Consider $\mathbb{X} = \{ u: [0,1] \to \mathbb{R} : u(0)=u(1)=0, u \in C^1([0,1]) \} $.
For any $n \geq 2$ let's consider the functional $F_n : \mathbb{X} \to \mathbb{R}$ such that $$F_n(u) = \int_0^1 (|u'|^n + (u-x)^{76} + 2^{-nu}) \space dx \space \space \forall u \in \mathbb{X}$$
- Prove that for any $n\geq 2$ , $F_n$ attains its minimum $m_n$ over $\mathbb{X}$.
- Find $\lim\limits_{n \to +\infty} m_n$
I am able to solve the first part, considering the appropriate Sobolev's Space and then proving the minimum is indeed $C^1$.
I am trying to solve the second part, but I am not able.
I think I have to find the right $\Gamma-$limit in $L^2$ and then prove there is a compact $\mathbb{K}$ in which all the $F_n$ attain their minima (which should lead to $\lim\limits_{n \to +\infty} m_n$ being the infimum/minimum of the limit problem).
Thus I have thought about some functionals that could be the $\Gamma-$limit, but I can't prove they are indeed what I look for.
Here is a rough sketch, probably with some small errors, also possible that it can be simplified. Caveat emptor.
We set on $W^{1,\infty}(0,1)$ $$F(u)=\begin{cases} \int_0^1 (u-x)^{76} dx & u\ge 0, |u'|\le 1\text{ almost everywhere}\\ +\infty & \text{ else}. \end{cases} $$
Let $u_n\in C^1$ be such that $F_n(u_n)\le C<\infty$. Then $(1+\frac1{\sqrt{n}})^n |\{|u_n'|\ge 1+\frac1{\sqrt n}\}|\int_{|u_n'|>1+\frac1{\sqrt n}} |u_n'|^n dx\le C$ so we have that $b_n=|\{|u_n'|\ge 1+\frac1{\sqrt n}\}|$ satisfies $b_n (1+\frac1{\sqrt n})^n \le C$, so (as $(1+\frac1{\sqrt n})^n\to\infty$) we have $b_n\to 0$. Choose a subsequence with $|u_n'|\le 2$. Then that subsequence is bounded in $W^{1,\infty}$, and converges weak* and uniformly to some $u\in W^{1,\infty}=C^{0,1}$. So we have compactness.
To show the lower bound, consider $u_n\overset{*}\rightharpoonup u$. We need to show that $F_n(u_n)\to\infty$ if $F(u)=\infty$, then the lower bound follows trivially because all we did is replace terms by zero if the conditions are not met. Wlog let $F_n(u_n)$ be convergent (possibly to infinity). Let $\delta>0$. If $|u'|>1+\delta$ on a set of positive measure then by weak lower semicontinuity in $W^{1,p}$, $\liminf_{n\to\infty}\int |u_n'|^p\ge (1+\delta)^p$. For every $p=1,2,\dots,$ iteratively choose subsequences such that $\int |u_{n,p}'|^p\ge (1+\delta)^p$. Then the diagonal sequence satisfies $\int |u_{n,n}'|^n dx\ge (1+\delta)^n$, so $F_n(u_n)\to\infty$. The argument for $u\ge 0$ is similar.
For the upper bound, consider $u\in C^{0,1}$ with $F(u)<\infty$. If $|u'|<1$ and $u>0$ outside a set of measure $0$, take $u_n=u$. Otherwise, increase $u$ slightly on intervals where it is zero (one zigzag with derivative $\pm\frac1n$) and otherwise set $u_n=(1-\frac1{\sqrt n})u$. Then we should have $F_n(u_n)\to F(u)$.