max(|s-t|,\|x-y\|,|g_n(s,x)-g_n(t,y)|\\&\;\;\;\;\;\;\;\;\;\;\;\;\le cL\max\left(|s-t|,\|x-y\|,|s-t|^{\frac\alpha2},\|x-y\|^{\alpha}\right)\end{split}\tag2\end{equation} for all $(s,x),(t,y)\in\overline I\times\overline\Omega$ and $n\in\mathbb N$.
But why does this yield the claim?
Best Answer
Since $I,\Omega$ are bounded, there is a constant $D$ such that if $s,t \in I$, $|s-t|\leq D|s-t|^{\alpha/2}$ and for all $x,y \in \Omega$, $\|x-y\| \leq D\|x-y\|^{\alpha}$. Here, $D$ is (say) $d_I^{1-\alpha/2}+d_{\Omega}^{1-\alpha}$ where $d_{\cdot}$ is the diameter.