I am working through the following text: http://homepage.ntlworld.com/ivan.wilde/notes/gf/gf.pdf
Several times the author cites 'Leibniz Formula' which I understand as the product rule. I am not sure how the results follow.
In particular the proof of Theorem 3.21 (i) on the page with page number 26, which corresponds to page 28 of the PDF.
Thank you
Yes, the Leibniz formula being used here is the product rule. However it is being used multiple times.
The product rule is $D(ab) = D(a)b +a D(b)$. But what then $D^n(ab)$? This is $D^{n-1}(D(a)b +a D(b))$ and we continue by applying the product rule again and again. One easily proves by induction that $$ D^n(ab) = \sum_{i=0}^n \binom{n}{i} D^i(a)D^{n-i}(b). $$
In response to your comments/edits, they are using the definition of convergence in $\mathcal D(\Omega)$ together with the product rule. We need to show that $\psi \phi_n - \psi \phi \to 0$.
But similarly to the above, by repeatedly applying the Leibniz rule, one sees that $D^\alpha (\psi \phi_n)$ can be written as a finite sum of products of derivatives of $\psi$ and derivatives of $\phi_n$, and so it follows that $D^\alpha (\psi \phi_n) - D^\alpha (\psi \phi)$ will be a finite sum of terms that look like $D^\beta(\psi) D^\gamma(\phi_n) - D^\beta(\psi) D^\gamma (\phi)$.
By assumption, $D^\gamma( \phi_n) - D^\gamma(\phi) \to 0$ uniformly for each $\gamma \in \mathbb Z^d$. By assumption/definition of convergence, we have that each $\phi_n$ is compactly supported (on the same compact set $K$) and thus so is each $D^\gamma(\phi_n)$. If we restrict to the support of $D^\gamma(\phi_n)$, $D^\beta(\psi)$ will be a bounded function on that support. So we get that $D^\beta(\psi) D^\gamma(\phi_n) - D^\beta(\psi) D^\gamma (\phi) \to 0$ uniformly. (Multiplying a uniformly convergent sequence by a bounded function yields another uniformly convergent sequence.) A finite sum of terms uniformly converging to $0$ will also converge uniformly to $0$. Thus we have that $\psi \phi_n \to \psi \phi$. A similar approach applies to part (ii) of the proof.