I found some trouble in solving this problem:
given $f$ a continuous function in $[0,1]$ and $u\in C^2([0,1])$, such that $u''(x)=f(x)$ on $(0,1)$ and $u(0)=u(1)=0$, prove, $\forall\epsilon > 0$, the inequality
$||u'||_2^2 \leq \epsilon||u||_2^2 + \frac{1}{4\epsilon}||f||_2^2$
where $||.||_2$ is the usual norm on $L^2 ([0,1])$.
First, I rewrote the equation in this way:
$-u(x)u''(x)=-u(x)f(x)$
in order to have
$||u'||_2^2 = -\int_{0}^{1}u(x)f(x)dx$.
Then I tried to use a Green function to represent $u$, i.e.
$F(x,y)$=\begin{cases} -x(1-y) & \text{if $0\leq x\leq y\leq 1$} \\ -y(1-x) & \text{if $0\leq y\leq x\leq 1$} \end{cases}
in order to have $u(x)=\int_{0}^{1}F(x,y)f(y)dy$ but I really don't know where that $\epsilon$ should come from. I also thought I could use some identity approximation or mollifiers, but I got stuck.
Hint:
$$ \epsilon u^2 + \frac{1}{4\epsilon} f^2 + uf = \left( \sqrt{\epsilon} u + \frac{1}{2\sqrt{\epsilon}}f \right)^2 $$