$\newcommand{\R}{\mathbb{R}}$ $\newcommand{\lam}{\lambda}$
I am trying to prove the following claim:
Let $\{u:[0,1]\to \mathbb{R} \mid u \, \, \text{ is differentiable a.e}, u(0)=u(1)=0 \}^{*} $. Consider the functional $I(u)=\int_0^1 ((u')^2-1)^2+u $
For every minimizing sequence $u_n$ of $I$, the sequence of derivatives $u_n'$ generates the Young measures $\nu_x =\frac{1}{2}(\delta_1+\delta_{-1})$
(This is taken from wikipedia).
So far I have succeeded in proving** this only for a specific minimizing sequence, which is a natural one to consider. My proof uses the specific structure of this sequence, so it is not applicable as is to the general case.
My question then, is how to prove this is true for every minimizing sequence?
(*) Perhaps the right space should be some Sobolev space (and then the boundary condition is in the trace sense).
(**) Here is my proof:
Define $A_1=[0,\frac{1}{2}],A_2=[0,\frac{1}{4}] \cup[\frac{2}{4},\frac{3}{4}],A_3=[0,\frac{1}{8}]\cup [\frac{2}{8},\frac{3}{8}]\cup [\frac{4}{8},\frac{5}{8}]\cup [\frac{6}{8},\frac{7}{8}]$, and so on for $n \in \mathbb{N}$. Define a sequence of ``zig-zag triangle'' functions, such that $|u_n| \le \frac{1}{2^n}$, and
$$ u'_n(x) = \begin{cases} 1, & \text{if $x\in A_n$} \\ -1, & \text{if $x \notin A_n$} \end{cases} $$
Then $(u'_n)^2=1$ a.e so $I(u_n) \le \frac{1}{2^{2n}} \rightarrow 0$.
Let $f:\R \to \R$ be a continuous function satisfying $\lim_{|\lambda|\to\infty}f(\lambda) =0$. Then
$$ (f \circ u'_n)(x) = \begin{cases} f(1), & \text{if $u_n'(x) =1$} \\ f(-1), & \text{if $u_n'(x) = -1$} \end{cases} $$
i.e,
$$ (f \circ u'_n)(x) = \begin{cases} f(1), & \text{if $x\in A_n$} \\ f(-1), & \text{if $x \in [0,1] \setminus A_n$} \end{cases} $$
So, denoting $f \circ u_n' =F_n$, we need to show $F_n$ converges in the weak* topology on $L^\infty([0,1])$ to the function: $F(x) = \int_{\R} f d\nu_x =\frac{1}{2}(f(1)+f(-1))$.
By definition of weak* convergence in $L^{\infty}=\left( L^1 \right)^*$ this means that for every $g \in L^{1}([0,1])$ :
$$(1) \lim_{n \to \infty} \int_0^1 F_n(x)g(x)\,dx= \int_0^1 F(x)g(x)\,dx= \frac{1}{2}(f(1)+f(-1)) \cdot \int_0^1 g(x)\,dx $$
Now, $$ \int_0^1 F_n(x)g(x)\,dx= f(1) \int_{A_n} g(x)\,dx + f(-1) \int_{[0,1] \setminus A_n} g(x)\,dx$$
Noting that $[0,1] \setminus A_n = A_n + \frac{1}{2^n}$, we get
$$(2) \int_0^1 F_n(x)g(x) \, dx= f(1) \int_{A_n} g(x) \, dx + f(-1) \int_{A_n + \frac{1}{2^n}} g(x)\,dx$$
So, by $(1),(2)$ it's enough to prove $$ (3) \lim_{n \to \infty} \int_{A_n} g(x)\,dx = \lim_{n \to \infty} \int_{A_n+\frac{1}{2^n}} g(x) \, dx= \frac{1}{2} \int_0^1 g(x)\,dx $$
We turn to handle the second integral:
First, note that we can extend $g$ to $\R$ by putting $g|_{\R \setminus \left[ 0,1 \right]}=0$. (Now we think of $g \in L^1(\R)$).
By variables changing: $y=x-\frac{1}{2^n}$ we get:
$$(4) \int_{A_n + \frac{1}{2^n}} g(x)dx = \int_{A_n}g\left(y+\frac{1}{2^{n}}\right) \,dy=\int_{A_n} g_{\frac{1}{2^n}}(y)dy $$
where $g_h(y)=g(y+h)$ (This is part of the reason we needed to extend $g$ to all $\R$, so now we do not need to worry about what happens when we ``translate'' outside the limits where the original $g$ is defined).
By the $L^1$-continuity of translation,we know: $$ \lim_{h \to 0} \int_\R |g_h(x)-g(x)| \, dx =0, $$ so in particular:
$$ \lim_{n \to \infty} \int_\R |g_{\frac{1}{2^n}}(x)-g(x)|\,dx = 0, $$
and since $$ \int_\R \left|g_{\frac{1}{2^n}}(x)-g(x)\right|\,dx \ge \int_{A_n} \left|g(x)_{\frac{1}{2^n}}-g(x)\right|\,dx \ge 0 $$
we get: $$ \lim_{n \to \infty} \int_{A_n} \left|g_{\frac{1}{2^n}}(x)-g(x)\right|\,dx = 0, $$
so
$$ \lim_{n \to \infty} \left| \int_{A_n} g_{\frac{1}{2^{n}}}(x)-g(x)\,dx \right| = 0, $$
which by $(4)$ is equivalent to:
$$ \lim_{n \to \infty} | \int_{A_n+\frac{1}{2^{n}}} g(x) -\int_{A_n}g(x)dx \, | = 0, $$
However, it is also holds that:
$$ \int_{A_n+\frac{1}{2^{n}}} g(x) + \int_{A_n}g(x)dx = \int_0^1 g(x)dx$$
So, after denoting $a_n = \int_{A_n+\frac{1}{2^{n}}} g(x)dx \, , \, b_n =\int_{A_n}g(x)dx, c=\int_0^1 g(x)dx $, we are in the following situation:
$$a_n+b_n=c, \lim_{n \to \infty}|a_n-b_n|=0 $$
This implies: $2a_n=(a_n+b_n)+(a_n-b_n)=c +(a_n-b_n) \Rightarrow \lim_{n \to \infty} a_n =\frac{c}{2} $, so actually $ \lim_{n \to \infty} b_n =\lim_{n \to \infty} a_n =\frac{c}{2} $
In our context, this translates into:
$$ \lim_{n \to \infty} \int_{A_n} g(x)\,dx = \lim_{n \to \infty} \int_{A_n+\frac{1}{2^{n}}} g(x)\,dx= \frac{1}{2} \int_0^1 g(x)\,dx $$
Which is exactly $(3)$ we wanted to show.