Explicit characterization of dual of $H^1$ Let's start by some well-known facts:


*

*$H^1(\mathbb{R})$ is a Hilbert space, hence there holds the Riesz representation theorem, stating that any linear functional on it can be represented as $L = \langle \cdot , v\rangle$, where $v \in H^1(\mathbb{R})$, and $\langle \cdot, \cdot \rangle$ is the pairing in $H^1(\mathbb{R})$:
$$
\langle u, v \rangle = \int_{\mathbb{R}} (u_x v_x + uv) d x.
$$
Here, subscript indicates derivative.

*Let $w \in L^2(\mathbb{R})$. Since the $H^1$-norm is stronger than the $L^2$-norm, the functional $v \mapsto L(v) := \int_{\mathbb{R}^2} w(x) v(x) d x$ is a continuous linear functional on $H^1(\mathbb{R})$.
My question is: is it possible to give the representative explicitly? In other words, find $z \in H^1(\mathbb{R})$ such that $L(v) = \langle v,z\rangle$.
I tried mimicking the proof of the Riesz theorem but it did not help. I am sure there is something really easy I don't see, but I seem to have some misconception here.
Any help would be appreciated.
 A: (i) $H^s(\mathbb{R}^n):=\lbrace u \in \mathcal{S}'(\mathbb{R}^n) : \Lambda^s u \in L^2(\mathbb{R}^n) \rbrace$ with $\omega_s(\xi):=(1+|\xi|^2)^{s/2}$, $\Lambda^s u := \mathcal{F}^{-1}(\omega_s \widehat{u})$ $\forall u \in \mathcal{S}'(\mathbb{R}^n)$, and by Plancherel theorem we have the norm
$$\left \| u \right \|_{H^s}= \left \| \Lambda^s u \right \|_{L^2}= \left\| \mathcal{F}(\Lambda^s u) \right\|_{L^2}=\left \| \omega_s \widehat{u} \right \|_{L^2}=\left( \int_{\mathbb{R}^n} |\widehat{u}(\xi)|^2 (1+|\xi|^2)^s d\xi \right)^{1/2}$$
Note that you can also define (ii) $H^s(\mathbb{R}^n):=\lbrace u \in L^2(\mathbb{R}^n) : \Lambda^s u \in L^2(\mathbb{R}^n) \rbrace$, the definition (i) is only a little more general. In particular, we have the scalar product
$$ (u,v)_{H^s}:=(\Lambda^s u, \Lambda^s v)_{L^2}= \int_{\mathbb{R}^n} \widehat{u}(\xi) (1+|\xi|^2)^s \overline{\widehat{v}(\xi)} d\xi.$$
Now $H^s(\mathbb{R}^n)$ and $H^{-s}(\mathbb{R}^n)$ are one of the dual isometric of the other. In other words for the space (iii) $H^s(\mathbb{R}^n)^* = \lbrace u \in H^s(\mathbb{R}^n)\longmapsto (u,v)_* : v \in H^s(\mathbb{R}^n) \rbrace$ we have $H^s(\mathbb{R}^n)^* \cong H^{-s}(\mathbb{R}^n)$, and likewise it is shown that $H^{-s}(\mathbb{R}^n)^* \cong H^{s}(\mathbb{R}^n)$. In particular, the application in (iii) are isometries, and $|(u,v)_*| \le \left \| u \right \|_{H^s} \left \| v \right \|_{H^{-s}}$. 
Proof.
let $v \in H^{-s}(\mathbb{R}^n)=\lbrace v \in \mathcal{S}'(\mathbb{R}^n) : \Lambda^{-s}v \in L^2(\mathbb{R}^n) \rbrace$ and $u \in H^s(\mathbb{R}^n)$. Consider the scalar product
$$ (u,v)_*:=(\Lambda^s u , \Lambda^{-s} v)_{L^2}=\int_{\mathbb{R}^n} \widehat{u}(\xi)(1+|\xi|^2)^{s/2} \overline{\widehat{v}}(\xi) (1+|\xi|^2)^{-s/2} d\xi$$
by Schwartz inequality $|(u,v)_*| \le \left \| u \right \|_{H^s} \left \| v \right \|_{H^{-s}}$.
We show that $\forall v \in H^{-s}(\mathbb{R}^n)$, the maps $u \in H^{s}(\mathbb{R}^n) \longmapsto (u,v)_* \in \mathbb{C}$ is an element of the dual space $H^s(\mathbb{R}^n)^*$, infact if $T \in H^s(\mathbb{R}^n)^*$, by Riesz rapresentation therem:
$$
\exists h \in H^s(\mathbb{R}^n): T(u)=(u,h)_{H^s} , \forall u \in H^s(\mathbb{R}^n)
$$
and $\left \| T \right \|_*=\left \| h \right \|_{H^s}$. Define $v=\mathcal{F}^{-1}(\omega_{2s} \widehat{h})$, by Plancherel theorem we have
$$
\left \| v \right \|_{H^{-s}}
= \left \| \Lambda^{-s}v \right \|_{L^2}
= \left \| \mathcal{F}\Lambda^{-s}v \right \|_{L^2}
= \left \| \omega_{-s} \mathcal{F}v \right \|_{L^2}
= \left \| \omega_{-s} \omega_{2s} \mathcal{F}h \right \|_{L^2}
= \left \| \omega_s \mathcal{F}h \right \|_{L^2}
< \infty
$$
therefore $v \in H^{-s}(\mathbb{R}^n)$ with $h \in H^s(\mathbb{R}^n)$, and for the uniqueness 
$$
T(u)=(u,h)_{H^s}
= \int_{\mathbb{R}^n} \widehat{u}(\xi) (1+|\xi|^2)^s \overline{\widehat{h}(\xi)} d\xi
= (u,v)_{*}
$$
and $H^s(\mathbb{R}^n)^* \cong H^{-s}(\mathbb{R}^n)$. In the end
$$
\left \| T \right \|_{*}
=  \left \| h \right \|_{H^s}
= \left \| \omega_s \mathcal{F}h \right \|_{L^2}
=\left \| v \right \|_{H^{-s}}. 
$$
