# Nonconvex functional show inf is zero but inf is not attained

This is Exercise 8.22 from John Hunters Applied Analysis books. The question says:

Consider the non-convex functional:

$f:W^{1,4}([0,1])\rightarrow&space;\Re$

Defined by

$f(v)=\int_{0}^{1}[v^{2}+(1-v'^{2})^{2}]dx$

Where W is the Sobolev space of functions that belong to $L^4([0,1])$ and whose weak derivatives belong to $L^4([0,1])$. Show that the infimum of f on W is equal to zero, but that the infimum is not attained

I have to show the following:

1. Prove that f is not convex
2. What is the infimum of f on W
3. Is the infimum attained and explain

For part 1) I know the definition of what a convex function is, but this is a functional and I'm not sure on how to show this.

For part 2) I'm trying to use a method used in another similar problem stated here:

Show that the infimum of a functional is zero, but this infimum is never achieved.

However, I can't seem to get it to work. I'm trying to use a triangle function, but the problem is that I can't make the integral equal to zero as easily as they have done in their problem. I can almost do it except for the $v^2$ term in my integrand.

Also if I pick my $v(x)$ to be a constant such as $v(x)=c$, it's true that I can find a value of c that makes the overall integral zero, yet c will be c=i which is an imaginary answer and I don't think this is allowed in my domain.

Any help would be great.

• For 1), the definition of a convex functional is the same than for functions, i.e. $f(tv+(1-t)u)\leq tf(v)+(1-t)f(u)$, for any $t\in[0,1]$ and $u,v\in W^{1,4}$. For 2), can you find a function $v_n$ such that $\sup |v_n| = 1/n$ and $|v_n'|=1$ almost everywhere? – zuggg May 15 '16 at 14:24
• For 3), if the infimum were attained at $v$, what would $v$ verify? You're right about $v(x)=i$ not being allowed here, because we're only talking about real functions. – zuggg May 15 '16 at 14:27
• For the 2) part is Vn a sequence? If not I tried to come up with a function that has derivative as 1, but all I can think of is a linear function of the form y=mx+b, but this doesn't work. If it looks like 1/n then maybe its y=exp(-x) but derivative is not 1. – Erock Brox May 16 '16 at 2:04
• @ErokBrox I was hinting at a $v_n$ that is piecewise linear, so that it's derivative is +1 or -1 almost everywhere. Think of a saw: you can get the teeth as small as you want them to be, their slope always being 1. – zuggg May 16 '16 at 14:15