Prove that $\int_0^1[f''(x)]^2dx\ge4.$ Let $f$ be a $C^2$ function on $[0,1]$ such that
$f(0)=f(1)=f'(0)=0,f'(1)=1.$ Prove that
$$\int_0^1[f''(x)]^2dx\ge4.$$
Find all $f$ for equality to occur.
 A: First a variational argument.  Assume that this expression is minimal for some smooth function $f$.  Then let $\delta: [0,1] \to \mathbb{R}$ be twice differentiable and such that $\delta(0) = \delta'(0) = \delta'(1) = \delta(1) = 0$ and consider $f + t \cdot \delta$ for a real number $t$.  This function also satisfies the boundary conditions, so
$$
\int_0^1 \left(f''(x) + t\cdot \delta''(x)\right)^2 dx
$$
is minimal for $t = 0$. This implies that
$$
\int_0^1 f''(x) \delta''(x) dx = 0.
$$
Apply partial integration twice to get
$$
\int_0^1 f^{(4)}(x) \delta(x) dx = 0.
$$
Since this must hold for any such function $\delta$ it follows that $f^{(4)}$ is identically zero on $[0,1]$ and so must be a polynomial of degree at most three.  The only such polynomial that satisfies the boundary conditions is
$$
f(x) = x^2 (x - 1).
$$
For this $f$ you obtain the lower bound
$$
\int_0^1 \left(f''(x)\right)^2 dx = 4.
$$
This argument has produced a nice candidate minimal function.  Once this candidate is found the claim follows easily.  Let $f$ be the polynomial above and take $g \in C^2[0,1]$ satisfying the boundary conditions as stated in the problem.  Then $g = f + (g - f)$ and if we define $\delta = g - f$ then $\delta$ has the properties assumed above.  In particular, by partial integration, we know that
$$
\int_0^1f''(x)\delta''(x) dx = 0.
$$
Then
$$
\int_0^1\left(g''(x)\right)^2 dx = \int_0^1\left(f''(x) + \delta''(x)\right)^2 dx = 4 + \int_0^1 \left(\delta''(x)\right)^2 dx \geq 4
$$
with equality only for $\delta'' = 0$. The latter implies $\delta = 0$ since $\delta'(0) = \delta(0) = 0$. So equality only holds when $g = f$.
