Given $\mathbf{f}$ with $f_1,...,f_N\in L^2(\Omega)$ $$\int_\Omega \mathbf{f} \cdot \nabla v = 0 \quad\forall v \in H_0^1(\Omega)$$ we have $\mathbf{f} = \mathbf{0}$ a.e. since $\mathbf{f} \in H^{-1}(\Omega)$.
Without using functional analysis, can we explicitly show the above result by cleverly choosing the test function?
For example for $f\in L^2(\Omega)$ and $\int fg = 0 \;\;\forall g\in L^2$, then choosing $g= f$, we have $$\int f^2 = 0 \quad \Rightarrow \quad f=0 \text{ a.e. }$$
But here $\mathbf{f}$ might not be the weak gradient of a function.
Thank you very much!
I think your "result" is not true. Take $\Omega = (0,1)$ for simplicity. Then, $$\int_0^1 f \, v' \, \mathrm{d} x = 0\quad\forall v \in H_0^1(\Omega)$$ just implies that $f$ is constant.