Leibniz's Rule/Real Valued Functions

Theorem
Let $f, g : \R^n \to \R$ be real valued functions, $k$ times differentiable on some open set $U \subseteq \R^n$.

Let $\alpha = \left({\alpha_1, \ldots, \alpha_n}\right)$ be a multiindex indexed by $\left\{{1, \ldots, n}\right\}$ with $\left\vert{\alpha}\right\vert \le k$.

For $i \in \left\{1,\ldots,n\right\}$ let $\partial_i$ denote the partial derivative $\partial_i = \dfrac{\partial}{\partial{x_i}}$.

Let $\partial^\alpha$ denote the partial differential operator:
 * $\partial^\alpha = \partial_1^{\alpha_1} \partial_2^{\alpha_2} \cdots \partial_n^{\alpha_n}$

Then as functions on $U$, we have:
 * $\displaystyle \partial^\alpha\left({f g}\right) = \sum_{\beta \mathop \le \alpha} \binom \alpha \beta \left({\partial^\beta f}\right)\left({\partial^{\alpha - \beta} g}\right)$

Proof
First, inserting the definitions, the statement of the theorem reads:


 * $\displaystyle \partial_1^{\alpha_1} \partial_2^{\alpha_2} \cdots \partial_n^{\alpha_n} \left({f g}\right) = \sum_{\beta_1 = 0}^{\alpha_1} \cdots \sum_{\beta_n \mathop = 0}^{\alpha_n} \binom {\alpha_1} {\beta_1} \cdots \binom {\alpha_n}{\beta_n} \left({\partial_1^{\beta_1} \cdots \partial_n^{\beta_n} f}\right) \left({\partial_1^{\alpha_1 - \beta_1} \cdots \partial_n^{\alpha_n - \beta_n} g}\right)$

We prove this by induction over $n \ge 1$.

Basis for the Induction
If $n = 1$, the result is a simple restatement of Leibniz's Rule in one variable:


 * $\displaystyle \left({f \left({x}\right) g \left({x}\right)}\right)^{\left({n}\right)} = \sum_{k \mathop = 0}^n \binom n k f^{\left({k}\right)} \left({x}\right) g^{\left({n - k}\right)} \left({x}\right)$

This is the basis for the induction.

Induction Step
Suppose now that the result is true for functions of $n - 1$ variables.

In particular, we suppose that:
 * $\displaystyle \partial_2^{\alpha_2} \cdots \partial_n^{\alpha_n} \left({f g}\right) = \sum_{\beta_2 = 0}^{\alpha_2} \cdots \sum_{\beta_n \mathop = 0}^{\alpha_n} \binom {\alpha_2} {\beta_2} \cdots \binom {\alpha_n}{\beta_n} \left({\partial_2^{\beta_2} \cdots \partial_n^{\beta_n} f}\right) \left({\partial_2^{\alpha_2 - \beta_2} \cdots \partial_n^{\alpha_n - \beta_n} g}\right)$

Now let us apply $\partial_1^{\alpha_1}$.

Using the linearity of derivatives:


 * $\displaystyle \partial_1^{\alpha_1} \partial_2^{\alpha_2} \cdots \partial_n^{\alpha_n} \left({f g}\right) = \sum_{\beta_2 = 0}^{\alpha_2} \cdots \sum_{\beta_n \mathop = 0}^{\alpha_n} \binom {\alpha_2} {\beta_2} \cdots \binom {\alpha_n} {\beta_n} \partial_1^{\alpha_1} \left({\left({\partial_2^{\beta_2} \cdots \partial_n^{\beta_n} f}\right) \left({\partial_2^{\alpha_2 - \beta_2} \cdots \partial_n^{\alpha_n - \beta_n} g}\right)}\right)$

Applying Leibniz's Rule in one variable we have:


 * $\displaystyle \partial_1^{\alpha_1} \partial_2^{\alpha_2} \cdots \partial_n^{\alpha_n} \left({f g}\right) = \sum_{\beta_2 = 0}^{\alpha_2} \cdots \sum_{\beta_n \mathop = 0}^{\alpha_n} \binom {\alpha_2} {\beta_2} \cdots \binom {\alpha_n} {\beta_n} \sum_{\beta_1 \mathop = 0}^{\alpha_1} \binom {\alpha_1} {\beta_1} \left({\partial_1^{\beta_1} \partial_2^{\beta_2} \cdots \partial_n^{\beta_n} f}\right) \left({\partial_1^{\alpha_1 - \beta_1} \partial_2^{\alpha_2 - \beta_2} \cdots \partial_n^{\alpha_n - \beta_n} g}\right)$

Now, all the sums are finite, thus trivially absolutely convergent.

By Manipulation of Absolutely Convergent Series and Summation is Linear we can move the inner sum to the far left, giving:


 * $\displaystyle \partial_1^{\alpha_1} \partial_2^{\alpha_2} \cdots \partial_n^{\alpha_n} \left({f g}\right) = \sum_{\beta_1 = 0}^{\alpha_1} \cdots \sum_{\beta_n \mathop = 0}^{\alpha_n} \binom {\alpha_1} {\beta_1} \cdots \binom {\alpha_n} {\beta_n} \left({\partial_1^{\beta_1} \cdots \partial_n^{\beta_n} f}\right) \left({\partial_1^{\alpha_1 - \beta_1} \cdots \partial_n^{\alpha_n - \beta_n} g}\right)$

The result now follows by the Principle of Mathematical Induction.