Existence of minimum of a functional on $L^p (0,1)$ I'm asked to solve the following problem.
Let $1 < p < \infty$ and let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a convex function satisfying $\alpha |\xi|^p \le f(\xi) \le \beta (|\xi|^p + 1), \forall \xi \in \mathbb{R}$.
Define $F: L^p (0,1) \rightarrow \mathbb{R}$, $F(u) := \int_{0}^{1} (1+x^2) f(u(x)) dx$ for all $u \in L^p(0,1)$.
Let $K := \{u \in L^p(0,1) : \int_{0}^{1} u(x) dx = 1 \}$.
Prove that:

*

*$K$ is weakly sequentially closed.

$\textit{Proof}$: Let $(u_n)_n \in K$ (thus, $\forall n \in \mathbb{N}$, we have that $\int_{0}^{1} u_n(x) dx = 1$). Our aim is to prove that if $u_n$ converges weakly to $u$ in $L^p(0,1)$ then $u \in K$. By definition of weak convergence in $L^p(0,1)$, $u_n$ converges weakly to $u$ in $L^p(0,1)$ $\Leftrightarrow$ $\forall v \in L^q (0,1), \frac{1}{p} + \frac{1}{q} = 1$, $\int_{0}^{1} u_n(x) v(x) dx \rightarrow \int_{0}^{1} u(x)v(x) dx$. Therefore, it suffices to consider $v = 1$ ($v \in L^q(0,1)$ for any $1<q<\infty$) in the preceding expression, to conclude ($\int_{0}^{1} u_n(x) dx = 1 \rightarrow \int_{0}^{1} u(x) dx$ for weak convergence, thus $\int_{0}^{1} u_n(x) dx = 1$, which implies $u \in K$).


*$F$ is sequentially lower semi-continuous with respect to the weak convergence.

$\textit{Proof}$: We have to show that, if $(u_n)_n \in L^p(0,1)$ is such that $u_n$ converges weakly to $u$ $\Rightarrow$ $F(u) \le \lim \inf_n F(u_n)$. I know that $f$, being convex on $(0,1)$, is also continuous, so my idea was to apply somehow Fatou's lemma to the sequence $(u_n)$, but I was not able to conclude.


*I have already proved that the problem $min \{F(u) : u \in K\}$ has a solution and that solution is unique if $f$ is strictly convex. I'm struggling to find the explicit solution for $p = 2$, $f(\xi) = |\xi|^2$. I have only found that the minimum is the $L^2$ norm of a function $\overline{u}$ which I am not able to determine (this follows from the reflexivity of $L^2(0,1)$ and the convexity of $K$). Any suggestion would be greatly appreciated.

 A: Thanks to the hint by PhoemueX, I've come up with a satisfying solution to the last part of my question.
Let $F(u) := \int_{0}^{1} dx (1+x^2) |u(x)|^2$, for all $un \in L^2(0,1)$.
Observe that $F(u)^{1/2}$ defines an equivalent norm on $L^2(0,1)$. In fact, $F(u) = \int_{0}^{1} dx (1+x^2) |u(x)|^2 \ge ||u||_{L^2}$ and $F(u) = \int_{0}^{1} dx (1+x^2) |u(x)|^2 \le 2 ||u||_{L^2}$. We denote $F(u)^{1/2} = ||u||$.
Thus, we can define on $L^2(0,1)$ the following scalar product
$(u,v) := \int_{0}^{1} (1+x^2) u(x)v(x) dx, \forall u,v \in L^2(0,1)$ (which is well defined since $uv \in L^1(0,1), (1+x^2) \in L^{\infty}(0,1)$.
Let $u \in K$. Then
$1 = |\int_{0}^{1} u(x) dx| = |\int_{0}^{1} dx (1+x^2) \frac{u(x)}{1+x^2} = |(u,v)|$ where $v(x) = \frac{1}{1+x^2}$.
Using Cauchy-Schwarz inequality, we can state that
$1 = |(u, v)| \le ||u||^2 ||v||^2 = F(u) F(v)$. (*)
At this point, we can easily calculate $F(v) = \int_{0}^{1} \frac{1}{1+x^2} dx = \frac{\pi}{4}$. Hence, $F(u) \ge \frac{4}{\pi}$, for all $u \in K$. The equality in (*) holds when $u(x)$ is proportional to $\frac{1}{1+x^2}$. Then for $\overline{u}(x) = \frac{4}{\pi} \frac{1}{1+x^2}$, we have that $F(\overline{u}) = \frac{4}{\pi}$, which is the sought minimum.
A: Taking the functional derivative wrt $u$ shows that the optimal $u^*$ satisfies $f'(u^*(x))=c/(1+x^2)$ for some $c$ to be chosen to satisfy the integral constraint.
