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Help me please with this question:

Is it true that for all $N>0$ exists $u \in C^{1}[0,1]$ such that $u(0)=0$ and $\frac{\int_{0}^{1}u^2(x)dx}{\int_{0}^{1}\left [ u'(x) \right ]^2dx}>N$; what happens without assuming that $u(0)=0$?

Thanks a lot!

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The Poincaré inequality is $\int |u|^2 \leq C \int |u'|^2$... –  Siminore Jul 5 '12 at 10:09
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I don't really understand the way the inequality is phrased in the question. However, it would probably be helpful to consider that without some sort of tie-down condition (like $u(0)=0$), constant functions can be added to any $u$, increasing the norm without affecting the norm of the derivative. –  Nick Alger Jul 5 '12 at 10:24
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1 Answer 1

up vote 1 down vote accepted

If $u\in C^1[0,1]$ satisfies $u(0)=0$ then $$\int_0^1u(x)^2dx=\int_0^1\left(\int_0^xu'(t)dt\right)^2dx\leq \int_0^1x\int_0^x(u'(t))^2dtdx\leq \frac 12\int_0^1(u'(t))^2dt,$$ hence for $K>1/2$ the inequality you want doesn't hold.

If we don't require $u(0)=0$, then take functions like $u(x)=x+C$ to get want we want.

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It's clear that @Lilly is a bit confused about the inequality. She wants to prove that $\inf_{\|u\|_2^2=1} \int |u'|^2 =0$, which is false, because of the correct Poincaré inequality (with the correct assumptions on $u$, of course). –  Siminore Jul 5 '12 at 11:17
    
thanks for example and explanation! –  Lilly Jul 7 '12 at 10:06
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