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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.2012-08-05
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    @DavideGiraudo I tried but it doesn't seem to help unfortunately.2012-08-07

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