About peakon kernel

Question

Do you have any reference which explain following

For differential operator $L=(I-\partial^2_x)$ and peakon kernel $Q=\frac{e^{-|x|}}{2}$

$$L^{-1}f(x)=Q*f=\int_\mathbb{R}Q(x-y)f(y)dy$$

Answer 1

To solve $Lu(x)=u(x)-u''(x)=f(x)$ for $u(x)$ (with suitable boundary conditions of decay at infinity), take the Fourier transform of both sides: $$ \hat{u}(\xi) - (i\xi)^2 \hat{u}(\xi) = \hat{f}(\xi) . $$ This gives $$ \hat{u}(\xi) = \frac{1}{1+\xi^2} \, \hat{f}(\xi) . $$ Here we recognize the standard Fourier transform $\hat{Q}(\xi)=1/(1+\xi^2)$, so $\hat{u}=\hat{Q}\hat{f}$, and therefore $u=Q*f$ (since multiplication of Fourier transforms corresponds to convolution of the original functions).