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Let $\Omega$ be a bounded Lipschitz domain in $\mathbb R^{d=3}$. Consider the convex functional $F:V\to \mathbb R\cup \{+\infty\}$ defined by $F(v)=\int\limits_{\Omega}{e^vdx}$, where $V$ can be $H_0^1(\Omega)$ or $L^2(\Omega)$.

Question 1. If $v\in V$ is s.t. $F(v)=+\infty$, i.e $\int\limits_{\Omega}{e^vdx}=+\infty$, then is it true that $F(\lambda v)= \int\limits_{\Omega}{e^{\lambda v}dx}=\|e^v\|_\lambda^{\frac{1}{\lambda}}<\infty$ for small enough $\lambda>0$ ? In other words, if $e^u\notin L^1(\Omega)$ then is there $1\ge\lambda_0>0$ s.t. for all $0<\lambda<\lambda_0$ it holds $e^v\in L^\lambda(\Omega)$?

(the question is for $V=H_0^1(\Omega)$, but examples or answers for $V=L^2(\Omega)$ are also welcome)

Question 2. If $u\in V$ is s.t. $F(u)<\infty$, then is $F$ bounded in a neighborhood of $u$ ? In other words, is there $M>0$ s.t. $F(u+v)<M,\,\forall v\in B(0,r)\subset V$ for some positive $r$ ? (Here I am interested in both cases for $V$)

Note that the first question is weaker than the second and the proposition in it would follow from the proposition in the second question with $u=0$. Also, for dimension $d=2$ it holds that $\int\limits_{\Omega}{e^udx}<\infty$ for all $u\in H_0^1(\Omega)$ (see this). However, in $d=3$ there are functions like $u=\ln{\frac{1}{|x|^\alpha}}\in H_0^1(B(0,1)),\,\alpha>3$ for which $\int\limits_{\Omega}{e^udx}=\infty$

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Let $\Omega=B(0,2)$. Fix a smooth function $\phi$ with $\phi(x)=1$ inside $B(0,1)$ and $\phi(x)=0$ outside $\Omega$. Define $$v(x)=\phi \cdot |x|^{-1/3}$$ Then $v$ is in $H^1_0$, but $\exp(\lambda v)$ is not integrable for any $\lambda>0$ because $\exp(\lambda v)>|x|^{-3}$ for sufficiently small $|x|$.

This answers (1) negatively, hence also answers (2) negatively. For $d=1,2$ we still have $v\in L^2$ but $|\nabla v|$ is $\tfrac{1}{3}|x|^{-4/3}$ inside $B(0,1)$, so is not square-integrable.

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