2
$\begingroup$

I have some doubts about the proof of this theorem. Essentially a proof of it can be found in the book "Real Analysis Modern Techniques And Their Applications" by Folland, but I followed also "Linear Funcional Analysis" by J. Cerda. A point of the demonstration I changed, following a lemma of another book. To better expose the demonstration, do a series of recalls.

Main notations and preliminary results:

$\mathcal{E}(\mathbb{R}^n)$ is the space of $C^\infty$-functions.

$\mathcal{E}'(\mathbb{R}^n)$ is the space of distributions with compact support.

$\mathcal{S}'(\mathbb{R}^n)$ is the space of tempered distributions.

$H^s(\mathbb{R}^n):=\lbrace u \in \mathcal{S}'(\mathbb{R}^n) : \Lambda^s u \in L^2(\mathbb{R}^n) \rbrace$, where $\Lambda^s u = \mathcal{F}^{-1}(\omega_s \widehat{u})$ $\forall u \in \mathcal{S}'(\mathbb{R}^n)$, is the Hilbert-Sobolev spaces with $s \in \mathbb{R}$.

$H_{loc}^s(\Omega):=\lbrace u \in \mathcal{D}'(\Omega) : \psi u \in H^s(\mathbb{R}^n), \forall \psi \in \mathcal{D}(\Omega)\rbrace$

$Theorem (1)$. If $s-k > n/2$, then we have continuous inclusion $H^s(\mathbb{R}^n) \hookrightarrow \mathcal{E}^k(\mathbb{R}^n)$. In particular $\bigcap_{s \in \mathbb{R}} H^s(\mathbb{R}^n) \subset \mathcal{E}(\mathbb{R}^n)$.

$Theorem (2)$. If $m -k > n/2$, then we have continuous inclusion $H^m(\Omega) \hookrightarrow \mathcal{E}^k(\Omega)$.

$Theorem (3)$. If $k \in \mathbb{N}$ and $s \in \mathbb{R}$, then $H^s(\mathbb{R}^n)= \lbrace u \in \mathcal{S}'(\mathbb{R}^n) : D^\alpha u \in H^{s-k}(\mathbb{R}^n), \forall |\alpha| \leq k \rbrace$ and $\left \| u \right \|_{H^s}$, $\sum_{|\alpha|\leq k} \left \| D^\alpha u \right \|_{H^{s-k}}$ are two equivalent norms.

$Lemma (2)$. Each distribution with compact support $u \in \mathcal{E}'(\mathbb{R}^n)$ is an element of set $H^s(\mathbb{R}^n)$ for some $s \in \mathbb{R}$, i.e. $\mathcal{E}'(\mathbb{R}^n) \subset \bigcup_{s \in \mathbb{R}} H^s(\mathbb{R}^n)$

$Lemma (3)$. A differential operator with constant coefficients $P(D)=\sum_{|\alpha| \leq m} a_\alpha D^\alpha$ of order $m \in \mathbb{N}$ is elliptic if and only if $\forall |\xi| \geq R >0$ there is a constant $C >0$ such that $|P(\xi)| \geq C |\xi|^m$.

$Lemma (4)$. Let $P(D)$ an elliptic operator with constant coefficients of order $m \in \mathbb{N}$. If $u \in H^s(\mathbb{R}^n)$ and $P(D)u \in H^s(\mathbb{R}^n)$, then $u \in H^{s+m}(\mathbb{R}^n)$.

$Elliptic-Regularity-Theorem$. Let $L=P(D)$ an elliptic operator with constant coefficients of order $m \in \mathbb{N}$, and $u \in \mathcal{D}'(\Omega)$ a distribution. If exists $s \in \mathbb{R}$ such that $Lu \in H_{loc}^s(\Omega)$, then $u \in H^{s+m}_{loc}(\Omega)$. If $Lu \in \mathcal{E}(\Omega)$ then $u \in \mathcal{E}(\Omega)$.

$Proof$. Note that if $Lu \in \mathcal{E}(\Omega)$, then $\varphi Lu \in H^s(\mathbb{R}^n)$ for all $\varphi \in \mathcal{D}(\Omega)$ and for all $s \in \mathbb{R}$, i.e. $Lu \in H_{loc}^s(\Omega)$. Now, since $L$ is of order $m \in \mathbb{N}$ we want to apply the $lemma (4)$ to verify that $\varphi u \in H^{s+m}(\mathbb{R}^n)$ $\forall \varphi \in \mathcal{D}(\Omega)$, and then $u \in H^{s+m}_{loc}(\Omega)$ by definition. With an application of the $theorem(2)$ follow that $\varphi u \in \mathcal{E}(\Omega)$ $\forall \varphi \in \mathcal{D}(\Omega)$, and then we can assume that $u \in \mathcal{E}(\Omega)$.

Let $\mathrm{supp}(\varphi) \subset U$, where $U$ is a open set with compact closure $\overline{U} \subset \Omega$, by regular version of the Urysohn lemma there is $\psi \in \mathcal{D}(\Omega)$ such that $\overline{U} \prec \psi \prec \Omega$ (this notation means that $\psi(x)=1$ $\forall x \in \overline{U}$ and $0\leq \psi(x) \leq 1$). Therefore $\psi u \in \mathcal{E}'(\mathbb{R}^n)$ is a distribution with compact support and by $lemma (2)$ exists $\sigma \in \mathbb{R}$ such that $\psi u \in H^{\sigma}(\mathbb{R}^n)$.

If $\sigma$ decreases we can assume that $s+m-\sigma=k \in \mathbb{N}$. (why this assumption? in the sense of the $theorem(1)$ or $theorem(2)$?)

Consider $\psi_0=\psi$, $\psi_k=\varphi$ and define $\psi_1,...,\psi_{k-1}$ by recurrence, so that \begin{align*} (\star) \displaystyle \mathrm{supp}(\psi_{j+1}) \prec \psi_j \prec U_j \subset \lbrace \psi_{j-1} =1 \rbrace \end{align*} Then it will be sufficient to prove that $\psi_j u \in H^{\sigma +j}(\mathbb{R}^n)$, since for $j=k$, we have that $\varphi u := \psi_k u \in H^{\sigma + k} (\mathbb{R}^n) = H^{s+m}(\mathbb{R}^n)$.

By definition of regular Urysohn functions as in $(\star)$ (is it correct this statement?) it will be sufficient to prove that if $\varphi, \psi \in \mathcal{D}(\Omega)$ satisfy $\mathrm{supp}(\varphi) \prec \psi$ and $\psi u \in H^\sigma(\mathbb{R}^n)$, then $\varphi u = \psi_k u \in H^{\sigma + 1}(\mathbb{R}^n)$. Now, by induction on $|\alpha|$ you can prove that \begin{align*} \displaystyle [L, \varphi]u:=L(\varphi u) - \varphi Lu \end{align*} is a differential operator of order $|\alpha| \leq m-1$, so that in general for distributional derivative $D^\alpha$ and by $theorem(3)$ this means that $f \in H^{\sigma}(\mathbb{R}^n)$ imply $D^\alpha f \in H^{\sigma-(m-1)}(\mathbb{R}^n)$. Therefore in terms of differential operator and since $\mathrm{supp}(\varphi) \prec \psi$, we have that \begin{align*} \displaystyle [L,\varphi]u=[L, \varphi](\psi u) \in H^{\sigma-(m-1)}(\mathbb{R}^n) \end{align*} Moreover, by hypothesis $\varphi L u \in \mathcal{D}(\mathbb{R}^n)$ and then we can conclude that $L(\varphi u)=[L,\varphi](\psi u) + \varphi Lu \in H^{\sigma-(m-1)}(\mathbb{R}^n)$, and by $lemma(4)$ follow that $\varphi u \in H^{\sigma + 1}(\mathbb{R}^n)$.

Thanks for any help or suggestions.

$\endgroup$

1 Answer 1

1
$\begingroup$

About the doubts wich I had, we can assume that $s+m-\sigma= k \in \mathbb{N}$ essentially to have an inductive argument, and if $\sigma$ decreases then $-\sigma$ increase and $k > 0$.

Concerning the second question, in fact to how we choose our Urysohn functions we have that if $\mathrm{supp}(\varphi) \prec \psi$ (means that $\psi(x)=1$ $\forall x \in \mathbb{supp}(\varphi)=\mathbb{supp}(\psi_k)$ by definition) and $\psi u \in H^{\sigma}(\mathbb{R}^n)$ therefore it will be sufficient to prove that $\varphi u = \psi_k u \in H^{\sigma + 1}(\mathbb{R}^n)$ because by $(\star)$ and $\forall j=1,...,k$ we have that $\psi_j u \in H^{\sigma + j}(\mathbb{R}^n)$.

It should be said better, but the sense is to pass from a form of the inductive argument to another, thanks to the particular choice of Urysohn functions.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.