This is Exercise 4.10 on Brezis and I am having trouble understanding Brezis' solution for part 3 of the problem:
This is the part of the solution where I am lost:
My question is two-folds: Why do we know that there exists a unique maximum for the given set above? I am trying to show that the function $f(x)u - j(u)-\frac{1}{n}|u|^p$ is strictly concave and approaches $-\infty$ as $|u| \to \infty$. However, I am skeptical about $f(x)u - j(u)-\frac{1}{n}|u|^p \to -\infty$ as $|u| \to \infty$. In particular, since we do not know the behavior of $j$, how can we conclude the value of the limit? My second questions, how do we know $$ j(u(x)) + \frac{1}{n}|u(x)|^p - f(x)u(x) \leq j(0) $$ implies $u \in L^p(\Omega)$ and $j(u) \in L^1(\Omega)$? I am trying to show this through integrating both sides, but this approach needs us to have $\int_\Omega f(x)u(x) \,dx < \infty$. This does not seem to be obvious to me. What am I missing here?
Let $a=j'_+(0)$ denote the right derivative of $j$ at $0$, which exists by convexity. Then $j(u) \ge au+b$ for all real $u$, where $b=j(0)$. (In other words, a convex function has a subgradient at every point on its graph. ) Therefore, $$\Psi(u):=f(x)u-j(u)-\frac{1}{n}|u|^p \le [f(x)-a]u-b-\frac{1}{n}|u|^p \to -\infty \quad \text{as} \; |u| \to \infty \,,$$ because $p>1$. $\;$ Since $\Psi$ is strictly concave, there is a unique $u=u(x)$ where its maximum is attained.
Next, from $$ j(u(x)) + \frac{1}{n}|u(x)|^p - f(x)u(x) \leq j(0)=b \,, \quad(*) $$ we infer that for all $x$, $$ \frac{1}{n}|u(x)|^p \le [f(x)-a] \cdot u(x) \, \,, $$ so $$ |u(x)| ^{p-1} \le n|f(x)-a| \,.$$ Thus, if $\mu$ is the measure that $\Omega$ is equipped with, then $$ \int_\Omega |u(x)|^p \, d\mu=\int_\Omega |u(x)|^{(p-1)p'} \, d\mu \le n^{p'}\int_\Omega |f(x)-a|^{p'} \, d\mu <\infty \,, $$ where we used finiteness of $\mu(\Omega)$ and the assumption $f \in L^{p'}$ in the rightmost inequality. Hence $u \in L^p$.
Observe that $(*)$ also yields $$0 \le j(u(x))-au(x)-b \le |f(x)-a| \cdot |u(x)| \,.$$ Since $f-a \in L^{p'}$ and $u \in L^p$, Holder's inequality implies that $|f(x)-a| \cdot |u(x)| \in L^1$, so $j \circ u-au-b \in L^1$ as well. Finally $u \in L^p$ so $u \in L^1$, and the triangle inequality gives that $j \circ u \in L^1$.