TITLE: Total Derivative and Multilinear Functions
QUESTION [12 upvotes]: So I'm starting to work through Spivak's Calculus on Manifolds and I'm having a little trouble verifying some of the claims made in the book problems. To review:

Given a function $\mathbf{f}:\mathbb{R}^{n}\to\mathbb{R}^m$, we say that $\mathbf{f}$ is differentiable at a point $\mathbf{a}=(a^{1},\ldots,a^{n})\in\mathbb{R}^n$ (considered as a $1\times n$ matrix) if there exists a linear transformation, $D\mathbf{f}(\mathbf{a}):\mathbb{R}^n\to\mathbb{R}^m$, (considered as an $m\times n$ matrix) such that $$\lim\limits_{\mathbf{h}\to\mathbf{a}}\dfrac{||\mathbf{f}(\mathbf{a}+\mathbf{h})-\mathbf{f}(\mathbf{a})-D\mathbf{f}(\mathbf{a})(\mathbf{h})||}{||\mathbf{h}||}=0.$$
$D\mathbf{f}(\mathbf{a})$ is called the total derivative or Jacobian matrix of $\mathbf{f}$ at $\mathbf{a}$ and is unique.

We are then asked to prove that if $\mathbf{f}:\mathbb{R}^n\times\mathbb{R}^m\to\mathbb{R}^p$ is bilinear, then $$\lim\limits_{(\mathbf{h},\mathbf{k})\to\mathbf{0}}\dfrac{||\mathbf{f}(\mathbf{h},\mathbf{k})||}{||(\mathbf{h},\mathbf{k})||}=0.$$
The best approach I could think of was that $||\mathbf{f}(\mathbf{h},\mathbf{k})||=||\mathbf{h}||\cdot||\mathbf{k}||\cdot||\mathbf{f}(\hat{\mathbf{h}},\hat{\mathbf{k}})||$, where $\hat{\mathbf{x}}=\frac{\mathbf{x}}{||\mathbf{x}||}$. We then have that $||\mathbf{h}||\cdot||\mathbf{k}||\leq||\mathbf{h}||^{2}+||\mathbf{k}||^{2}=||(\mathbf{h},\mathbf{k})||^{2}$, which gives us $$\dfrac{||\mathbf{f}(\mathbf{h},\mathbf{k})||}{||(\mathbf{h},\mathbf{k})||}\leq||(\mathbf{h},\mathbf{k})||\cdot||\mathbf{f}(\hat{\mathbf{h}},\hat{\mathbf{k}})||,$$ where $||\hat{\mathbf{h}}||=||\hat{\mathbf{k}}||=1$. And since $||\mathbf{f}(\mathbf{x},\mathbf{y})||\leq||\mathbf{f}(\hat{\mathbf{x}},\mathbf{y})||\cdot||\mathbf{x}||$ (similarly for $\mathbf{y}$) then $f$ is a bounded (continuous) linear transformation and
$$||\mathbf{f}||_{x}=\sup\{||\mathbf{f}(\mathbf{x},\mathbf{y})||:||\mathbf{x}||=1\}<\infty$$
$$||\mathbf{f}||_{y}=\sup\{||\mathbf{f}(\mathbf{x},\mathbf{y})||:||\mathbf{y}||=1\}<\infty,$$
which I was hoping would imply that $||\mathbf{f}(\hat{\mathbf{h}},\hat{\mathbf{k}})||<\infty$, and so we'd have the result we're looking for. But I am stuck here.
Assuming this result, I was able to complete the problem and prove that $$D\mathbf{f}(\mathbf{a},\mathbf{b})(\mathbf{x},\mathbf{y})=\mathbf{f}(\mathbf{a},\mathbf{y})+\mathbf{f}(\mathbf{x},\mathbf{b}).$$

The author then proceeds to ask us to prove the following:

Given a multilinear function $\mathbf{f}:\mathbb{R}^{n_1}\times\cdots\times\mathbb{R}^{n_k}\to\mathbb{R}^p$, show that for $\mathbf{h}=(\mathbf{h}_1,\dots,\mathbf{h}_k)$, with $\mathbf{h}_i\in\mathbb{R}^{n_i}$, we have
$$\lim\limits_{\mathbf{h}\to\mathbf{0}}\dfrac{||\mathbf{f}(\mathbf{a}_1,\ldots,\mathbf{a}_{i-1},\mathbf{h}_i,\mathbf{a}_{i+1}\ldots,\mathbf{a_{j-1}},\mathbf{h}_j,\mathbf{a}_{j+1},\ldots,\mathbf{a}_k)||}{||\mathbf{h}||}=0,$$ for $i\neq j$. Use this to prove that
$$D\mathbf{f}(\mathbf{a}_1,\ldots,\mathbf{a}_k)(\mathbf{x}_1,\ldots,\mathbf{x}_k)=\sum_{i=1}^{k}{\mathbf{f}(\mathbf{a}_1,\ldots,\mathbf{a}_{i-1},\mathbf{x}_i,\mathbf{a}_{i+1}\ldots,\mathbf{a}_k)}.$$

Using the hint from the book of considering the bilinear function, $\mathbf{g}(\mathbf{x},\mathbf{y})=\mathbf{f}(\mathbf{a}_1,\ldots,\mathbf{x},\ldots,\mathbf{y},\ldots,\mathbf{a}_k)$, I was able to show the first part of the problem assuming the above result (which I still can't prove). However I'm now also stuck on how to prove the rest of the question.
For example, consider the specific case with three arguments. Using the fact that $\mathbf{f}$ is multilinear, we have
$$\mathbf{f}(\mathbf{a}+\mathbf{h_1},\mathbf{b}+\mathbf{h_2},\mathbf{c}+\mathbf{h_3})-\mathbf{f}(\mathbf{a},\mathbf{b},\mathbf{c})=$$
$$\mathbf{f}(\mathbf{a},\mathbf{b},\mathbf{h_3})+\mathbf{f}(\mathbf{a},\mathbf{h_2},\mathbf{c})+\mathbf{f}(\mathbf{h_1},\mathbf{b},\mathbf{c}) +$$
$$+\mathbf{f}(\mathbf{a},\mathbf{h_2},\mathbf{h_3})+\mathbf{f}(\mathbf{h_1},\mathbf{b},\mathbf{h_3})+\mathbf{f}(\mathbf{h_1},\mathbf{h_2},\mathbf{c})+$$
$$+\mathbf{f}(\mathbf{h_1},\mathbf{h_2},\mathbf{h_3}).$$
The first three terms are what should be $D\mathbf{f}(\mathbf{a},\mathbf{b},\mathbf{c})$ while the next three will disappear in the limit given the first part of the proof. The trouble is how do I control the limit $\lim\limits_{\mathbf{h}\to\mathbf{0}}\frac{||\mathbf{f}(\mathbf{h_1},\mathbf{h_2},\mathbf{h_3})||}{||(\mathbf{h_1},\mathbf{h_2},\mathbf{h_3})||}$? This trouble arises in the general case as well, any time the number of $\mathbf{h_i}$ is more than two in any single term. E.g. how would I control the term $\mathbf{f}(\mathbf{a_1},\mathbf{a_2},\mathbf{h_3},\mathbf{h_4})$, which has two "$h$" and "$a$" terms each?
Thanks for any help. My analysis skills are a little rusty as I've been focusing on passing my algebra qual.

REPLY [4 votes]: Since the other two answers are more or less just hints towards the complete answer, I am presenting my solution below to clarify my own ideas.
The OP's questions are from Problems 2-12 and 2-14 of Spivak's Calculus on Manifolds.

We want to show that if $f \colon \mathbb{R}^n \times \mathbb{R}^m \to \mathbb{R}^p$ is bilinear, then
$$
\lim_{(h,k) \to 0} \frac{|f(h,k)|}{|(h,k)|} = 0.
$$
So, we take a cue from Spivak's proof of Theorem 2-3(5) (given on page 21): there he computes the derivative of $p \colon \mathbb{R}^2 \to \mathbb{R}$ defined as $p(x,y) = x \cdot y$. Since this is actually a bilinear function, the idea of the proof carries over.
So, let $h = (h^1,\dots,h^n) \in \mathbb{R}^n$ and $k = (k^1,\dots,k^m) \in \mathbb{R}^m$. Let $e_1,\dots,e_n$ be the standard basis of $\mathbb{R}^n$ and let $\tilde{e}_1,\dots,\tilde{e}_m$ be the standard basis of $\mathbb{R}^m$. Then,
$$
f(h,k) = \sum_{i=1}^n \sum_{j=1}^m h^i k^j f(e_i,\tilde{e}_j).
$$
Let $M = \max\{ |f(e_i,\tilde{e}_j)| : 1 \leq i \leq n, 1 \leq j \leq m \}$. Then, for $(h,k) \neq (0,0)$, we have
$$
\begin{align}
\frac{|f(h,k)|}{|(h,k)|} &= \frac{\left|\sum_{i=1}^n \sum_{j=1}^m h^i k^j f(e_i,\tilde{e}_j)\right|}{|(h,k)|} \\
&\leq \sum_{i=1}^n \sum_{j=1}^m \frac{|h^i k^j f(e_i,\tilde{e}_j)|}{|(h,k)|}\\
& \leq M \sum_{i=1}^n \sum_{j=1}^m \frac{|h^i k^j|}{|(h,k)|}.
\end{align}
$$
Note that
$$
|h^i k^j| \leq
\begin{cases}
|h^i|^2, & k^j \leq h^i;\\
|k^j|^2, & h^i \leq k^j.
\end{cases}
$$
Hence,
$$
|h^i k^j| \leq |h^i|^2 + |k^j|^2 \leq |h|^2 + |k|^2.
$$
Therefore, for $(h,k) \neq (0,0)$, we have
$$
\frac{|h^i k^j|}{|(h,k)|} \leq |(h,k)|.
$$
Hence,
$$
\frac{|f(h,k)|}{|(h,k)|} \leq Mmn|(h,k)|.
$$
Taking $\lim_{(h,k) \to 0}$ on both sides, we get the desired result.

Now, let $k \geq 2$. To check that the derivative of the multilinear function $f \colon E_1 \times \dots \times E_k \to \mathbb{R}^p$ at the point $(a_1,\dots,a_k)$ is given by
$$
Df(a_1,\dots,a_k)(x_1,\dots,x_k) = \sum_{i=1}^k f(a_1,\dots,a_{i-1},x,a_{i+1},\dots,a_k),
$$
where the $E_i$'s are Euclidean spaces of dimension $n_i$, we need to check that
$$
\lim_{h \to 0} \frac{\left|f(a_1+h_1,\dots,a_k+h_k) - f(a_1,\dots,a_k) - \sum_{i=1}^k f(a_1,\dots,a_{i-1},h_i,a_{i+1},\dots,a_k)\right|}{|h|} = 0.
$$
The expression inside the limit can be simplified in the following manner. 
Since $h_i \in E_i$ for each $1 \leq i \leq k$, let $h_i = (h_i^1,\dots,h_i^{n_i})$ for each $1 \leq i \leq k$. For each $2 \leq l \leq k$, let $I = (i_1,\dots,i_l)$ be an $l$-tuple such that $1 \leq i_1 < \dots < i_l \leq k$. We shall call such an $I$ an $l$-shuffle (note that this terminology is non-standard). For each $1 \leq i \leq k$, let $e_i^1,\dots,e_i^{n_i}$ be the standard basis of $E_i$.
Now,
$$
\begin{align}
f(a_1+h_1,\dots,a_k+h_k) &= f(a_1,\dots,a_k) + \sum_{i = 1}^k f(a_1,\dots,a_{i-1},h_i,a_{i+1},\dots,a_k) \\
& \qquad {}+ \sum_{l=2}^k \sum_{\substack{I=(i_1,\dotsc,i_l) \\ l\mathrm{-shuffles}}} f(a_1,\dots,a_{i_1 - 1},h_{i_1},a_{i_1 +1},\dots,a_{i_l - 1},h_{i_l},a_{i_l + 1},\dots,a_k).
\end{align}
$$
Consider the term inside the last summation.
$$
\begin{align}
&\ \ f(a_1,\dots,a_{i_1 - 1},h_{i_1},a_{i_1 +1},\dots,a_{i_l - 1},h_{i_l},a_{i_l + 1},\dots,a_k)\\ = \ &\sum_{j_1=1}^{n_{i_1}} \cdots \sum_{j_l = 1}^{n_{i_l}} h_{i_1}^{j_1} \cdots h_{i_l}^{j_l} f(a_1,\dots,a_{i_1 - 1},e_{i_1}^{j_1},a_{i_1 + 1},\dots,a_{i_l - 1},e_{i_l}^{j_l},a_{i_l + 1},\dots,a_k).
\end{align}
$$
Let
$$
M_I = \max\{ |f(a_1,\dots,a_{i_1 - 1},e_{i_1}^{j_1},a_{i_1 + 1},\dots,a_{i_l - 1},e_{i_l}^{j_l},a_{i_l + 1},\dots,a_k)| : 1 \leq j_1 \leq n_{i_1},\dots,1 \leq j_l \leq n_{i_l} \}.
$$
Then,
$$
\begin{align}
& |f(a_1,\dots,a_{i_1 - 1},h_{i_1},a_{i_1 +1},\dots,a_{i_l - 1},h_{i_l},a_{i_l + 1},\dots,a_k)|\\ \leq \ &M_I \sum_{j_1=1}^{n_{i_1}} \cdots \sum_{j_l = 1}^{n_{i_l}} \left| h_{i_1}^{j_1} \cdots h_{i_l}^{j_l} \right|.
\end{align}
$$
Since we are interested in the limit as $h$ goes to $0$, we can assume that $0 \leq \left|h_{i_1}^{j_1}\right|,\dots,\left|h_{i_l}^{j_l}\right| \leq 1$. In particular, for $(h_{i_1}^{j_1},\dots,h_{i_l}^{j_l}) \neq (0,\dots,0)$, we get that
$$
\left|h_{i_1}^{j_1}\cdots h_{i_l}^{j_l}\right| \leq \left|h_{i_1}^{j_1} h_{i_2}^{j_2} \right|\\
\implies \frac{\left|h_{i_1}^{j_1}\cdots h_{i_l}^{j_l}\right|}{\left(\left| h_{i_1}^{j_1} \right|^2 + \dots + \left| h_{i_l}^{j_l} \right|^2 \right)^{1/2}} \leq \frac{\left|h_{i_1}^{j_1} h_{i_2}^{j_2} \right|}{\left(\left| h_{i_1}^{j_1} \right|^2 + \dots + \left| h_{i_l}^{j_l} \right|^2 \right)^{1/2}} \leq \frac{\left|h_{i_1}^{j_1} h_{i_2}^{j_2} \right|}{\left(\left| h_{i_1}^{j_1} \right|^2 + \left| h_{i_2}^{j_2} \right|^2 \right)^{1/2}}
$$
By what we have proved for bilinear functions, it is clear that
$$
\lim_{\substack{h \to 0\\ (h_{i_1}^{j_1},\dots,h_{i_l}^{j_l}) \neq (0,\dots,0)}} \frac{\left|h_{i_1}^{j_1}\cdots h_{i_l}^{j_l}\right|}{|h|} = 0.
$$
Moreover, the above limit is quite obviously zero even without the condition  $(h_{i_1}^{j_1},\dots,h_{i_l}^{j_l}) \neq (0,\dots,0)$. Hence,
$$
\lim_{h \to 0} \frac{\left|h_{i_1}^{j_1}\cdots h_{i_l}^{j_l}\right|}{|h|} = 0.
$$
This is true for every $l$-shuffle $I$, for each $2 \leq l \leq k$. Hence, by using the triangle inequality, we get that
$$
\lim_{h \to 0} \frac{\left|f(a_1+h_1,\dots,a_k+h_k) - f(a_1,\dots,a_k) - \sum_{i=1}^k f(a_1,\dots,a_{i-1},h_i,a_{i+1},\dots,a_k)\right|}{|h|} = 0.
$$