How do I prove $\lambda_1 \geq c |U|^{\frac{-2}{n}}$?

14 Views Asked by At

Let $U$ be an open, bounded and connected set in $\mathbb{R}^n$ with smooth boundary $n \geq 3$. Let $\lambda_1$ be the principle eigenvalue of $-\Delta$ with zero boundary condition on $U$.

Show that there exist $c>0$ such that $\lambda_1 \geq c |U|^{\frac{-2}{n}}$?

My idea: We know that $\|u\|_{L^{\frac{2n}{n-2}}(U)} \leq C \|Du\|_{L^2(U)}$.

Let $w_1$ is the Eigen function associated with $\lambda_1$, then by using the above inequality

$\|w_1\|_{L^{\frac{2n}{n-2}}(U)} \leq C \|Dw_1\|_{L^2(U)}$

$\|w_1\|_{L^{\frac{2n}{n-2}}(U)} \leq \int_{U} -w_1 \Delta w_1 dx$

$\|w_1\|_{L^{\frac{2n}{n-2}}(U)} \leq \lambda_1 \int_U w_1^2 dx.$

I stuck in the L.H.S term. Can anyone suggest how do I do last step?