Let $I=(a,b)$ with $a 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
Let $I=(a,b)$ with $a 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
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.