I have the following **Question**:


1) Is it true that
if $\Omega\subset\mathbb R^3$, $\Omega$ - bounded, $v\in H\_0^1(\Omega)$ but $v\notin L\_\infty(\Omega)$ implies $\int\limits\_{\Omega}{e^vdx}=+\infty$ or $\int\limits\_{\Omega}{e^{-v}dx}=+\infty$, (or both)? In other words, does it imply that $\int\limits\_{\Omega}{cosh(v)dx}=+\infty$ 


Here is a **[paper](http://www.mi.uni-koeln.de/mi/Forschung/Kawohl/kawohl/Kawohl_LuciaRev.pdf)** which basically gives me that $\int\limits\_{\Omega}{e^udx}< +\infty$ for functions $u\in W\_0^{1,d}(\Omega)$. So, for $d=3$ it might be the case that: if a function $v$ is in $H\_0^1(\Omega)$ but not in $L\_\infty(\Omega)$, then no matter how weak singularity it has, the exponent is already not summable.


My question is motivated by the following problem that I am trying to solve:


Let $F: H\_0^1(\Omega)\rightarrow \overline{\mathbb R}, $ $F(v)=\int\limits\_{\Omega}{f(x,v(x))dx}=\int\limits\_{\Omega}{k(x)cosh(v(x)+g(x))dx}$, where $k(x)\ge 0$ in $\Omega$, $k(x)$ is not identically 0 in $\Omega$ (it can be taken for positive constant) and $g\in S:=L\_\infty(\Omega)\cap H\_0^1(\Omega)$ (dense in $H\_0^1(\Omega)$) is given. Let $V\_\alpha :=\{v\in S | F(v)\leq \alpha\}$ for some arbitrary $\alpha\in\mathbb R$, is the lower level set of $F$ with elements in $S$. I would want to show that $V\_\alpha$ is closed in the $H\_0^1$ norm.
**Here is what I have done:**


First, note that $F$ is convex. Then, I show that $F$ is sequentially weakly lower semicontinuous in $H\_0^1$ by using this **[Theorem 3.20](https://books.google.at/books?id=6Z6zfhAc2tYC&lpg=PP1&pg=PA94#v=onepage&q&f=false)** and the fact that each weakly convergent sequence in $H\_0^1$ is strongly convergent in $L\_2$ and therefore also weakly convergent in $L\_2$. Then proceed as usual: if $\{v\_n\}\_{n=1}^{\infty}\subset V\_\alpha$ and $v\_n\rightarrow v$ in $\|.\|\_{H\_0^1(\Omega)}$ norm, because $F$ is sequentially w.l.s.c $\Rightarrow$ $F(v)\leq \alpha \Rightarrow$ it remains to show that $v\in S=L\_\infty(\Omega)\cap H\_0^1(\Omega)$ (obviously $v\in H\_0^1(\Omega)$). Now, I will be finished if I show one of the following 


1) (the main question from above) - because if I assume $v\notin L\_\infty\Rightarrow F(v)=+\infty$ and by w.l.s.c. of $F\Rightarrow F(v\_n)\rightarrow +\infty$ which is a contradistion with $F(v\_n)\leq \alpha$ for each $n$


or


2) (less restrictive than 1) ) Again assume $v\notin L\_\infty(\Omega)$. We have $\{v\_n\}\subset V\_\alpha, \|v\_n-v\|\_{H\_0^1(\Omega)}\rightarrow 0$. Is it true that $F(v\_n)\rightarrow +\infty$ . 


Also, as a consequence of $v\_n\rightarrow v$ in $\|.\|\_{H\_0^1(\Omega)}$ and $v\notin L\_\infty$, we have $\|v\_n\|\_{L\_\infty(\Omega)}\rightarrow +\infty$ and 2) is like coercivity of $F$ w.r.t $\|.\|\_{L\_\infty}$ norm.


Any help or a reference on 1) or 2) will be very appreciated. Thanks.



Is it true that for dimension d=3, if $v\in H_0^1(\Omega)$ but $v\notin L_\infty(\Omega)$ then exponent of v or -v is not summable?
If it is true that $\int e^u<\infty$ whenever $u\in W^{1,3}$, for $d=3$, **and** $W^{1,3}$ is not a subset of $L^\infty$, then, since $W\_0^{1,3}\subset H\_0^1$, there certainly exist essentially unbounded functions $v$ for which both $e^v$ and $e^{-v}$ are integrable ... Or did I miss something?


Example : $v=\log\log\frac1r$ on $B(0,\frac1e)$


