# 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.

• This fact you can see it clearly if you are familiar with the characterization of the norm $H^1$ (or more generally $H^s$ with $s \in \mathbb{R}$ ) through the Fourier transform, and using the representation Riesz theorem. – user288972 Mar 12 '16 at 21:09

(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 $$\left \| T \right \|_{*} = \left \| h \right \|_{H^s} = \left \| \omega_s \mathcal{F}h \right \|_{L^2} =\left \| v \right \|_{H^{-s}}.$$