Prove $\int_0^1 ((g'(x))^2-1)^2dx \geq 1$ for smooth $g$ with $g(0)=g(1)=0$ This came up in an optimization problem.
How do you prove that $\int_0^1 ((g'(x))^2-1)^2dx \geq 1$ for any $g$ which is twice continuously differentiable on $[0,1]$ and such that $g(0)=g(1)=0$?
 A: The functional $\int_0^1 ((g'(x))^2-1)^2dx  $ can be made arbitrarily close to $0$ subject to the stated conditions $g(0)=g(1)=0$ and $g\in C^2$. 
Begin with the triangle $x\mapsto \frac12- |x-1/2|$; formally, the functional is zero for this function but of course this is not a differentiable function. Redefine it in $\epsilon$-neighborhood of $1/2$, replacing the vertex of triangle with a smooth cap. This can be done so that $|g'|$ does not exceed $1$. Hence, 
$$\int_0^1 ((g'(x))^2-1)^2dx  \le \int_{1/2-\epsilon}^{1/2+\epsilon} 1\, dx  =2\epsilon$$ 
A: Let $g(x)=x(x-1)$.  Then
$$\int_0^1((g'(x))^2-1)^2\,dx={8\over15}$$
Remark:  Even without doing a specific example, it's possible to show that the inequality cannot hold in general.  By expanding $((g'(x))^2-1)^2$ to $(g'(x))^4-2(g'(x))^2+1$, we find
$$\int_0^1((g'(x))^2-1)^2\,dx\ge1\iff\int_0^1(g'(x))^4\,dx\ge2\int_0^1(g'(x))^2\,dx$$
Since the only constraint on $g$ is $g(0)=g(1)=0$, we can multiply $g$ by any constant we like, and this would imply
$$r^4\int_0^1(g'(x))^4\,dx\ge2r^2\int_0^1(g'(x))^2\,dx$$
for any $r$, no matter how small, and that's obvious nonsense for any non-constant function $g$.
A: This is a comment on Woodface's answer that is too long for a comment :-).
It is instructive to look at the form of the functional
$J(g) = \int_0^1 L(g')$, where $L(x) = (x^2-1)^2$.
Note that $L$ looks like:

If we deal with absolutely continuous functions $g$ on $[0,1]$, it is clear that $g^*(x) = {1 \over 2} - |x - {1 \over 2}|$ results in $J(g^*) = 0$, and since $J(g) \ge 0$, we see that
$g^*$ is a minimiser (this is from Woodface's answer).
As Woodface noted, we can construct $C^2$ functions $g_n$ such that $
\lim_n J(g_n) = 0$, and, referring to the above plot, we see that
the $g_n'$s 'spend' more and more time at the $\pm 1$ points and quickly cross
over from $+1$ to $-1$ near $x={1 \over 2}$.
A little work shows that if $g$ is differentiable, then $J(g) >0$, so
we can never attain the infimising value with a differentiable (everywhere) function.
To emphasise, there is no minimiser if we require that our $g$s be
differentiable.
We can also see from the above plot, that the function $g = 0$ is a local
maximiser of $J$, and small perturbations from $g$ will result in a
smaller value of $J$! This is the extremal that is found using the Euler
Lagrange equations. It is the only stationary point of $J$.
The cautionary tale is that we should first establish the existence of
a minimiser.
