Equality in dual space

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 $H^*$??

• In an Hilbert space $H$ is isomorphic to $H^\star$. So what you want to show is enough if you take $v=u-u'$ (which is an element in $H$ under the identification). – Yanko May 3 '18 at 20:37
• @Yanko $H^1_0$ is Hilbert ,right? – Math1995 May 3 '18 at 20:40
• I'm not sure what you mean by $H_0^1$. Is it the sobolev space? – Yanko May 3 '18 at 20:41
• Yes it is $H^1$ with zero on the boundary – Math1995 May 3 '18 at 20:42
• In here en.wikipedia.org/wiki/Sobolev_space they says it is indeed an Hilbert space in the special case where $H$ is $L^2$ of the Torus. – Yanko May 3 '18 at 20:46

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.
• What if I know that $u \in H$, can I write the duality pairing? – Math1995 May 4 '18 at 8:35
• I am not sure what you mean. If $u$ and $v$ are both in $H$ then the dual pairing between $u$ and $v$ only makes sense by identifying $u$ with an element of the dual space. You do that by considering the element $\tilde{u} \in H'$ which is the functional $$v \mapsto ( u, v).$$ – pcp May 4 '18 at 8:58
• in case of Hilbert space ,can we always identify $H$ by $H^*$? – Math1995 May 4 '18 at 8:59
• So I can deduce that if $u \in H^*$ and $u \in H$ we can show that $\langle u,v \rangle =(u’,v)$ – Math1995 May 4 '18 at 9:03