Infimum of the $L^2$ norm of the derivative of functions of $H^1_0$ of $L^2$ norm $1$

Let $I=(a,b)$ with $a<b$ real numbers. Define $\lambda_0$ by $$\lambda_0=\inf\{\|u'\|_{L^2(I)}^2:\ u\in H_0^1(I),\ \|u\|_{L^2(I)}=1 \}.$$

How can I prove that $\lambda_0\geq\displaystyle\frac{\pi^2}{(b-a)^2}$?

Thanks

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some description of notation will be really helpful here –  dexter04 Nov 24 '12 at 19:06
Well, what do you need? @dexter04 –  Tomás Nov 24 '12 at 19:10
what do $L^2(I), H_0^1(I)$ mean?? –  dexter04 Nov 24 '12 at 19:23
$L^2(I)$ square integrable functions on $I$. $H_0^1(I)$ Sobolev Space. –  Tomás Nov 24 '12 at 19:40
It is an instance of the Poincaré-Wirtinger inequality: if $f$ has mean zero over $[0,2\pi]$ and both $f$ and $f'$ belong to $L^2([0,2\pi])$, then $\|f'\|_2^2\geq\|f\|_2^2$, so the statement follows from a simple rescaling argument: take $f(x)=u\left(a+\frac{b-a}{2\pi}x\right)$. –  Jack D'Aurizio Nov 24 '12 at 19:58
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Considering $v\left(\frac{x-a}{b-a}\right):=u(x)$, we have to show the result when $a=0$ and $b=1$.
Let $u(x):=\sin(\pi x)$ for $x\in (0,1)$. As $u(0)=u(1)=0$, $u\in H^1_0(0,1)$; $\lVert u\rVert_2^2=1/2$ and $\lVert u'\rVert_2^2=\pi^2/2$.
To get the converse, consider the problem $$\left\{\begin{array}{ll}-u''(x)=\sin(\pi x),&x\in (0,1),\\ u(0)=u(1)=0. \end{array}\right.$$ So in the case of an interval, we can compute the best constant in Poincaré inequality.