0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ thank you, fixed it. $\endgroup$ – Tesla May 15 '18 at 10:59
1
$\begingroup$

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 $u \mapsto -u''$.

$\endgroup$
0
$\begingroup$

I now found out that we get equality by chosing $u(x)=\sin(\frac{x-a}{b-a}π)$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.