A constrained variational problem

We proceed in several steps.

Step 1. Coercivity of $\mathcal{F}$.

To prove the coercivity of $\mathcal{F}$, we need to show that $\mathcal{F}(u) \to \infty$ as $\|u\|_{W^{1,4}_0(\Omega)} \to \infty$. Consider the functional $\mathcal{F}(u)$:
\[ \mathcal{F}(u) = \int_\Omega |\nabla u|^4 \, dx - \int_\Omega fu \, dx. \]
By applying the Hölder inequality with conjugate exponents $p = 4$ and $q = \frac{4}{3}$, we obtain
\[ \left| \int_\Omega fu \, dx \right| \leq \| f \|_{L^{\frac{4}{3}}(\Omega)} \| u \|_{L^4(\Omega)}. \]
By Gagliardo–Nirenberg–Sobolev inequality (https://en.wikipedia.org/wiki/Sobolev_inequality) there exists a positive constant $C$ such that
\[ \| u \|_{L^4(\Omega)} \leq C \| \nabla u \|_{L^\frac{12}{7}(\Omega)}. \]
Using this inequality, we can estimate the second term of $\mathcal{F}(u)$:
\[ \left| \int_\Omega fu \, dx \right| \leq \| f \|_{L^{\frac{4}{3}}(\Omega)} C \| \nabla u \|_{L^\frac{12}{7}(\Omega)}. \]
It follows that
\[ \mathcal{F}(u) \geq \| \nabla u \|_{L^4(\Omega)}^4 - C \| f \|_{L^{\frac{4}{3}}(\Omega)}\| \nabla u \|_{L^\frac{12}{7}(\Omega)}. \]
As $\| u \|_{W^{1,4}_0(\Omega)} \to \infty$, so must $\| \nabla u \|_{L^4(\Omega)}$ since it contributes to the Sobolev norm. Therefore, the term $\| \nabla u \|_{L^4(\Omega)}^4$ increases more rapidly than the term $\| f \|_{L^{\frac{4}{3}}(\Omega)}\| \nabla u \|_{L^\frac{12}{7}(\Omega)}$. Hence,
\[ \lim_{\| u \|_{W^{1,4}_0(\Omega)} \to \infty} \mathcal{F}(u) = \infty \]
and the functional $\mathcal{F}(u)$ is coercive over $W^{1,4}_0(\Omega)$.

Step 2. The set $A=\{u \in W^{1,4}_0(\Omega): \int_{\Omega}g(u)dx=0 \}$ is weekly closed. 

Suppose $u_n$ converges to some $u$ weekly in $W^{1,4}_0(\Omega)$. It follows from Soboleve inequalities that $W^{1,4}_0(\Omega) \subset C^{0, \frac{1}{4}}(\Omega)$, and hence $u_n$ converges to $u$ also in $C^{0, \frac{1}{4}}(\Omega)$. 


Let $\{u_n\}$ be a sequence of differentiable functions on $\Omega \subset \mathbb{R}^3$, and $\Omega$ is bounded. Suppose $u_n$ converges to some $u$ in $C^2(\Omega)$. Then for any continuous function $g$, the following limit holds:
\[ \lim_{n \to \infty} \int_\Omega g(u_n) \, dx = \int_\Omega g(u) \, dx. \]



Since $u_n$ converges to $u$ in $C^{0, \frac{1}{4}}(\Omega)$, it implies uniform convergence on the compact set $\overline{\Omega}$, as $C^{0, \frac{1}{4}}(\Omega)$ implies continuity of $u$ as well as its first and second derivatives. Thus, for every $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that for all $n \geq N$,
\[ \sup_{x \in \Omega} |u_n(x) - u(x)| < \epsilon. \]

Given that $g$ is continuous, it is also uniformly continuous on the compact image of $u$ (and $u_n$ for large enough $n$ due to uniform convergence). Therefore, for every $\epsilon > 0$, there exists a $\delta > 0$ such that for all $x, y$ with $|x - y| < \delta$,
\[ |g(x) - g(y)| < \epsilon. \]

By the uniform convergence of $u_n$ to $u$, we can choose $n$ large enough such that $|u_n(x) - u(x)| < \delta$, which implies $|g(u_n(x)) - g(u(x))| < \epsilon$ for all $x \in \Omega$.

Now, consider the integrals:
\[ \left| \int_\Omega g(u_n) \, dx - \int_\Omega g(u) \, dx \right| = \left| \int_\Omega [g(u_n) - g(u)] \, dx \right|. \]

Since $\Omega$ is bounded, let $V$ denote its volume. By the uniform continuity of $g$ and the uniform convergence of $u_n$ to $u$, we can make $|g(u_n) - g(u)|$ as small as needed for all $x \in \Omega$, hence:
\[ \left| \int_\Omega [g(u_n) - g(u)] \, dx \right| \leq \int_\Omega |g(u_n) - g(u)| \, dx < \int_\Omega \epsilon \, dx = \epsilon V. \]

As $\epsilon$ was arbitrary, we can make it as small as we wish, and thus the integral on the left-hand side can be made arbitrarily small for sufficiently large $n$. This leads to the conclusion that:
\[ \lim_{n \to \infty} \int_\Omega g(u_n) \, dx = \int_\Omega g(u) \, dx, \]
completing our proof.

Step 3. Week lower semi continuity of $\mathcal{F}$. Note that 
\[ \mathcal{F}(u) = \int_\Omega |\nabla u|^4 \, dx - \int_\Omega fu \, dx \leq \|u\|_{W^{1,4}_0(\Omega)}+\| f \|_{L^{\frac{4}{3}}(\Omega)} \| u \|_{L^4(\Omega)} \]
\[\leq \|u\|_{W^{1,4}_0(\Omega)} + C \|u\|_{W^{1,4}_0(\Omega)}=C' \|u\|_{W^{1,4}_0(\Omega)}.\]
Hence $\mathcal{F}$ is continuous on $\|u\|_{W^{1,4}_0(\Omega)}$. Since $\mathcal{F}$ is also convex, it is weakly lower continuous. 

Now we are ready to prove the result. Consider minimizing $\mathcal{F}$ on the set $A=\{u \in W^{1,4}_0(\Omega): \int_{\Omega}g(u)dx=0 \}$.  Suppose $u_n$ is a minimizing sequence, since $\mathcal{F}$ is coercive, it should have a converging subsequence, denoted by $u_n$ again, that converges to some $u \in W^{1,4}_0(\Omega)$. Moreover, since $A$ is weakly closed, $u \in A$. Finally since $\mathcal{F}$ is weakly lower semi continuous, 


$$\mathcal{F}(u) \leq \liminf_{n \to \infty} \mathcal{F}(u_n),$$
and thus $u\in A$ is a minimizer. 


Refer to : https://en.wikipedia.org/wiki/Sobolev_inequality for Sobolev inequalities 

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We use the Sobolev inequality 
\[W^{k,p}(\Omega)\subset C^{r, \alpha} (\Omega)\]
with $k=1$, $p=4$, $r=0$, $\alpha=\frac{1}{4}$, and $n=3$
so that
\[r+\alpha=k-\frac{n}{p}.\]

Also note that $\Omega$ is bounded and hence $\bar{\Omega}$ is compact, and we can use all argument above are fine.