Question 12.23: (Malgrange - Ehrenpreiss) Every linear partial differential ......

Let us define

K(x)=\int_{ \mathbf{R} ^{n-1}} \int_{\mathfrak{J}\left(\xi_n\right)=\Phi\left(\xi^{\prime}\right)} \exp (2 \pi \imath x \cdot \xi)(P(\xi))^{-1} d \xi_n d \xi^{\prime}

where, as before, \xi=\left(\xi_1, \ldots, \xi_{n-1}, \xi_n\right)=\left(\xi^{\prime}, \xi_n\right) \text {, while } \Phi was defined in Example 12.22.
We shall show that K is well defined, and, moreover, is the desired fundamental solution of P. To that end, put first

P_N(\xi)=P(\xi)\left(1+4 \pi^2 \overset{n}{\underset{j=1}{\sum}} \xi_j^2\right)

where N is a (sufficiently big) natural number. Next, put

K_N(x)=\int_{ \mathbf{R} ^{n-1}} \int_{\mathfrak{J}\left(\xi_n\right)=\Phi\left(\xi^{\prime}\right)} \exp (2 \pi \imath x \cdot \xi)\left(P_N(\xi)\right)^{-1} d \xi_n d \xi^{\prime}

where the function Φ was chosen as in Example 12.22. Since then on the domain of integration it holds \left|P_N(\xi)\right| \geq C \cdot\left(1+|\xi|^2\right)^N, the last integral converges for N > n/2.
Next, let us prove

P_N(D) K_N=\delta

Assume \varphi \in \mathcal {D}( \mathbf{R}^n). Since the adjoint operator to P_N(D) \text { is } P_N(-D), it holds

Note that we used the equality

\varphi(0)=\int_{ \mathbf{R}^n} \mathcal {F} \varphi(\xi) d \xi

which follows using the inverse Fourier transformation.
So we obtain

\delta=P_N(D) K_N=P_N(D)(1-\Delta)^N K_N

Now K=(1-\Delta)^N K_N \text {, hence } is the fundamental solution of the operator P(D).