Proof of Peeling the Innermost Variable from a Multi-Index Partial Derivative

Throughout, "clause 1" and "clause 2" refer to the corresponding clauses of Partial Derivative of a Multi-Index on a Euclidean Open Set, and we use that those clauses constitute a definition: for a multi-index $\beta\ne 0$ of length $n$, a function $g:U\to\mathbb{R}$, and $p$ the least index $j$ with $1\le j\le n$ and $1\le\beta_j$, the statement that $\partial^{\beta}g$ exists on $U$ holds precisely when $\partial^{\beta-e_p}g$ exists on $U$ and the partial derivative of $\partial^{\beta-e_p}g$ with respect to the $p$th variable exists at every point of $U$; in that case $\partial^{\beta}g$ is the function on $U$ whose value at $x$ is that partial derivative at $x$, which we abbreviate as $\partial_p(\partial^{\beta-e_p}g)$.

Since the order of a multi-index is the sum of its entries and all entries are nonnegative, a multi-index $\beta$ with $1\le\beta_i$ for some $i$ has $1\le|\beta|$, and $|\beta|=|\beta-e_i|+1$, because $\beta-e_i$ differs from $\beta$ only in the $i$th entry, which is smaller by one.

We prove by induction the following statement $P(k)$, for every natural number $k$: for every $f:U\to\mathbb{R}$, every multi-index $\alpha$ of length $n$ with $|\alpha|=k$, and every index $i$ satisfying the hypotheses of the lemma for that $f$ and $\alpha$, the conclusion of the lemma holds. Since a multi-index is nonzero exactly when its order is a natural number, this proves the lemma.

Base case $k=1$. Here $1\le\alpha_i$ and the entries of $\alpha$ are nonnegative with sum $1$, so $\alpha_i=1$ and $\alpha_l=0$ for every $l\ne i$; that is, $\alpha=e_i$ and $\alpha-e_i=0$. By clause 1, $\partial^{\alpha-e_i}(\partial_i f)=\partial_i f$. On the other hand $i$ is the least index at which $\alpha$ is nonzero, $\partial^{\alpha-e_i}f=f$ exists on $U$ by clause 1, and the partial derivative of $f$ with respect to the $i$th variable exists at every point of $U$ by hypothesis. So by clause 2, $\partial^{\alpha}f$ exists on $U$ and is the function whose value at $x$ is that partial derivative at $x$, that is, $\partial^{\alpha}f=\partial_i f=\partial^{\alpha-e_i}(\partial_i f)$.

Inductive step. Let $k$ be a natural number for which $P(k)$ holds, and let $f$, a multi-index $\alpha$ with $|\alpha|=k+1$, and an index $i$ be as in the hypotheses of the lemma. Write $h=\partial_i f$ and $\gamma=\alpha-e_i$, so that $|\gamma|=k$ and $\partial^{\gamma}h$ exists on $U$ by hypothesis. Let $p$ be the least index $j$ with $1\le j\le n$ and $1\le\alpha_j$; such an index exists because $\alpha\ne 0$, and $p\le i$ because $1\le\alpha_i$.

Case 1: $p=i$. Then $\alpha_l=0$ for every $l<i$ by minimality of $p$, and $\alpha_l=0$ for every $l>i$ by the hypothesis on $i$; hence $\alpha_i=|\alpha|=k+1$, so $\gamma_i=k$ and $\gamma_l=0$ for $l\ne i$. In particular $\gamma\ne 0$, and $i$ is the least index at which $\gamma$ is nonzero. Applying clause 2 to $\partial^{\gamma}h$, which exists on $U$: $\partial^{\gamma-e_i}h$ exists on $U$, the partial derivative of $\partial^{\gamma-e_i}h$ with respect to the $i$th variable exists at every point of $U$, and $\partial^{\gamma}h=\partial_i(\partial^{\gamma-e_i}h)$.

Now $P(k)$ applies to the function $f$, the multi-index $\gamma$ and the index $i$: indeed $|\gamma|=k$, $1\le\gamma_i$, $\gamma_l=0$ for every $l>i$, the partial derivative of $f$ with respect to the $i$th variable exists at every point of $U$, and $\partial^{\gamma-e_i}h$ exists on $U$. It yields that $\partial^{\gamma}f$ exists on $U$ and $\partial^{\gamma}f=\partial^{\gamma-e_i}h$. Consequently the partial derivative of $\partial^{\gamma}f$ with respect to the $i$th variable exists at every point of $U$. Since $p=i$ is the least index at which $\alpha$ is nonzero and $\alpha-e_p=\gamma$, clause 2 gives that $\partial^{\alpha}f$ exists on $U$ and

Case 2: $p<i$. Put $\alpha'=\alpha-e_p$, so $|\alpha'|=k$. Since $\gamma$ differs from $\alpha$ only in the $i$th entry and $p\ne i$, we have $\gamma_p=\alpha_p$, so $1\le\gamma_p$, while $\gamma_l=\alpha_l=0$ for every $l<p$; thus $p$ is the least index at which $\gamma$ is nonzero. Also $\gamma-e_p=\alpha'-e_i$. Applying clause 2 to $\partial^{\gamma}h$: $\partial^{\gamma-e_p}h$ exists on $U$, the partial derivative of $\partial^{\gamma-e_p}h$ with respect to the $p$th variable exists at every point of $U$, and $\partial^{\gamma}h=\partial_p(\partial^{\gamma-e_p}h)$.

Now $P(k)$ applies to the function $f$, the multi-index $\alpha'$ and the index $i$: indeed $|\alpha'|=k$; $\alpha'_i=\alpha_i$ because $p\ne i$, so $1\le\alpha'_i$; $\alpha'_l=\alpha_l=0$ for every $l>i$; the partial derivative of $f$ with respect to the $i$th variable exists at every point of $U$; and $\partial^{\alpha'-e_i}h=\partial^{\gamma-e_p}h$ exists on $U$. It yields that $\partial^{\alpha'}f$ exists on $U$ and $\partial^{\alpha'}f=\partial^{\gamma-e_p}h$. Consequently the partial derivative of $\partial^{\alpha'}f$ with respect to the $p$th variable exists at every point of $U$, and since $\alpha-e_p=\alpha'$, clause 2 gives that $\partial^{\alpha}f$ exists on $U$ and

In both cases the conclusion of the lemma holds, so $P(k+1)$ holds. By induction, $P(k)$ holds for every natural number $k$, which is the assertion of the lemma.