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!
 A: 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
$$
A \leq 2B + 2(w_0^2 + B) = 2w_0^2 + 4B.
$$
A: 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.
