I posted this question first in Math.StackExchange one week ago [here](https://math.stackexchange.com/q/1596096/254733), but I didn't get an answer or a helpful comment so I repost it here:


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. 



$\int\limits_{\Omega}{uvdx}<\infty,\forall v\in H_0^1(\Omega)$ implies $u\in L^{6/5}(\Omega)$
That doesn't work because $H\_0^1$ functions are small near the boundary, so testing against them won't detect bad behavior of $u$ near $\partial\Omega$.


For a concrete example, take $\Omega$ as the unit ball and $u(x)=1/(1-|x|)\notin L^1$. Then
$$
\int |uv|\, dx \le \left( \int \frac{v^2\, dx}{(1-|x|)^{3/2}} \int \frac{dx}{(1-|x|)^{1/2}} \right)^{1/2} .
$$
If $v\in H\_0^1$ is also smooth, then we can estimate the first integral in the same way as in this [related question](https://mathoverflow.net/questions/226965/convergence-of-sobolev-functions-near-the-boundary/226985#226985) (by just integrating the gradient, starting from the boundary, to bound $v$). This gives $\int v^2/(1-|x|)^{3/2}\lesssim \|v\|^2\_{H^1}$, so $\int |uv| \lesssim \|v\|\_{H^1}$ for all such $v$, and by density of the smooth functions, this also holds for arbitrary $v\in H\_0^1$.


