most recent 30 from http://mathoverflow.net 2013-06-18T23:07:00Z http://mathoverflow.net/feeds/question/47314 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/47314/bounding-a-smooth-function-near-the-endpoint henryreed 2010-11-25T09:33:30Z 2011-01-06T14:22:13Z

<p>Let $\Omega=(a,b)$ a finite interval, $g\in \mathcal{H}^k(\Omega)$ some integer $k$, with $g(a)=0$ and let $\epsilon>0$. Is there an $\alpha\geq 1+k$ such that: </p> <p>$ \left\|g\right\|_{L_2(a,a+\epsilon)}\leq C\epsilon^{\alpha}\left\|g\right\|_{\mathcal{H}^k(\Omega)} $ where $C$ and $\alpha$ do not depend on $g$ or $\epsilon>0$?</p> <p>I am thinking of $\epsilon>0$ small so the bound only has to hold for sufficiently small $\epsilon$. </p> <p>If the above is not possible are there any extra conditions I can put on $g$ at the end point $a$ or extra smoothness of $g$ in $\Omega$ I can impose?</p> <p>Poincare's inequality for $k=1$: I get $ \left\|g\right\|_{L_2(a,a+\epsilon)}\leq (1+C)\left\|g\right\|_{\mathcal{H}^1(a,a+\epsilon)}$ How do I continue from here?</p>

http://mathoverflow.net/questions/47314/bounding-a-smooth-function-near-the-endpoint/47321#47321 Michael Renardy 2010-11-25T11:51:09Z 2010-11-25T11:51:09Z

<p>You need to make the stronger assumption $g(a)=g'(a)=...=g^{(k-1)}(a)=0$. Then your statement is true with $\alpha=k$. You can see this by using the Cauchy-Schwarz inequality in $g^{(k-1)}(x)=\int_a^x g^{(k)}(y)\,dy$ to obtain $|g^{(k-1)}(x)|\le C(x-a)^{1/2}$, and then integrating repeatedly to get $|g(x)|\le C(x-a)^{k-1/2}$.</p> <p>This is essentially optimal, since the function $(x-a)^{k-1/2}/\ln(x-a)$ satisfies all the hypotheses. In particular, you cannot get $\alpha=k+1$. </p>