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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:

$ \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 $\epsilon>0$?

I am thinking of $\epsilon>0$ small so the bound only has to hold for sufficiently small $\epsilon$.

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?

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?

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Assuming that by $\|g\|_{H^k(\Omega)}$ you mean $\sum_{0\le j\le k}\|g^{(j)}\|_{L^2(\Omega)}$ then this is trivially true for $\alpha=0$. I think you forgot some extra condition. –  Florian Nov 25 '10 at 9:50
    
Yes. I forgot the important condition on $\alpha$: $\alpha>1+k$!! –  alext87 Nov 25 '10 at 9:58
    
This requirement is too strong: it's not even true for $k=0$. But $\alpha=k$ seems reasonable to me. –  Florian Nov 25 '10 at 10:20
    
Thanks, sure. I was hoping to get $\alpha=1+k$ with the $1$ coming from the restricted domain $(a,a+\epsilon)$. Does $\alpha=k$ work? –  alext87 Nov 25 '10 at 10:25
    
It's true at least for $\alpha=k=1$ (Poincaré's inequality). But in order to iterate the inequality, you'd need $g^{(j)}(a)=0$ for $j<k$. –  Florian Nov 25 '10 at 10:32

1 Answer 1

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 $|g(x)|\le C(x-a)^{k-1/2}$.

This is essentially optimal, since the function $(x-a)^{k-1/2}/\ln(x-a)$ satisfies all the hypotheses. In particular, you cannot get $\alpha=k+1$.

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