I'm following the book "Introduction to the theory of distributions" by Friedlander and Joshi. There is the following result p. 109

Theorem (8.6.1). Let $X \subset \mathbb{R}^n$ be an open set, and let $P$ be an elliptic operator with constant coefficients. Then


As an observation after the demonstration says:

"This principle, applied to Schwartz kernels and backed by an appropriate construction, gives the elliptic regularity theorem for differential operators with variable coefficients."

Would you give me references for this more general case? Is the theory of pseudo-differential operators necessary?


I was able to find the right reference. This construction is present in the book "Introduction to pseudo-differential and Fourier Integral volume 1" by J.F. Treves. More specifically the sections are as follows

  1. Parametrices of Elliptic Equations

  2. Definition and Continuity of the "Standard" Pseudodifferential Operators in an Open Subset of Euclidean Space. Pseudodifferential Operators Are Pseudolocal

up to page 12, there is the following result:

Lemma (2.2). Let $P$ denote a differential operator with variable coefficients in $\Omega$. Suppose that there is a very $K: \mathcal{E}'(\Omega) \longrightarrow \mathcal{D}'(\Omega)$ regular operator such that $KP-I$ is regularizing (which is sometimes expressed by saying that $K$ is a left parametrix of $P$). Then $P$ is hypoelliptic, i.e., it has the following property: Given any open set $U$ of $\Omega$, then every distribution $u$ in $U$ such that $Pu \in C^{\infty}(U)$ is a $C^\infty$ function in $U$.

In other worlds this lemma says that $$\mathrm{singsupp}(u)=\mathrm{singsupp}(Pu)$$ The theory of pseudo-differential operators is not necessary, in fact this construction is a particular introduction to it.

PS. To understand this construction, it is necessary to study Chapter 6 (Schwartz kernels and kernel theorem) of the book "Introduction to the theory of distributions" by Friedlander and Joshi.

| cite | improve this answer | |

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.