# $\int\limits_{\Omega}{uvdx}<\infty,\forall v\in H_0^1(\Omega)$ implies $u\in L^{6/5}(\Omega)$

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 $$a(u,v)+\int\limits_{\Omega}{f(u)vdx}=0,\forall v\in H_0^1(\Omega)$$

or

$$(2)$$ 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.

• @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. – Svetoslav Jan 9 '16 at 19:28