4
$\begingroup$

If we have a smooth symbol $r(x,\xi)$ of order $d$ the corresponding pseudodifferential operator $P$ is integral operator with kernel given by $$K(x,y)=\int_{\mathbb{R}^n}e^{i(x-y)\cdot \xi} r(x,\xi)d\xi$$ and $K$ is $C^k$ in $(x,y)$ if $d+m<-n$. ( see for instance Gilkey’s book, Invariance theory...1984, lemma1.2.2 and lemma 1.2.5). For example if the symbol is smooth and of order $-\infty$ then $K(x,y)$ is smooth.

My question is: Is there any similar statement in the case of pseudo differential operators whose symbol is singular. Some results that can give the kernel of pseudodifferential operator in terms of its symbol and determines how regular such a kernel is, i.e. for which $k$, $K(x,y)\in C^k$.

$\endgroup$
1
$\begingroup$

Let $a\in \mathscr S'(\mathbb R^{2n})$: you can quantize that tempered distribution, i.e. associate linearly to $a$ a linear operator $\text{Op} a$ from $\mathscr S(\mathbb R^{n})$ into $\mathscr S'(\mathbb R^{n})$ with the formula, where $u,v\in \mathscr S(\mathbb R^{n})$, $$ \langle (\text{Op} a) u,v\rangle_{\mathscr S'(\mathbb R^{n}), \mathscr S(\mathbb R^{n})}=\langle a,\mathcal H(u,v)\rangle_{\mathscr S'(\mathbb R^{2n}), \mathscr S(\mathbb R^{2n})}, \tag{$\sharp$}$$ where the Wigner function $\mathcal H(u,v)$ is defined on $\mathbb R^{2n}$ by $$ \mathcal H(u,v)(x,\xi)=\int e^{-2iπ z \xi} u(x+\frac{z}{2}) \bar v(x-\frac{z}{2}) dz. $$ It is easy to prove that $\mathcal H(u,v)$ belongs to $S(\mathbb R^{2n})$ when $u,v$ are in $S(\mathbb R^{n})$ and this gives a meaning to $(\sharp)$. Now when you have the symbol, and if you use the classical quantization formula, you get the kernel by the formula $$ k(x,y)=F_2a(x, y-x), $$ where $F_2$ is the Fourier transformation with respect to the second variable, which makes sense on $S'(\mathbb R^{2n})$. This also implies that the symbol $a$ can be expressed in terms of the kernel with
$$ a(x,\xi)=\int F_2a(x, z) e^{2iπ z \xi} d\xi=\int k(x, x+z) e^{2iπ z \xi} d\xi, $$ where the integrals should be understood in a weak sense on $S'(\mathbb R^{2n})$. As a result, that reversible relationship between kernel and symbol always holds. Symbols are in general nicer to manipulate than kernels: the kernel of the identity is $\delta_0(x-y)$ whereas its symbol is 1. The Hilbert transform has the kernel ($y,x\in \mathbb R$) $$ \frac{1}{iπ(y-x)} $$ i.e. is the convolution with $pv(i/π x)$ and has symbol $\text{sign} (\xi)$, and for instance the $L^2$ boundedness of that operator is trivial knowing the symbol since we have a Fourier multiplier by a bounded function, whereas that property is not obvious when you look at the kernel. A parametrix of the Laplace operator in $d\ge 3$ dimensions has kernel $$ c_d\vert x-y\vert^{2-d}, $$ a locally integrable function which is also a temperate distribution, and its symbol is $\vert \xi\vert^{-2}$.

| cite | improve this answer | |
$\endgroup$

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.