Multivariable Multiindex Leibniz Rule

Question

Question states: "Let $f,g: \Bbb R^n \to \Bbb R$ be of class $C^{\infty}$ and let $\alpha \in \Bbb N_0^n$ be a multiindex. Show that $$\partial^\alpha(fg) = \sum_{\beta+\gamma=\alpha}\frac{\alpha!}{\beta!\gamma!}(\partial^\beta f)(\partial^\gamma g)".$$ I tried solving this using induction. I first proved the case for $\lvert \alpha \rvert = 1$ (or, equivalently, when $\alpha$ has only $1$ in an arbitrary position and all other terms are $0$). However, I couldn't really deduce that if the case is true for $\lvert \alpha \rvert = k$, then $\lvert \alpha \rvert = k+1$ is true. Any help? Also, if there is other any to prove this statement that would be welcome.

Answer 1

When $|\alpha| = k + 1$, write $\alpha = (\alpha_1, \dots, \alpha_n)$ and choose some $1 \leq i \leq n$ with $\alpha_i > 0$. Split $\alpha = \beta + \gamma$ where

$$ \beta = (\alpha_1, \dots, \alpha_{i-1}, \alpha_i - 1, \alpha_{i+1}, \dots, \alpha_n),\\ \gamma = (0, \dots, 0, 1, 0, \dots, 0). $$

Then

$$ \partial^{\alpha}(fg) = \partial^{\gamma + \beta}(fg) = \frac{\partial}{\partial x_i} \left( \partial^{\beta} (fg) \right) $$

and you can apply the induction hypothesis to the $\partial^{\beta}(fg)$ term and then use linearity and apply $\frac{\partial}{\partial x^i}$ to each of the resulting terms (you know what should be the result as this is the base case for the induction).