# 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 $C^{0, \alpha}$.

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According to your previous post, it seems that $u$ satisfies a particular equation. Maybe it's easier in this context. –  Davide Giraudo Aug 5 '12 at 19:28
@DavideGiraudo I tried but it doesn't seem to help unfortunately. –  TagWoh Aug 7 '12 at 17:06
It's the equation $u_t=au\cdots$, or not. In any case, you should write it in the OP. –  Davide Giraudo Aug 7 '12 at 17:08
@DavideGiraudo I added it. –  TagWoh Aug 7 '12 at 17:17

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 $$|\Delta_t({\tau})u_x(x,t)|\le C\tau^{1/2}.$$
Thank you! Could you just explain me what you mean by "finite difference approximation with step $\tau^{\frac{1}{2}}$"? I can't find anything about it. All I know about finite difference is, $$u_x(x,t) = \frac{u(x+h)-u(h)}{h}$$. I have never encountered this step size. Thanks. –  TagWoh Aug 8 '12 at 8:19
@TagWoh It is put here $h=\tau^{1/2}$, I've edited the answer. –  Andrew Aug 8 '12 at 8:40