# 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?

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 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 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.

• regarding 1). What about the case of $v in H_0^1(\Omega)$ not bounded but also negative. Then no issue with $\int_\Omega e^v dx$... – Math604 Jul 12 '15 at 23:58
• @Math604 What do you mean by $vinH_0^1(\Omega)$ ? – Svetoslav Jul 13 '15 at 8:23
• Aha, I understand - a typo. Yes, you are right, so I will modify a bit the question, because I initially meant weather $\int\limits_{\Omega}{cosh(v)dx}$ is summable. So I need both $e^v$ and $e^{-v}$ to be summable at the same time. – Svetoslav Jul 13 '15 at 9:14

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 $B(0,\frac1e)$