2010-11-10
(series of three presentations)
These operators show up in many areas of mathematics. We will concentrate here on regularity properties of u.
We study $Lu = f$ on $\Omega$. Assuming we know $f$ and we know $L$, we want to understand $u$? The approach described here is different from what appears in Evans, which I view as more mechanical.
If we can find an operator $L^{-1}$ then $L^{-1}[Lu] = u = L^{-1}f$ and we then know $u$. But, in general, no such $L^{-1}$ exists. So, we want to find an operator $P$ satisfying $P[L[\cdot]] = Id + error$ and same for $L[P[\cdot]]$. We look at $Pf$ to deduce regularity of $u$.
Example:
In General:
Definition: Let $m \in {\mathbb{R}}$. We say that a smooth function $\sigma(x,\xi)$ on $\mathbb{R}^n \times \mathbb{R}^n$ is a symbol of order m if $supp (\sigma) \subset K \times {\mathbb{R}}^n.
Definition: $H^s$ norm defined using Fourier transform…
Definition: For $p \in S^m$ we define a pseudodifferential operaator $P$ associated to symbol $p$ by $$P \phi(x) = \int \widehat{\phi}(\xi) p(x, \xi) e^{-i x \cdot \xi} d\xi.$$
(Some discussion spawned by Arthur Huang about how it would be more natural to write the phase with +…)
Theorem: With all the notation placed above, then $P: H^s \rightarrow H^{s-m}$.
If $m>0$, the theorem tells us that we basically lose smoothness under $P$. For $m<0$, we observe a gain in regularity.
The proof is lengthy, and a bit technical.
Lemma 1: (Complex Variables Lemma) For any complex $a,b$ we have $\frac{1 + |a|}{1+|b|} \leq 1 + |a-b|.$
Proof: $ 1 + |a| \leq 1 + |a-b| + |b| \leq 1 + |a-b| + |b| + |b| |a-b| = (1+ |a-b|)(1+|b|).$
Lemma 2: If $p \in S^m$ then for any multiindex $\alpha$ and any integer $k$, we have where ${\mathcal{F}}$ denotes Fourier transform in the $x$-variable.
$$ |{\mathcal{F}} (D_x^\alpha p(x,\xi)) (\eta)| \leq C_{k, \alpha} \frac{(1+|\xi|)^m}{(1+ |\eta|)^k}. $$
Proof: Let $\gamma$ be some multiindex such that $|\gamma| = k$. Let’s calculate $|{\mathcal{F}} (D_x^\alpha p(x,\xi)) (\eta)|$. He does some calculations…
Lemma 3: $(H^s)^* = H^{-s}$.
We proved this in class so we omit the proof here.
Lemma 3 implies that to estimate the $H^s$ norm of $\phi$, it is enough to check for every $\psi \in H^s$ that $$ |\int \phi(x) \overline{\psi}(x)dx| \leq C \| \psi \|_{H^{-s}}. $$ So we proceed by duality.
Proof of Theorem: Fix a $\phi \in H^s$. We want to study the $H^{s-m}$ norm of $ P\phi.$ We have a representation formula: $$ P \phi (x) = \int \widehat{\phi}(\xi) p(x, \xi) e^{-i x \cdot \xi} d\xi. $$ Notice that this is well-defined because $x$ is restricted to a compact set. (Q? What about in $\xi$?) Take the Fourier transform in $x$ and denote the dual variable by $\eta$. After some manipulations, he finds $$ {\widehat{P \phi}} (\eta) = \int \int \widehat{\phi} (\xi ) p(x,\xi) e^{i (\xi - \eta)\cdot x} d\xi dx. $$ $$ = \int {\widehat{S}}{x} (\eta - \xi, \xi) \widehat{\phi}(\xi) d\xi. $$ where ${\widehat{S}}{x} (\lambda, \xi) = \int e^{i x \lambda} p(x, \xi)d\xi.$
We need to show that for test functions $\psi$ that we have the required duality bound. Using Plancherel follwed by the manipulations above we find, $$ |\int {\widehat{P \phi}} (\xi ) {\widehat{\overline{\psi}}} (\xi) d\xi = |\int \int {\widehat{S}}_x (\xi - \eta, \eta) (1+|\eta|)^{-s} (1+|\xi|)^{s-m} {\widehat{\overline{\psi}}}(\xi) {\widehat{\phi}}(\xi) (1+|\xi|)^{m-s} (1+|\eta|)^s d\eta d\xi.$$
Let us define $K(\xi, \eta) = {\widehat{S}}_x (\xi - \eta, \eta) (1+|\xi|)^{s-m} (1+|\eta|)^{-s}.$
The main thrust of the theorem reduces to proving the claim: Claim: $$ \int |K| d\xi \leq C.$$ $$ \int |k| d\eta \leq C.$$
Assuming the claim we find using Cauchy-Schwarz the upper bound $$ \leq ( \int \int K(\xi, \eta) (1+|\xi|^2)(m-s)|\widehat{\overline{\psi}} (\xi)|^2 d\xi d\eta )^{1/2} ( \int \int K(\xi, \eta) (1+|\eta|^2)(s)|\widehat{\phi} (\xi)|^2 d\xi d\eta)^{1/2}. $$ and this unravels to prove the theorem assuming the claim.
So, I have 4 minutes left and I will now quickly prove the claim. Some calculations reveal that $$ | K(\xi, \eta)| \leq C_k (\frac{1+|\eta|}{1+|\xi|})^{m-s} (1+ |\xi - \eta|)^{-k}. $$ Since $m,s$ are fixed, we can choose a large $k$.
2010-11-12
Assume $n < p \leq \infty$. Then $\| u \|{{C^{0,\gamma}}({\mathbb{R}}^n)} \leq C \| u \|{W^{1,p}} ~\forall~ u \in C^1 (\mathbb{R}^n).
proof: We start with a
claim: $$ \frac{1}{|B|} \int_{B(x,r)} |u(x) - u(y)| dy \leq C_n \int_{B(x,r)} \frac{|\nabla u(u)|}{|x-y|^{n-1}} dy.$$
The prooof of the claim follows from an argument using the Fundamental Theorem of Calculus, the triangle inequality. The resulting estimate reads $$ |u(x + s \omega) - u(x)| \leq \int_0^s |\nabla u(x+ t \omega)| dt. $$ This inequality is then integrated over $\omega \in \partial B(0,1)$. We recognize that $|x-y| = |t\omega|$ and then convert the double integration into a solid integration over the ball. The Jacobian introduces a $t^{n-1}$ related to the volume element in spherical coordinates and this is the source of the denominator on the right side of the claim. Similar ideas unravel into a proof of the claim.
Morrey’s estimate follows quite easily from the claim.
The $C^{0,\gamma}$ norm is defined as $\sup_{x \in R^n} |u(x)| + \sup_{x \neq y} \frac{|u(x) - u(y)|}{|x-y|^\gamma}$.
Consider the first term: $$ |u(x)| \leq | \frac{1}{B}\int_B u(x) dy| \leq \frac{1}{B} \int |u(x) - u(y)|dy + \frac{1}{B}\int |u(y)| dy. $$ We then use the claim. Notice that the right side of the claim has the $|\nabla u|$ term appearing to power 1. We can use Hölder inquality to raise the $p$ to find the $W^{1,p}$ norm multiplied by an integral of the denomiator factor raised to the conjugate exponent. The assumptions we make at the beginning implies that the integral over that apparently singular denominator is actually bounded. The estimate we obtain is valid for arbitrary $x$ so the supremum is also bounded.
Next, we estimate the second term in the definition of the $C^{0,\gamma}$ norm. We first study the numerator. At the end, we extract dependence upon a scale which will cancel the denominator. Notice that $u(x) - u(t) = \frac{1}{|W|} \int_W |u(x) - u(y)| dz$ for any measurable set $W$. With this, and the triangle inequality, we consider $r = |x-y|$ and $W = B(x,r) \cap B(y,r)$. We use the fact that $W \subset B(x,r)$ to replace the integral over $W$ by the integral over $B$ since the integrand is positive. Then we quote the claim and proceed via Hölder inequality as before. We keep track of the dependence upon $r$ emerging from the calculation and recall that $r = |x-y|$. The dependence upon $r$ is then exactly cancelled away by the denominator in the definition of the second term in the definition of the $C^{0,\gamma}$.
Extension Lemma (Evans Section 5.4): Assume $1 \leq p \leq \infty$. Let $U$ denote a bounded set with $C^1$ boundary. Then there exists a bounded linear operator $E: W^{1,p} (U) \rightarrow W^{1,p} ({\mathbb{R}}^n)$ such that for a bounded open set $U \subset \subset V$, the function $Eu$ has support inside $V$.
Sobolev Inequality: $\forall ~ u \in C^1 (R^n)$, we have control of the $L^{p^*}$ norm by the $W^{1,p}$ norm.
Sobolev Embedding: For $U$ bounded, open with $C^1$ boundary, we have
$$ W^{1,p}(U) \subset L^{p^*} (U) $$
and also $$W^{1,p} (U) \subset L^q (U)$$ for all $1 \leq q \leq p^*$.
Lemma (Rellich-Kondrachov Compactness): Assume $U$ is bounded, open with $C^1$ boundary. Assume $1 \leq p < n$. Then $W^{1,p} (U) \subset \subset L^q (U)$ for all $1 \leq q < p^*$.
It suffices to prove the compactness. This is based on the Arzela-Ascoli theorem and approximation by a mollification sequence. (I got a little lazy typing……sorry -Jim)
2010-11-15
(mostly following 9.1 in Evans)
We consider a PDE of the form $-\nabla \cdot a(\nabla u)=f$ in $U$ with (vanishing) Dirichlet boundary conditions. Here $f \in L^2$ and $a: R^n \rightarrow R^n$ We want to find $v \in H^1_0 (U)$.
We’ll find weak solutions $u \in H^1_0 (U)$ satisfying $$ \tag{WeakFormulation} \int_U a(\nabla u) \cdot \nabla v dx = \int_U f v dv, ~ \forall v \in H^1_0 (U). $$
We will need to make some conditions on $a$. We will require that $a$ be monotonoe $$ (a(p) - a(q))\cdot (p-q) \geq 0. $$ We might want to compare this with the ellipticity condition $$ \sum_{i,j} a_{ij} \xi_i \xi_j \geq 0, $$ By setting $\xi = p-q$ and making some manipulations, we can show that ellipticity implies monotonicity.
Question: What is the relationship between monotone and elliptic operators? Some discussion about strict ellipticity versus nonnegative quadratic form condition. A remark about uniqueness….not so clear to me.
The space $H^1_0 (U)$ is a Hilbert space with inner product $(u,v) = \int_U \nabla u \cdot \nabla v dx.$ This is a separable Hilbert space since $U$ is assumed to be a $C^1$ open bounded set. Therefore, we have an orthonormal basis ${ w_k }$ of $H^1_0$.
Theorem: $\forall ~m~\exists d_m^k$ such that $$ u_m = \sum_{k=1}^m d^k_m w_k \int_U a (\nabla u_m) \cdot \nabla w_k = \int_U f w_k dx. $$
Lemma: If $v: R^n \rightarrow R^n$ continouns with $v(x)\cdot x \geq 0 ~\forall ~x$ with |x| = r$ then $\exists ~ x \in B(0,r)$ with $v(x) = 0$.
(Based on Brouwer’s Fixed Point Theorem; proof omitted)
Proof (of theorem assuming Lemma): Define $v: R^n \rightarrow R^n$ and he does some calculations with the expressions $u$ in the weak formulation of the PDE replaced by explicit finite sums over these vectors in the basis. …. trouble keeping up with the typing. OH! Now, I see. He defines a function $v$ whose components consist of the integral conditions asserted in the theorem. He then proves (based on the monotonicity assumption) that the expression $v(d) \cdot d$ satisfies the hypotheses of the Lemma. He then asserts, based on the lemma, that there exists a choice of argument $d$ for which $v(d) = 0$. That choice of $d$ corresponds with the collection $d^k_m$ whose existence is the prinicpal assertion of the theorem.
Next, we want to extend our projected solutions to the whole space. He recalls the weak formulation of the PDE.
He makes some remarks about bounded sequences in $H^0_1$ being weak-* compact, but this is a Hilbert space, so in fact we have weak compactness….
Theorem: $\exists C$ with the $H^1_0 (U)$ norm of $u_m$ bounded (uniformly in $m$) by $C$.
$\exists ~u$ and a subsequence $m_j$ with $u_{mj} \rightarrow u$ weakly in $H^1_0$. This implies that $u_{mj}\rightarrow u$ weakly in $L^2$ and $\nabla u_{mj} \rightarrow \nabla u$ weakly in $L^2$. We want to conclude that $a(\nabla u_{mj}) \rightarrow a(\nabla u)$ but we can’t make that assertion. The next result tells us what goes wrong.
Proposition: If $a: R \rightarrow R$ is a continuous nonlinear ($a(z) \neq \alpha z + \beta$) then $\exists ~f_k, f \in L^2$ with $f_k \rightarrow f$ weakly in $L^2$ but $a(f_k)$ does not converge weakly to $a(f)$.
Example illustrating the idea: Let $a(x) = x^2$ and consider the Haar basis then it can be shown that $f_n \rightarrow \frac{1}{2}$ weakly but $a(f_n) = f_n$ so $a(f_n)$ converges weakly to $\frac{1}{2}$ but this is inconsistent with the hoped for result since $a(1/2) = 1/4.$
Theorem: There exists $u \in H^1_0 (U)$ which satisfies the WeakFormulation condition.
We already know that $u_{mj} \rightarrow u$ weakly in $L^2, H^1_0 (U)$. We also have $\nabla u_{mj} \rightarrow \nabla u$ weakly in $L^2$.
…..he starts calculating the $L^2$ norm of $a(\nabla u_{mj})$, uses the montonicity condition and eventually bounds this expression. OK, by Banach-Alaogolu there exists a further subsequence such that $a(\nabla u_{mk}) \rightarrow \xi$
$$ \int a (\nabla u_m) \cdot \nabla w_k dx = \int f w_k dx $$ $ \implies $ $$ \int \xi \cdot \nabla w_k dx = \int f w_k dx. $$ Therefore, we have $\forall ~v \in H^1_0$ that $\int \xi \cdot \nabla v dx = \int f v dx.$
He reformulates things taking advantage of the Lemma, quotes the weak convergences, refers to an equation called *.
He recalls a few things from Jeon’s talk.
What’s a symbol? To a given symbol $p$, we associate an operator sometimes denoted $T_p$ or $Op (p) $ or by $P$.
There are 3 main facts.
Let’s try to understand what we mean here by “essentially.” We will interpret this in terms on an asymptotic expansion for $\psi$DOs.
Definition: (Asymptotic Expansion of $\psi$DOs) Let ${ a_j }_1^\infty$ be a set of symbols. We say that the symbol $a$ satisfies $a \thicksim \sum_1^\infty a_j$ if $\forall ~ L \in {\mathbb{R}}^+ ~\exists ~ M \in {\mathbb{Z}}^+$ such that $a - \sum_1^M a_j \in S^{-L}$.
Notation: $\alpha ! = \alpha_1 ! \alpha_2 ! \cdots \alpha_n !$, etc.
Definition: Let $K \subset {\mathbb{R}}^n$ be compact. Let $\Psi_K$ be the set of symbols $p(x, \xi)$ with $supp (p) \subset K \times {\mathbb{R}}^n$.
Theorem 1: (Representation of the Adjoint Operator) Fix $K$ compact subset of $ {\mathbb{R}}^n$. Let $p \in S^m \cap \Psi_K$ and let $P$ denote the associated operator. Then $P^*$ has symbol in $S^m \cap \Psi_K$ and has the asymptotic expansion $$ P^* \thicksim \sum_\alpha D_x^\alpha (\frac{\partial}{\partial \xi})^\alpha {\overline{p(x,\xi)}} \frac{1}{\alpha !} $$
Definition: (Hörmander Class of Symbols) $r(x,\xi, y)$ is smooth on ${\mathbb{R}}^N \times {\mathbb{R}}^N\times {\mathbb{R}}^N$ is said to be in $T^m$ if $\exists ~K $ compact such that
For $\alpha, \beta, \gamma ~ \exists ~ C_{\alpha, \beta}$ such that $$ | (\frac{\partial}{\partial \xi})^\alpha (\frac{\partial}{\partial x})^\beta (\frac{\partial}{\partial y})^\gamma r(x,\xi, y)| \leq C_{\alpha, \beta, \gamma} |\xi|^{m - |\alpha|}. $$ The correspoinding operator $R$ is defined by the formula $$ R \phi (x) = \int \int e^{i (y-x) \cdot \xi} r(x, \xi, y) \phi (y) dy d\xi. $$
Proposition 1: Let $r \in T^m$ have $x,y$ supports in $K$ compact. Then $R$ as defined above defines a $\psi$DO with symbol $p \in \Psi_K$ with asymptotic expansion $$ p(x,\xi) = \sum_\alpha \frac{1}{\alpha !} \partial_\xi^\alpha \partial_y^\alpha r(x,y\xi, y) \|_{y=x}.$$
**Proof of Theorem 1:$$ Let $p \in \Psi_K \cap S^m, ~ P$ the corresponding $\Psi$DO. Let $\phi, \psi \in C^\infty_c$. Then $$\langle \phi, P^* \psi \rangle = \langle P\phi, \psi \rangle = \int P\phi (x) \overline{\psi}(x)dx.$$ We can then observe this equals $$ = \int \int e^{- i x \cdot \xi} p(x, \xi) \widehat{\phi} (\xi) d\xi {\overline{\psi}}(x)dx. $$
Claim: If $p$ has compact support in $\xi$ then $p^* (x,\xi)$ (the symbol associated to $P^*$) has asymptotic expansion $$ \thicksim \sum_\alpha \frac{1}{\alpha !} \partial_\xi^\alpha D_x^\alpha {\overline{p(x,\xi)}}. $$
Assuming this for the moment, we return to the proof of the theorem.
Let $\phi$ be a standard smooth cutoff equaling one on the unit ball and supported on the ball of radius two. Let $p_j (x, \xi) = \phi(\frac{\xi}{j}) p(x,\xi)$. We then have $p_j \rightarrow p$ as $j \rightarrow \infty$. We can then calculation $$ (Op (p_j))^{*} \thicksim \sum_\alpha \frac{1}{\alpha !} \partial_\xi^\alpha D_x^\alpha {\overline{p_j (x, \xi)}}$$ $$ \thicksim \sum_\alpha \frac{1}{\alpha !} \partial_\xi^\alpha D_x^\alpha {\overline{ \phi(\frac{\xi}{j} p(x,\xi))}}.$$ this gives us a representation of $(Op (p))^*$ by taking a limit as $j \rightarrow \infty$.
Proof of Claim:
If $p$ has compact support in $\xi$ then our integral is abolutely convergent. So, we can calculate (recalling from above) $$ \langle \phi, P^{*} \psi \rangle = \langle [\int e^{- i x \cdot \xi} p(x, \xi) \widehat{\phi} (\xi) d\xi], \psi \rangle. $$ We can rewrite this by putting in some conjugates and rearranging to eventually realize $$ P^* \psi (x) = \int \int e^{i (x-y) \cdot \xi} {\overline{p(x, \xi)}} \psi (x) d\xi dx. $$ Let $\rho$ denote an nice cutoff function looking like a smoothed version of the characteristic fucntion of $K$. Set $r(x,\xi, y) = \rho(x) p(y, \xi)$. Then $P^* \psi (y) = R \psi (y)$ where $r (x, \xi, x) = {\overline{p(x,\xi)}} \rho(y).$
Some discussion:* The name $R$ here denotes an operator in the Hörmander class. The name $P^*$ denotes the same operator in the standard operator class associated to symbols of class $S^m$.
Proposition 1 $\implies P^{*}$ has symbol $p^*$ where $$p^{*} \thicksim \sum_\alpha \frac{1}{\alpha !} \partial_\xi^\alpha D_x^\alpha {\overline{p(x, \xi)}}.$$ This completes the proof of Theorem 1.
Theorem 2 (Kohn-Nirenberg): Let $p \in \Psi_K \cap S^m, ~ q \in \Psi_K \cap S^n$. Let $P, Q$ denote the corresponding operators. The $P \circ Q = T_\sigma$ where
Proof: Write $Q = (Q^{*})^*$. Then $$Q\phi (x ) = {\mathcal{F}}^{-1} ( \int e^{i x \cdot \xi} \phi (x) {\overline{q(y, \xi)}}) dy).$$ Now: $$ (P \circ Q) \phi (x) = \int e^{-i x \cdot \xi} p(x, \xi) {\widehat{Q \phi}}(\xi) d\xi. $$ which can be rewritten $$ = \int \int e^{-i (x - y ) \cdot \xi} p(x, \xi) {\overline{q^{*} (y, \xi)}} \phi (y) dy d\xi. $$ We define $r(x,\xi, y) = p(x, \xi ) {\overline{q^{*}(x, \xi)}}$.
(HOMEWORK EXERCISE: Check that $r \in T^{m+n}$)
$R = P \circ Q$. By the proposition, there exists a symbol $\sigma (x, \xi)$ such that
$$ \sigma (x, \xi) \thicksim \sum_\alpha \frac{1}{\alpha !} \partial_\xi^\alpha D_y^\alpha r(x, \xi, y) |_{y=x}. $$
Some further manipulations…..Leinbniz rule…..
….more calculations and manipulations with the sum…..
(The discussion on $\psi$DOs in this course is often based on a book by Steven Krantz.)
Definition: A $\psi$DO $P$ is smoothing if $\sigma(P) = p \in S^{-\infty}= \cap_{m>0} S^{-m}$. Recall that $S^m: H^s \rightarrow H^{s-m}$.
Proposition: If ${ p_j }_{j=0}^\infty, ~p_j \in S^{m_j}, m_j \searrow - \infty$ then $\exists ~ p \in S^{m_0}$ such that $p \thicksim \sum p_j$ uniquely modulo smoothing operators.
Definition: $p \in S^m$ is elliptic on an open set $U$ if there exists a continuous $c > 0$ on $U$ such that $$ |p(x, \xi)| \geq c(x) |\xi|^m ~{\mbox{for}}~ |\xi| ~{\mbox{large}}. $$
Definition: A left (respectively right) parametrix for $P$ on an open set $U$ is a $\psi$DO $Q$, such that $\exists ~ \psi \in C^\infty_c, ~ \psi = 1$ on $U$ and $QP - \psi I \in S^{-\infty}$ (respectively $PQ - \psi I \in S^{-\infty}$).
Thus, we “essentially” have an inverse for $P$.
Theorem: If $P \in S^m$ where $m>0$ is elliptic on $U$ and $L \subset \subset U$ then $\exists$ a two-sided parametrix $Q$ for $P$ on $L$. Moreover, $Q \in S^{-m}$.
Andrew Question: Why is this elliptic? Can you show how this compares with elliptic PDE. ….some discussion….
proof: $\exists ~ c_0 > 0 $ so that $c(x) > c_0$ on $L$. Consider $|\xi| > M$. $$ |p(x,\xi)| \geq c_0 |\xi|^m $$ Introduce a cutoff $\psi$ adapted to look like a smooth characteristic function of $L$ supported inside on $U$. (Text had a $K$….some discussion…). Introduce another cutoff $\phi$ adapted to equal 1 in the region $|\xi| \geq M+2$ and supported inside $|\xi| \geq M+1$.
Claim: $q_0 (x, \xi) = \psi(x) \frac{\phi(\xi)}{p(x,\xi)} \in S^{-m}.$
(I got distracted and did not type details here…but you basically follow your nose.)
Using the Kohn-Nirenberg calculus, we have $$ \sigma (Q_0 \circ P) \thicksim \sum_\alpha \frac{1}{\alpha !} \partial_\xi^\alpha D_x^\alpha p = q_0 p + \dots = \psi (x) p(\xi) + \dots . $$ You can see that all the omitted terms are in $S^{-1}$. The proposition implies that $$\sigma (Q_0 \circ P) = \psi(x) p(\xi) + {\tilde{r}}{-1}$$ where ${\tilde{r}}{-1} \in S^{-1}$. (Ack…please read some of the -1’s appearing around here as subscripts…) Let’s define $r_{-1}$ by $$ r_{-1} + \psi (x) = \psi(x) \phi(\xi) + {\tilde{r}}^{-1}. $$ With this object, we can improve our parametrix. Let ${\tilde{\psi}}$ be a bump function adapted to the $x$-support of $r_{-1}$ and define then $$ q_1 = - \tilde{psi}(x) \frac{\phi(\xi)}{p(x,\xi)} r_{-1} (x, \xi) \in S^{-m - 1}. $$ We then calculate $\sigma( (Q_0 + Q_1)\circ P) = \dots$ to find this equals $\psi(x) + r_{-2}$ where $r_{-2} \in S^{-2}$.
He then described the inductive argument that allows us to move the remainder to an arbitrary smoothing operator.
He then describes $\sigma(Q \circ P)$ and comments on the infinite smoothing claim.
The basic structure of the argument above constructs a left parametrix. A similar construction builds the right inverse. But there is a simpler argument by working modulo smoothing operators.
I want to prove a theorem about regularity. But I first need to define the action of $\psi$DOs on distributions. You do it using the same ideas we used to define derivatives of distributions.
If $\psi, \phi \in {\mathcal{D}}$ then \langle P^t \phi, \psi \rangle = \langle \phi, P \psi \rangle = \int \phi (P \psi). Then for $f \in {\mathcal{D}}’$ we define $\langle Pf , \phi \rangle = \langle f , P^t \phi \rangle$ and this defines an action on distributions.
Theorem: Let $U \subset {\mathbb{R}}^n$ be open and $P$ an elliptic $\psi$DO on $U$ of order $m>0$. If $f \in H^s$ and $u \in {\mathcal{D}}’$ with $Pu = f$ then we can show that $u \in H^{s+m}_{loc}$.
To prove this completely, you need to put in some cutoff functions and be a little careful. As a simplification and to illustrate the basic idea, suppose that $f$ and $u$ have compact support. Let $\psi \in C_c^\infty$ be a smooth cutoff adapted to $supp (u)$. We know $\exists$ a 2-sided parametrix $Q$ for $P$. We can therefore write $$ QP = \psi (x) I + S $$.
We write $u = \psi u = (QP - S)u = Q f - Su$. Then $Su \in C^\infty_c$. Since $Q$ is the parametrix, we know that $Q \in S^{-m}$ so we know that $Qf \in H^{s+m}$. Therefore, $u \in H^{s+m}$.
Without the compact support property, you use cutoffs….