# Improving the constant in the Poincaré inequality to $\frac{b-a}{\pi}$

Let $u \in H_0^1(a,b)$, then I know that $$||u||_{0,2}\leq\frac{b-a}{\sqrt{2}} |u|_{1,2}^2$$

How can I show that I can improve the constant in the inequality to $\frac{b-a}{\pi}$?

I feel like proving the inequality above where I first show that $|u(x)|^2 \leq (x-a)|u|_{1,2}^2$ by using the Cauchy-Schwartz inequality and then conclude $$||u||_{0,2}^2=\int_a^b|u(x)|^2 \ dx \leq \int_a^b (x-a) \ dx |u|_{1,2}^2 =\frac{(b-a)^2}{2} |u|_{1,2}^2$$ does not hint me to any improvements of the constant. Any suggestions?

• thank you, fixed it. – Tesla May 15 '18 at 10:59

The constant can be improved from $\frac1{\sqrt 2}$ to $\frac12$ by using $$\|u\|_{0,2}^2 = \int_a^b \min(x-a,b-x)\ dx\cdot |u|_{1,2}^2.$$ To get $\frac1\pi$, you need to know the eigenfunctions of $u \mapsto -u''$.
I now found out that we get equality by chosing $u(x)=\sin(\frac{x-a}{b-a}π)$