The Wodzicki residue

Now we return to the symbol expansion of a classical $\Psi$DO $P$, of integral order $d \in \bZ$, with

$$p(x,\xi) = \sum_{j=0}^{N-1} p_{d-j}(x,\xi) + r_N(x,\xi),$$

where $r_N \in S^{d-N}(U)$, and $p_{d-j}(x, t\xi) = t^{d-j} p_{d-j}(x,\xi)$. Now apply $\sF_2^{-1}$, the inverse Fourier transform in the second variable, to this sum, to get the integral kernel

$$k_P(x,y) = \sum_{j=0}^{N-1} h_{j-d-n}(x,x-y) + (\sF_2^{-1}r_N)(x,x-y).$$

If $N > n + d$, then $r_N \in S^{d-N}(U)$ is integrable in $\xi$, so the term $\sF_2^{-1}r_N(x, z)$ is bounded as $z \to 0$. For the terms $h_{j-d-n}(x, z)$, there are 3 cases, which may give singularities. So before applying $\sF_2^{-1}$ to $p_{d-j}(x,\xi)$, we must regularize $p_{d-j}(x,\xi)$ to $R_f p_{d-j}(x,\xi)$ by using a suitable cutoff:

$$p(x,\xi) = \sum_{j=0}^{N-1} R_f p_{d-j}(x,\xi) + s_N(x,\xi),$$

with $s_N$ integrable. Now take $h_{j-d-n} := \sF_2^{-1}(R_fp_{d-j})$.