# Poincaré inequality 1D

I am trying to prove the following result that I find in a book and which is denoted as Poincaré inequality.

Let $w$ be continuously differentiable on $(0,1)$. Then,

$\int_0^1 w^2(x)dx\leq 2 w^2(0)+4\int_0^1 w^2_x(x)dx$

where $w_x$ denotes the derivative with respect to $x$.

Thanks!

## 2 Answers

To simplify the notation, let $A := \int_0^1 w^2$, $B := \int_0^1 w_x^2$, $w_0 := w(0)$.

Since $w\in C^1$ and $(w^2)_x = 2 w w_x$, we have that $$w^2(x) = w_0^2 + \int_0^x 2 w(t) w_x(t)\, dt,$$ hence $$A := \int_0^1 w^2(x)\, dx = w_0^2 + 2\int_0^1 dx \int_0^x dt\, w(t) w_x(t).$$ If we change the order of integration in the last integral we get $$\begin{split} A & = w_0^2 + 2\int_0^1 dt \int_x^1 dx \, w(t) w_x(t) = w_0^2 + 2\int_0^1 (1-x) w(t) w_x(t) dt \\ & \leq w_0^2 + 2\int_0^1 |w(t)| \, |w_x(t)| dt \leq w_0^2 + 2 \sqrt{A} \sqrt{B}. \end{split}$$ Then $$(\sqrt{A} - \sqrt{B})^2 \leq w_0^2 + B,$$ so that $$\sqrt{A} \leq \sqrt{B} + \sqrt{w_0^2 + B}.$$ Since $(a+b)^2 \leq 2(a^2 + b^2)$ we finally get and finally $$A \leq 2B + 2(w_0^2 + B) = 2w_0^2 + 4B.$$

Seems to me that you may do slightly better: $$|w(x)-w(0)| = \left|\int_0^1 w'(t) 1_{[0,x]}(t) dt\right| \leq \sqrt{ \int_0^1 (w'(t))^2 dt \times 1}$$ and then $$|w(x)|^2 \leq \left(|w(0)| + \sqrt{ \int_0^1 (w'(t))^2 dt} \right)^2 \leq 2 \left(|w(0)|^2 + \int_0^1 (w'(t))^2 dt \right)$$ which you may then integrate to get your inequality with a factor 2 instead of 4 on the last term.