> [!theorem] > > Let $\alpha, \beta, \gamma \in \nat_0^n$ be [[Multi-Index|multi-indices]] such that $\alpha = \beta + \gamma$, then > $ > \frac{\alpha!}{\beta!\gamma!} = \prod_{k \in [n]}\frac{\alpha_k!}{\beta_k!\gamma_k!} = \prod_{k \in [n]}{\alpha_k \choose \beta_k} > $ > *Proof*. > $ > \begin{align*} > \frac{\alpha!}{\beta!\gamma!} &= \frac{\prod_{k \in [n]}\alpha_k!}{\prod_{k \in [n]}\beta_k! \cdot \prod_{k \in [n]}\gamma_k!} \\ > &= \prod_{k \in [n]}\frac{\alpha_k!}{\beta_k!\gamma_k!} \\ > &= \prod_{k \in [n]}\frac{\alpha_k!}{\beta_k!(\alpha_k - \beta_k)!} \\ > &= \prod_{k \in [n]}{\alpha_k \choose \beta_k} > \end{align*} > $ > [!theorem] Lemma A > > Let $e_j$ be the vector with $1$ on its $j$-th entry and $0$ everywhere else, $\gamma \le \alpha$ be multi-indices, then > $ > \begin{align*} > {\alpha + e_j \choose \gamma } &= {\alpha \choose \gamma - e_j} + {\alpha \choose \gamma} > \\ > \frac{(\alpha + e_j)!}{\gamma!(\alpha + e_j - \gamma)!} &= \frac{\alpha!}{(\gamma - e_j)!(\alpha - \gamma + e_j)!} + \frac{\alpha!}{\gamma!(\alpha - \gamma)!} > \end{align*} > $ > *Proof*. Expand each of the products as > $ > \begin{align*} > \frac{\alpha!}{(\gamma - e_j)!(\alpha - \gamma + e_j)!} &= {a_j \choose \gamma_j - 1} \cdot \prod_{k \ne j}{\alpha_k \choose \gamma_k} \\ > \frac{\alpha!}{\gamma!(\alpha - \gamma)!} &= {\alpha_j \choose \gamma_j} \cdot \prod_{k \ne j}{\alpha_k \choose \gamma_k} > \end{align*} > $ > Adding the first binomial coefficient leads to > $ > {\alpha_j \choose \gamma_{j} - 1} + {\alpha_j \choose \gamma_j} = {\alpha_j + 1 \choose \gamma_j} > $ > Adding the terms together yields > $ > \begin{align*} > &{a_j \choose \gamma_j - 1} \cdot \prod_{k \ne j}{\alpha_k \choose \gamma_k} + {\alpha_j \choose \gamma_j} \cdot \prod_{k \ne j}{\alpha_k \choose \gamma_k} \\ > &= {\alpha_j + 1 \choose \gamma_j} \cdot \prod_{k \ne j}{\alpha_k \choose \gamma_k} \\ > &= \frac{(\alpha + e_j)!}{\gamma!(\alpha + e_j - \gamma)!} > \end{align*} > $ > [!theorem] > > Let $\gamma, \alpha$ be multi-indices such that $\gamma \le \alpha$ and $\gamma < e_j$ ($\gamma_j = 0$), then > $ > \begin{align*} > {\alpha \choose \gamma} &= {\alpha + e_j \choose \gamma} > \\ > \frac{\alpha!}{\gamma!(\alpha - \gamma)!} &= \frac{(\alpha + e_j)!}{\gamma!(\alpha + e_j - \gamma)!} > \end{align*} > $ > *Proof*. > $ > \begin{align*} > \frac{\alpha!}{\gamma!(\alpha - \gamma)!} &= \prod_{k \in [n]}{\alpha_k\choose \gamma_k} \\ > &= {\alpha_j\choose \gamma_j}\prod_{k \ne j}{\alpha_k\choose \gamma_k}\\ > &= {\alpha_j\choose 0}\prod_{k \ne j}{\alpha_k\choose \gamma_k}\\ > &= {\alpha_j + 1\choose 0}\prod_{k \ne j}{\alpha_k\choose \gamma_k} \\ > &= \frac{(\alpha + e_j)!}{\gamma!(\alpha + e_j - \gamma)!} > \end{align*} > $ > [!theorem] > > Let $f, g \in C^{\text{enough}}$, then > $ > \begin{align*} > &\sum_{\gamma \le \alpha}{\alpha \choose \gamma} \cdot \frac{\partial }{\partial x_j}(\partial^\gamma f)(\partial^{\alpha - \gamma} g) \\ > &= \sum_{\gamma \le \alpha}{\alpha \choose \gamma }(\partial^{\gamma + e_j} f)(\partial^{\alpha - \gamma} g) + \sum_{\gamma \le \alpha}{\alpha \choose \gamma}(\partial^{\gamma}f)(\partial^{\alpha + e_j - \gamma}) \\ > &= \sum_{\gamma \le \alpha + e_j}{\alpha + e_j \choose \gamma}(\partial^{\gamma}f)(\partial^{\alpha + e_j - \gamma}) > \end{align*} > $ > *Proof*. First consider the left term. Shift it by $e_j$, then split it into two parts: $\gamma \le \alpha + e_j$, $\gamma = \alpha + e_j$. > $ > \begin{align*} > &\sum_{\gamma \le \alpha}{\alpha \choose \gamma }(\partial^{\gamma + e_j} f)(\partial^{\alpha - \gamma} g) \\ > &= \sum_{e_j \le \gamma \le \alpha + e_j}{\alpha \choose \gamma - e_j }(\partial^{\gamma} f)(\partial^{\alpha + e_j - \gamma} g) \\ > &= \pb{\sum_{e_j \le \gamma \le \alpha}{\alpha \choose \gamma - e_j }(\partial^{\gamma} f)(\partial^{\alpha + e_j - \gamma} g)} + \po{\underbrace{{\alpha \choose \alpha} (\partial^{\alpha + e_j} f)(\partial^{0} g)}_{\gamma = \alpha + e_j}} > \end{align*} > $ > Similarly, split the right term into two parts: $\gamma < e_j$ and $\gamma \ge e_j$: > $ > \begin{align*} > &\sum_{\gamma \le \alpha}{\alpha \choose \gamma}(\partial^{\gamma}f)(\partial^{\alpha + e_j - \gamma}) \\ > &= \pg{\underbrace{\sum_{\gamma \le \alpha, \gamma < e_j}{\alpha \choose \gamma}(\partial^{\gamma}f)(\partial^{\alpha + e_j - \gamma})}_{\gamma_j = 0}} + \pb{\sum_{e_j \le \gamma \le \alpha}{\alpha \choose \gamma}(\partial^{\gamma}f)(\partial^{\alpha + e_j - \gamma})} > \end{align*} > $ > The blue parts can be combined with Lemma A to produce > $ > \begin{align*} > &\pb{\sum_{e_j \le \gamma \le \alpha}{\alpha \choose \gamma - e_j }(\partial^{\gamma} f)(\partial^{\alpha + e_j - \gamma} g)} + \pb{\sum_{e_j \le \gamma \le \alpha}{\alpha \choose \gamma}(\partial^{\gamma}f)(\partial^{\alpha + e_j - \gamma})} \\ > &= \pb{\sum_{e_j \le \gamma \le \alpha}{\alpha + e_j \choose \gamma}(\partial^{\gamma}f)(\partial^{\alpha + e_j - \gamma})} > \end{align*} > $ > Since ${\alpha \choose \alpha} = 1 = {\alpha + e_j \choose \alpha + e_j}$, the orange part can simply be rewritten as > $ > \po{\underbrace{{\alpha \choose \alpha} (\partial^{\alpha + e_j} f)(\partial^{0} g)}_{\gamma = \alpha + e_j}} = \po{\underbrace{{\alpha + e_j \choose \alpha + e_j} (\partial^{\alpha + e_j} f)(\partial^{0} g)}_{\gamma = \alpha + e_j}} > $ > The green part can be rewritten with Lemma B to produce > $ > \pg{\underbrace{\sum_{\gamma \le \alpha, \gamma < e_j}{\alpha \choose \gamma}(\partial^{\gamma}f)(\partial^{\alpha + e_j - \gamma})}_{\gamma_j = 0}} = \pg{\sum_{\gamma \le \alpha, \gamma < e_j}{\alpha + e_j \choose \gamma}(\partial^{\gamma}f)(\partial^{\alpha + e_j - \gamma})} > $ > Adding all of the terms yields > $ > \begin{align*} > &\sum_{\gamma \le \alpha}{\alpha \choose \gamma }(\partial^{\gamma + e_j} f)(\partial^{\alpha - \gamma} g) + \sum_{\gamma \le \alpha}{\alpha \choose \gamma}(\partial^{\gamma}f)(\partial^{\alpha + e_j - \gamma}) \\ > &= \pg{\sum_{\gamma \le \alpha, \gamma < e_j}{\alpha + e_j \choose \gamma}(\partial^{\gamma}f)(\partial^{\alpha + e_j - \gamma})} \\ > &+ \pb{\sum_{e_j \le \gamma \le \alpha}{\alpha + e_j \choose \gamma}(\partial^{\gamma}f)(\partial^{\alpha + e_j - \gamma})} \\ > &+ \po{\underbrace{{\alpha + e_j \choose \alpha + e_j} (\partial^{\alpha + e_j} f)(\partial^{0} g)}_{\gamma = \alpha + e_j}} \\ > &= \sum_{\gamma \le \alpha + e_j}{\alpha + e_j \choose \gamma}(\partial^{\gamma}f)(\partial^{\alpha + e_j - \gamma}) > \end{align*} > $