Proposition 8.13 Brezis

Question

I'm trying to add all the missing details in the proof of Poincare's inequality given by Brezis in the Proposition 8.13. That's what I've done: Brezis says: Let $u\in W_0^{1,p}(I)$ where $I=(a,b)$ is bounded. Since $u(a)=0$, then we have that $$|u(x)|=|u(x)-u(a)|=\biggr|\int_a^x u'(t)dt\biggr|\leq ||u'||_1.$$ Now taking the supremum we end up with $||u||_\infty\leq ||u'||_1$. The result follow by Holder's inequality (how!?). I really don't know what's I'm missing but i cannot link these two things to solve the problem. I dont'w understand the hint given in the answer Poincaré's Inequality on Sobolev Spaces in One Dimension

Answer 1

To be clear, Proposition 8.13 asks you to prove that $\|u\|_{W^{1,p}(I)} \leq C\|u’\|_{L^p(I)}$ for $u\in W^{1,p}_0(I)$.

By definition it suffices to bound $\|u\|_{L^p(I)}$. This is done via

$$\begin{align}\|u\|_{L^p(I)} &\leq |I|^{1/p} \|u\|_\infty \\ &\leq |I|^{1/p} \int_I |u’(t)|\chi_{I}(t)dt \\ &\leq |I|^{1/p}(\|u’\|_{L^p(I)} \|\chi_I\|_{L^q(I)}) \\ &= |I|^{1/p+1/q}\|u’\|_{L^p(I)} \\ &= |I|\|u’\|_{L^p(I)}\end{align}$$

The first inequality follows from pulling the sup norm out of the integral, the second is the one you described in your post, the third is Holder (with $q$ the Holder dual of $p$), and the remaining are calculations.