A Hölder norm bound that I need help with Define the seminorm on the space $S=[0,1]\times[0,T]$
$$[u]_{\alpha} = \sup\frac{|u(x, t) - u(y,s)|}{(|x-y|^2 + |t-s|)^{\frac{\alpha}{2}}}.$$
Define the norms on the same space
$$\lVert u \rVert_{C^{0, \alpha}} = \lVert u \rVert_{C^0} + [u]_{\alpha}$$
and
$$\lVert u \rVert_{C^{2, \alpha}} = \lVert u \rVert_{C^0} +\lVert u_x \rVert_{C^0}+\lVert u_{xx} \rVert_{C^0}+\lVert u_t \rVert_{C^0}+ [u_{xx}]_{\alpha} + [u_t]_{\alpha}.$$
I want to show that $[u_x]_\alpha \leq C\lVert u \rVert_{C^{2, \alpha}}$ where $C$ doesn't depend on $u_{xt}$. Does anyone have any hints how to do this? I tried using the MVT but that gives me a $u_{xt}$ which I can't bound above. Or is there something I can do with $u_{xt}$?
Alternatively, is there anything I can do (as in bound above) with
$$\sup\frac{|u_x(x, t) - u_x(x,s)|}{|t-s|^{\frac{\alpha}{2}}}?$$
I can't seem to avoid getting a mixed derivative $u_{xt}$ here.
Thanks for any help.
ADDED: $u$ solves the equation
$$u_t = a_1u_{xx} + b_1u_x + c_1u + (f_1 + a_2v_{xx} + b_2v_x + c_2v)$$
where $v$ solves $$v_t = a_3v_{xx} + b_3v_x + c_3v + f_3$$ and the $a_i$, etc, are functions of $(x,t)$ in $C^{0, \alpha}$.
 A: The function from $C^{2,\alpha}(S)$ can be extended to a function of the same class on $\mathbb R^2$ so WLOG we can assume $u\in C^{2,\alpha}(\mathbb R^2)$. Denote $\tau=s-t$. It is enough to consider the case $0<\tau\le1$. 
Let's change the first derivative $u_x(x,t)$ on it's finite difference approximation with step $\tau^{1/2}$:
$$\tau^{-1/2}\Delta_x({\tau^{1/2}})u(x,t)=
\frac{u(x+\tau^{1/2},t)-u(x,t)}{\tau^{1/2}}.$$
Then instead of the required difference $\Delta_t({\tau})u_x(x,t)$ we'll have $\tau^{-1/2}\Delta_t({\tau})\Delta_x({\tau^{1/2}})u(x,t)$. Since the differences commute, for the last expression we obtain
$$
|\tau^{-1/2}\Delta_t({\tau})\Delta_x({\tau^{1/2}})u(x,t)|=
\tau^{-1/2}\left|\int_t^{t+\tau} \Delta_x({\tau^{1/2}}) 
u_t(x,\lambda)\,d\lambda\right|\le
$$
$$
\le \tau^{-1/2} C\int_t^{t+\tau} \tau^{\alpha/2}\,d\lambda=
C\tau^{(1+\alpha)/2}.
$$
The difference between that is needed and that was introduced can be divided into two similar summands,
$$
\Delta_t({\tau})u_x(x,t)-\tau^{-1/2}\Delta_t({\tau})\Delta_x({\tau^{1/2}})u(x,t)=
$$
$$
\Delta_t({\tau})(u_x(x,t)-\tau^{-1/2}\Delta_x({\tau^{1/2}})u(x,t))=
$$
$$
\tau^{-1/2}(\tau^{1/2}u_x(x,t+\tau)-\Delta_x({\tau^{1/2}})u(x,t+\tau))-
\tau^{-1/2}(\tau^{1/2}u_x(x,t)-\Delta_x({\tau^{1/2}})u(x,t)).
$$
Both can be estimated in the same way:
$$
|\tau^{-1/2}(\tau^{1/2}u_x(x,t)-\Delta_x({\tau^{1/2}})u(x,t))|=
\tau^{-1/2}\left|\int_x^{x+\tau^{1/2}}(u_x(x,t)-u_x(y,t))\,dy\right|\le
$$
$$
C\tau^{-1/2}\int_x^{x+\tau^{1/2}}(y-x)\,dy=C_1\tau^{1/2}.
$$
Summing up we get 
$$
|\Delta_t({\tau})u_x(x,t)|\le C\tau^{1/2}.
$$
