most recent 30 from http://mathoverflow.net 2013-05-25T01:47:19Z http://mathoverflow.net/feeds/question/103844 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/103844/how-to-show-this-holder-bound tagwoh 2012-08-03T09:17:50Z 2012-08-03T10:29:07Z

<p>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 $$\lVert u \rVert_{C^{2, \alpha}} = \lVert u \rVert_{C^0} +\lVert u_x \rVert_{C^0}+\lVert u_{xx} \rVert_{C^0}+\lVert u_t \rVert_{C^0}+ \mid u_{xx}\mid_{\alpha} + \mid u_t\mid_{\alpha}.$$</p> <p>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 $K$?</p> <p>Or equivalently, want to show that $$\sup_{\lVert u \rVert_{C^{2,\alpha}} \leq C_1}\lVert au_{xx} + bu_x + cu\rVert_{C^{0, \alpha}} \leq K_1$$</p> <p>(ALL the above norms are over the compact set $S$).</p> <p>Thanks for any help</p>

http://mathoverflow.net/questions/103844/how-to-show-this-holder-bound/103847#103847 Davide Giraudo 2012-08-03T09:52:19Z 2012-08-03T10:29:07Z

<p>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')|&amp;=|f(x,t)(g(x,t)-g(x',t'))+g(x',t')f(x,t)-f(x',t')g(x',t')|\\ &amp;\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]<em>{\alpha}\leq \lVert f\rVert</em>{\infty}[g]<em>{\alpha}+\lVert g\rVert</em>{\infty} [f]<em>{\alpha}. \end{equation} We deduce that \begin{align} \lVert au</em>{xx}+bu_x+cu\rVert_{C^{0,\alpha}}&amp;\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})\\ &amp;+\max(\lVert a\rVert_{\infty},\lVert b\rVert_{\infty},\lVert c\rVert_{\infty})([ u_{xx}]_{\alpha}+[u_x]<em>{\alpha}+[u]</em>{\alpha}) \\ &amp;+\max([a]<em>{\alpha},[b]</em>{\alpha},[c]<em>{\alpha})(\lVert u</em>{xx}\rVert_{\infty}+\lVert u_x\rVert_{\infty}+\lVert u\rVert_{\infty})\\ &amp;\leq \max(\max([a]<em>{\alpha},[b]</em>{\alpha},[c]<em>{\alpha}),\max(\lVert a\rVert</em>{\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]<em>{\alpha}+\lVert u\rVert</em>{\infty}+[u_t]_{\alpha}+\lVert u_t\rVert_{\infty}+[u_x]_{\alpha}+\lVert u_x\rVert_{\infty}+[u_{xx}]<em>{\alpha}+\lVert u</em>{xx}\rVert_{\infty}. \end{equation} This can be shown using mean value theorem. </p>