Let $\Omega\subset\mathbb{R}^N$ be a bounded domain and $p\in (1,2)$. Take $u\in W_0^{1,p}(\Omega)$ and suppose that $$\int_\Omega uv<\infty,\ \forall\ v\in W_0^{1,p}(\Omega)$$
Does this implies that $u=0$?
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Take $u\in C^\infty_c(\Omega)\subset W^{1,p}_0(\Omega)$. For every $v\in W_0^{1,p}(\Omega)$ you have $$ \left|\int_\Omega uv \right| ~\leq~ \int_\Omega|uv| ~\leq~ \|u\|_{L^\infty}\|v\|_{L^1} ~<~ \infty $$ where the last inequality comes from the fact that since $\Omega$ is bounded, then $L^p(\Omega)\subseteq L^1(\Omega)$, therefore if the $L^p$-norm of $v$ is finite, so is its $L^1$-norm.
To sum up, $u$ need not be zero.
AndreasT
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Anyway, the assumption of boundedness of $\Omega$ is not necessary, nor the fact that you are considering $W^{1,p}$ rather than $L^p$. A compact supported function $u$ would work anyway. – AndreasT Mar 04 '13 at 14:41