If $H$ is a Hilbert space and $H^* $ is its dual and $(\cdot ,\cdot)$ is the inner product defined on $H$, $\langle \cdot , \cdot \rangle$ is the duality pairing.
If I want to prove that $u=u’$ in $H^*$, is it enough to show that $$(u,v)=(u’,v) \forall v \in H$$
And does this depends on the space where we take the elements, whether they are in $H$ or in $H^*$??
First of all, if $$ \langle u, v\rangle = \langle u', v\rangle \forall v \in H, $$ then $u = u'$ per definition. What you are asking is if something similar holds if $$ ( u, v) = ( u', v) \forall v \in H. $$ The issue is that a priori this statement does not really make sense. It only gets some meaning if you identify $H$ with $H'$ with the Riesz map, i.e. the isomorphism between $H$ and $H'$ guaranteed by the Theorem of Frechet-Riesz.
If you make the identification between $H$ and $H'$ then the question becomes: For $u,u' \in H$ does $$ ( u, v) = ( u', v) \forall v \in H. $$ imply $u = u'$. The answer is yes, by taking $v = u-u'$ as pointed out in the first comment by Yanko.