Parametrix

Remarks

An operator $Q$ with only $QP - \id$ smoothing is a left parametrix, and one with only $PQ - \id$ smoothing is a right parametrix. Two parametrices of the same operator differ by a smoothing operator. For constant-coefficient operators on $\R^n$, a parametrix corresponds to a distribution $e$ with $P e = \delta + \omega$ for a smooth function $\omega$, an approximate fundamental solution. An elliptic operator of order $m$ has a parametrix that is a pseudo-differential operator of order $-m$.

Example: on the circle $S^1 = \R / 2\pi\Z$ with $d\mu = dx$, let $P = -d^2/dx^2$ on functions, and define $$Q\Bigl(\sum_{n \in \Z} c_n e^{inx}\Bigr) = \sum_{n \ne 0} \frac{c_n}{n^2} \, e^{inx}.$$ Then $PQu = QPu = u - c_0$, where $c_0 = \frac{1}{2\pi} \int_0^{2\pi} u(y) \, dy$. So $QP - \id$ and $PQ - \id$ are both the operator with the constant smooth kernel $K(x, y) = -\frac{1}{2\pi}$, and $Q$ is a parametrix.

Non-example: the zero operator is not a parametrix for $P = -d^2/dx^2$. Here $PQ - \id = -\id$, and the identity is not smoothing. An operator with a continuous kernel extends to a compact operator on $L^2(S^1)$, while the identity of an infinite-dimensional Hilbert space is not compact.