Total Derivative and Multilinear Functions

Solution 1:

this is just a comment to help you to see why $k$-linear functions on finite dimensional spaces are bounded.

to see that bilinear forms in finite dimensional spaces are bounded you can argue like that: (using your notation)

Let ${\bf x}=(x_1,\ldots,x_n)$ and ${\bf y}=(y_1,\ldots,y_m)$ be unit vectors in $\mathbb{R}^n$ and $\mathbb{R}^m$, respectively, then $$ \left\|{\bf f}({\bf x},{\bf y})\right\|=\left\|\sum_{i=1}^n\sum_{j=1}^m x_iy_j\ {\bf f}(e_i,e_j)\right\| \leq \max_{i,j}\ \|{\bf f}(e_i,e_j)\|\sum_{i=1}^n\sum_{j=1}^m (x^2_i+y^2_j). $$ If we call $M=\max_{i,j}\ \|{\bf f}(e_i,e_j)\|$ we have from the above inequality that $$ \|{\bf f}({\bf x},{\bf y})\| \leq M(m+n)\|{\bf x}\|^2\cdot \|{\bf y}\|^2 $$ Since ${\bf x}$ and ${\bf y}$ are unit vectors follows that ${\bf f}$ is bounded by $M(m+n)$. I hope you can extend this for any $k$-linear function.

Solution 2:

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 $$ \lim_{h \to 0} \frac{\left|f(a_1+h_1,\dots,a_k+h_k) - f(a_1,\dots,a_k) - \sum_{i=1}^k f(a_1,\dots,a_{i-1},h_i,a_{i+1},\dots,a_k)\right|}{|h|} = 0. $$

Solution 3:

@Leandro Thanks for the help; that got me going in the right direction. I had trouble verifying your exact inequality, but I don't think it really matters. I got that $$\left\| {\bf f}({\bf x},{\bf y}) \right\|=\left\| \sum_{i=1}^{n}\sum_{j=1}^{m}{x^{i}y^{j}{\bf f}({\bf e}_i,{\bf e}_j)} \right\| \leq \max_{i,j}\left\| {\bf f}({\bf e}_i,{\bf e}_j) \right\|\sum_{i=1}^{n}\sum_{j=1}^{m}{|x^{i}y^{j}|}$$

and that $|x^iy^j|\leq|x^i|^2+|y^j|^2$, hence

$$\left\| {\bf f}({\bf x},{\bf y}) \right\| \leq M \sum_{i=1}^{n}\sum_{j=1}^{m}{|x^{i}|^2+|y^{j}|^2}.$$

But $\sum_{i=1}^{n}\sum_{j=1}^{m}{|x^{i}|^2+|y^{j}|^2} = m\left\| {\bf x} \right\|^2 + n\left\| {\bf y} \right\|^2 = (n+m)$ if ${\bf x}$ and ${\bf y}$ are unit vectors. Hence I have that $\left\| {\bf f}({\bf x},{\bf y}) \right\| \leq M(n+m)$ which still is independent of both $\bf x$ and $\bf y$, so that $\bf f$ is bounded. Also, isn't this bound independent of the norm as well, apart from the value of $M$?

The process is more or less identical for higher-linear functions, with only the max $M$ (defined in the same way) and the product and sum of the spaces' dimensions contributing to the final bound on $\left\| {\bf f}({\bf x}_1,\ldots,{\bf x}_k) \right\|$. This gives me the final bit of information I needed to show that $D{\bf f}({\bf a}_1,\ldots,{\bf a}_k)$ is indeed as the book stated. (It was a bit misleading, I think, to have the first part of the proof only be concerned with the case for when there are exactly two $\mathbf{h}_i$ terms.)