Let $d=3$ and $\Omega\subset \mathbb R^d$ is a bounded Lipschitz domain and $u$ is a measurable function. A sufficient condition for the integral $\int\limits_{\Omega}{uvdx}<\infty,\forall v\in H_0^1(\Omega)$ is that $u\in L^{6/5}(\Omega)$ which follows from Holder's inequality and the (continuous) embedding $H^1(\Omega)\hookrightarrow L^6(\Omega)$.

Question: Is the opposite true, i.e is it true that $$\int\limits_{\Omega}{uvdx}<\infty,\forall v\in H_0^1(\Omega)$$ implies $u\in L^{6/5}(\Omega)$ or at least $u\in L^1(\Omega)$ ?

My thoughts: It is easy to see that $u\in L^1_{loc}(\Omega)$ by taking $v$ to be smooth cut-off functions equal to $1$ in compact subsets of $\Omega$ and $0$ in a neighborhood of the boundary $\partial \Omega$.

The motivation for this question is the "correct" weak formulation of a nonlinear problem - whether to formulate it as $(1)$ or as $(2)$:

$(1)$ Find $u\in H_0^1(\Omega)$ such that $f(u)\in L^{6/5}(\Omega)$ and $$a(u,v)+\int\limits_{\Omega}{f(u)vdx}=0,\forall v\in H_0^1(\Omega)$$


$(2)$ Find $u\in H_0^1(\Omega)$ such that $\int\limits_{\Omega}{f(u)vdx}<\infty,\forall v\in H_0^1(\Omega)$ and $$a(u,v)+\int\limits_{\Omega}{f(u)vdx}=0,\forall v\in H_0^1(\Omega)$$

where $a(.,.)$ is a bilinear form and $f(.)$ is in general a nonlinear function. If the answer to my question is affirmative then both formulations are equivalent.

Note that $(2)$ is less restrictive, because the set in which we search for a solution is bigger, so it might be easier to find such.

  • $\begingroup$ @Tomás I reposted the question in MathOverflow here mathoverflow.net/questions/227892/… and I have an answer. I just can not understand how the important inequality in the comment is derived. $\endgroup$ – Svetoslav Jan 9 '16 at 19:28

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