# A version of Poincaré's inequality in one dimension

Let $u\in W^{1,p}\big((0,1) \big)$ with $\int_{(0,1)} u=0$. Why is it true that $$\Vert u\Vert_{L^p} \leq \Vert u' \Vert_{L^p}?$$ This is given as exercise in a proof of a version of Poincaré's inequality for cubes which proceeds by induction on the dimension (the base case being the above one). I've managed to make a proof, but I am not sure if it is the intended one, and I get a constant $2$ in the inequality (although this is probably due to a crude estimate on the step following Fubini). The reference I'm using seems to imply that the inequality is very direct.

You have $$-u(t)=\int_0^1 \int_t^xu'(y)\,dy\,dx.$$ Hence, very greedily, $$|u(t)|\,\le\,\int_0^1\int_{[t,x]} |u'(y)|dy\,dx\,\le\,\int_0^1|u'(y)|\,dy = \|u'\|_1\,\le\,\|u'\|_p.$$ Now, you can take the $p$-th power, integrate over $[0,1]$, and get your inequality.
• Thanks! Just to be clear, in your second inequality you are expanding the integration interval to $[0,1]$ and then using Fubini, correct? Oct 15, 2017 at 16:40
• Yes and no. I am extending the inner integral to $[0,1]$ and then pull it out (ok, which is trivial Fubini ;-)). Oct 15, 2017 at 16:42
We have the equality $$u(x)-u(t)=\int_t^xu'(y)dy.$$ Integrating with respect to $x$ on $(0,1)$, we get by assumption (and using the fact that $\mu\big((0,1)\big)=1$) that $$-u(t)=\int_0^1 \int_t^xu'(y)dydx.$$ Since the integral $\int_t^x$ can be such that $x<t$, let's separate the integrals for safety. We have $$\int_0^1 \int_t^xu'(y)dydx=\int_0^1 \int_0^xu'(y)dydx-\int_0^1 \int_0^tu'(y)dydx.$$ By Fubini, $$\int_0^1 \int_0^xu'(y)dydx=\int_0^1\int_y^1 u'(y)dx dy=\int_0^1u'(y)(1-y)dy;$$ $$\int_0^1 \int_0^tu'(y)dydx=\int_0^t\int_0^1 u'(y)dxdy=\int_0^tu'(y)dy.$$ Hence, $$-u(t)=\int_0^1u'(y)(1-y)dy-\int_0^t u'(y)dy.$$ Therefore, $$|u(t)| \leq \int_0^1|u'(y)| |(1-y)| dy +\int_0^t |u'(y)|dy\leq 2\int_0^1 |u'(y)|dy.$$ By Holder's inequality, $$\Vert u'\Vert_{L^1} \leq \Vert u'\Vert_{L^p}.$$ It follows by integrating the previous inequality (after taking the $p$-th power) that $$\int_0^1 |u|^p\leq 2^p(\int_0^1 |u'|)^p \leq 2^p\Vert u'\Vert_{L^p}^p=2^p\int_0^1|u'|^p.$$ By taking the $p$-th root, we get that $$\Vert u\Vert_{L^p} \leq 2\Vert u' \Vert_{L^p},$$ as we wanted (not quite).
Since $u$ (has a representative that) is continuous and $\int_0^1 u = 0$, there exists a point $a\in (0,1)$ such that $u(a) = 0$.
Hence, by Hoelder's inequality, $$|u(x)| \leq \left| \int_a^x |u'(s)|\, ds \right| \leq \left|\int_a^x |u'(s)|^p\, ds\right|^{1/p} \, |x-a|^{(p-1)/p}$$ so that $$|u(x)|^p \leq \left|\int_a^x |u'(s)|^p\, ds\right| \, |x-a|^{p-1}.$$ Integrating the inequality in $(0,1)$ we get $$\begin{split} \int_0^1 |u(x)|^p\, dx & \leq \int_0^a dx \int_x^a |u'(s)|^p\, (a-x)^{p-1}\, ds + \int_a^1 dx \int_a^x |u'(s)|^p\, (x-a)^{p-1}\, ds \\ & = \int_0^a |u'(s)|^p \frac{a^p - (a-s)^p}{p}\, ds + \int_a^1 |u'(s)|^p \frac{(s-a)^p}{p}\, ds \\ & \leq \frac{a^p}{p}\int_0^a |u'(s)|^p \, ds + \frac{(1-a)^p}{p}\int_a^1 |u'(s)|^p \, ds. \end{split}$$ Since $a^p, (1-a)^p \leq 1$, we get $$\int_0^1 |u(s)|^p\, ds \leq \frac{1}{p} \int_0^1 |u'(s)|^p\, ds \leq \int_0^1 |u'(s)|^p\, ds.$$