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Define the seminorm on the space $S=[0,1]\times[0,T]$ $$\mid u\mid_{\alpha} = \sup\frac{|u(x, t) - u(y,s)|}{(|x-y|^2 + |t-s|)^{\frac{\alpha}{2}}}.$$ Define the norms on the same space $$\lVert u \rVert_{C^{0, \alpha}} = \lVert u \rVert_{C^0} + \mid u\mid_{\alpha}$$ and $$\lVert u \rVert_{C^{2, \alpha}} = \lVert u \rVert_{C^0} +\lVert u_x \rVert_{C^0}+\lVert u_{xx} \rVert_{C^0}+\lVert u_t \rVert_{C^0}+ \mid u_{xx}\mid_{\alpha} + \mid u_t\mid_{\alpha}.$$

Suppose that $\lVert u \rVert_{C^2, \alpha} \leq C$ where $C$ is a constant. Let $a, b, c \in C^{0, \alpha}$. How can I show that $$\lVert au_{xx} + bu_x + cu\rVert_{C^{0, \alpha}} \leq K\lVert u \rVert_{C^{2, \alpha}}$$ for some constant $K$?

Or equivalently, want to show that $$\sup_{\lVert u \rVert_{C^{2,\alpha}} \leq C_1}\lVert au_{xx} + bu_x + cu\rVert_{C^{0, \alpha}} \leq K_1$$

(ALL the above norms are over the compact set $S$).

Thanks for any help

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Thanks, I edited. –  user25266 Aug 3 '12 at 10:07
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1 Answer

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If $f,g\in C^{\alpha}(S)$, then for all $(x,t),(x',t')\in S$, we have \begin{align} |f\cdot g(x,t)-f\cdot g(x',t')|&=|f(x,t)(g(x,t)-g(x',t'))+g(x',t')f(x,t)-f(x',t')g(x',t')|\\\ &\leq \lVert f\rVert_{\infty}|g(x,t)-g(x',t')|+\lVert f\rVert_{\infty}|f(x,t)-f(x',t')|, \end{align} hence \begin{equation}\[f\cdot g\]_{\alpha}\leq \lVert f\rVert_{\infty}\[g\]_{\alpha}+\lVert g\rVert_{\infty} \[f\]_{\alpha}. \end{equation} We deduce that \begin{align} \lVert au_{xx}+bu_x+cu\rVert_{C^{0,\alpha}}&\leq \max(\lVert a\rVert_{\infty},\lVert b\rVert_{\infty},\lVert c\rVert_{\infty})(\lVert u_{xx}\rVert_{\infty}+\lVert u_x\rVert_{\infty}+\lVert u\rVert_{\infty})\\\ &+\max(\lVert a\rVert_{\infty},\lVert b\rVert_{\infty},\lVert c\rVert_{\infty})(\[ u_{xx}\]_{\alpha}+\[u_x\]_{\alpha}+\[u\]_{\alpha}) \\\ &+\max(\[a\]_{\alpha},\[b\]_{\alpha},\[c\]_{\alpha})(\lVert u_{xx}\rVert_{\infty}+\lVert u_x\rVert_{\infty}+\lVert u\rVert_{\infty})\\\ &\leq \max(\max(\[a\]_{\alpha},\[b\]_{\alpha},\[c\]_{\alpha}),\max(\lVert a\rVert_{\infty},\lVert b\rVert_{\infty},\lVert c\rVert_{\infty}))\lVert u\rVert_{C^{2,\alpha}}. \end{align} We can get an equivalent norm on $C^{2,\alpha}(S)$ defining \begin{equation}\lVert u\rVert:=\[u\]_{\alpha}+\lVert u\rVert_{\infty}+\[u_t\]_{\alpha}+\lVert u_t\rVert_{\infty}+\[u_x\]_{\alpha}+\lVert u_x\rVert_{\infty}+\[u_{xx}\]_{\alpha}+\lVert u_{xx}\rVert_{\infty}. \end{equation} This can be shown using mean value theorem.

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Thanks for the answer, but I don't see how you get the last inequality. The $C^{2,\alpha}$ norm of $u$ does not include the seminorms $[u_x]_{\alpha}$ and $[u]_{\alpha}$ that are present on the lhs unfortunately. –  user25266 Aug 3 '12 at 10:02
I hope it's not a problem, because with mean value theorem we get an equivalent norm when we include them. –  Davide Giraudo Aug 3 '12 at 10:14
Thanks. I believe you're right but I can't show how $[u]_\alpha$ and $[u_x]_{\alpha}$ are bounded above by something useful using the MVT. For $[u]_{\alpha}$, I get something like needing to show $\sup\left(\frac{(x-y)^2 + (t-s)^2}{(|x-y|^2 + |t-s|)^{\alpha}}\right)^{\frac{1}{2}} < \infty.$ Does that look about what you meant? –  user25266 Aug 3 '12 at 17:09
In your last expression (as in the first of the OP), a square is missing, I think ($|t-s|^{\color{red}2}$). This supremum can be bounded by a constant involving the diameter of the compact, namely, $\operatorname{diam}(S)^{1-\alpha}$. –  Davide Giraudo Aug 3 '12 at 18:16
I don't think that square should be there. From the source I'm using and from Krylov's book, the powers of |x-y| and |t-s| are different. I'm not really sure what the point of defining the space in such a way is though. –  user25266 Aug 4 '12 at 10:12
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