Uniform equicontinuity of a family of functions Consider $0<\beta<1$ and $\mathbb{T}\subset\mathbb{R}$ the torus. Assume that $\theta(t,x)\in C^{\beta}((0,T); C^{1,\beta}(\mathbb{T}))$ (this means that the function $\theta:(0,T)\to C^{1,\beta}(\mathbb{T})$ is $\beta$-Hölder with respect to norm of $C^{1,\beta}(\mathbb{T})$).
Let $0<2\alpha<\beta$ and define
$$f(t,x,h)=\frac{(\theta(t,x+h)-\theta(x,t))^2}{(\xi(t)^2+|h|^2)^{\alpha}}$$
with the convention $f(t,x,0)=0$ and $\xi:[0,\infty)\to[0,\infty)$ is $C^2$ and nonincreasing such that $\xi(t)=0$ for all $t>T^*$ for some $T^*\in(0,T)$.
When $t,s\geq T^*$, I have that
\begin{align*}
|f(t,x,h)-f(s,x,h)|&\leq 4|t-s|^{\beta}||\theta||_{C^{\beta}((0,T); C^{\beta}(\mathbb{T}))}.
\end{align*} 
Hence, $f(t,x,h)$ is uniformly equicontinuous in $[T_*,T)$.
My question: Is it $f(t,x,h)$ uniformly equicontinuous in the case $t<T*$ or $s<T^*$?  
 A: I believe I have found an answer. Define $$\delta_h\theta(t,x)=\theta(t,x+h)-\theta(t,x).$$
For all $s,t\in(0,T)$, we have
\begin{align}
f_{\lambda}(t)-f_{\lambda}(s)=\;&\frac{\delta_h\theta(t,x)^2-\delta_h\theta(s,x)^2}{(\xi^2(t)+|h|^2)^{\alpha}}\nonumber\\
&-\frac{\delta_h\theta(s,x)^2\Big[(\xi^2(t)+|h|^2)^{\alpha}-(\xi^2(s)+|h|^2)^{\alpha}\Big]}{(\xi^2(s)+|h|^2)^{\alpha}(\xi^2(t)+|h|^2)^{\alpha}}\nonumber\\
=&:I_1+I_2
\end{align}
where
\begin{align}
I_1&\leq \frac{1}{|h|^{2\alpha}}|\delta_h\theta(t,x)+\delta_h\theta(s,x)||\delta_h\theta(t,x)-\delta_h\theta(s,x)|\nonumber\\
&\leq4|t-s|^{\beta}||\theta||_{C^{\beta}(0,T;L^{\infty}(\mathbb{T}))}||\theta||_{L^{\infty}(0,T;C^{\beta}(\mathbb{T}))}
\end{align}
and using that $f(x)=x^{\alpha}$ is $\alpha$-Hölder and the mean value theorem, I obtain
\begin{align}
I_2&\leq C[\theta(s)]_{C^{2\alpha}}^2|\xi^2(t)-\xi^2(s)|^{\alpha}\nonumber\\
&\leq C(2\xi_0)^\alpha|\xi(t)-\xi(s)|^{\alpha}||\theta||_{L^{\infty}(0,T;C^{\beta}(\mathbb{T}))}^2\nonumber\\
&\leq C|t-s|^{\alpha}||\theta||_{L^{\infty}(0,T;C^{\beta}(\mathbb{T}))}^2.\label{3}
\end{align}
when $\xi_0,||\xi^{\prime}||_{L^{\infty}}<\infty$. Adding $I_1$ and $I_2$ estimates, I have that $\{f_{\lambda}\}_{\lambda\in\mathcal{K}}$ is equicontinuous.
