Proposition

A function $f: \mathbb{R}^n \times \mathbb{R}^m \rightarrow \mathbb{R}^p$ is bilinear if for $x, x_1, x_2 \in \mathbb{R}^n$, $y, y_1, y_2 \in \mathbb{R}^m$, and $a \in \mathbb{R}$ we have

\[\begin{align*} f(ax, y) &= af(x, y) = f(x, ay), \\ f(x_1 + x_2, y) &= f(x_1, y) + f(x_2, y), \\ f(x, y_1 + y_2) &= f(x, y_1) + f(x, y_2). \end{align*}\]

Prove that if $f$ is bilinear, then $\begin{align*} \lim\limits_{(h, k) \rightarrow 0} \frac{\abs{f(h, k)}}{\abs{(h, k)}} = 0. \end{align*}$
Prove that $Df(a, b)(x, y) = f(a, y) + f(x, b)$.
Show that the formula for $Dp(a, b)$ in Theorem 2-3 is a special case of (b).

Solution

1

Let $(h, k) = ((h_1, \cdots, h_n), (k_1, \cdots, k_m)) \in \mathbb{R}^n \times \mathbb{R}^m$ be given.

Let $u_i$’s and $v_j$’s denote the standard unit vectors in $\mathbb{R}^n$ and $\mathbb{R}^m$, respectively.

\[\begin{align*} \abs{f(h, k)} &= \abs{\sum_{i=1}^n \sum_{j=1}^m f(h_iu_i, k_jv_j)} \\ &\leq \abs{\sum_{i=1}^n \sum_{j=1}^m h_ik_jf(u_i, v_j)} \\ &\leq \sum_{i=1}^n \sum_{j=1}^m \abs{h_ik_jf(u_i, v_j)} \\ &= \sum_{i=1}^n \sum_{j=1}^m \abs{h_ik_j}\abs{f(u_i, v_j)}. \end{align*}\]

Let $M = \max\{ \abs{f(u_i, v_j)} \}$.

\[\begin{align*} \sum_{i=1}^n \sum_{j=1}^m \abs{h_ik_j}\abs{f(u_i, v_j)} &\leq M \sum_{i=1}^n \sum_{j=1}^m \abs{h_ik_j}. \end{align*}\]

For each $i$ and $j$, $\abs{h_ik_j} \leq \max\{\abs{h_i}^2, \abs{k_j}^2\} \leq \abs{h_i}^2 + \abs{k_j}^2$.

\[\begin{align*} \sum_{i=1}^n \sum_{j=1}^m \abs{h_ik_j} &\leq \sum_{i=1}^n \sum_{j=1}^m \abs{h_i}^2 + \abs{k_j}^2 \\ &= m \sum_{i=1}^n \abs{h_i}^2 + n \sum_{j=1}^m \abs{k_j}^2 \\ &< (m + n) \sum_{i=1}^n \abs{h_i}^2 + (m + n) \sum_{j=1}^m \abs{k_j}^2 \\ &= (m + n) (\sum_{i=1}^n \abs{h_i}^2 + \sum_{j=1}^m \abs{k_j}^2) \\ &= (m + n) \abs{(h, k)}^2. \end{align*}\]

By putting these together, we get $\abs{f(h, k)} < M(m + n)\abs{(h, k)}^2$ where $M(m + n)$ does not depend on the choice of $h, k$.

Therefore,

\[\begin{align*} \lim\limits_{(h, k) \rightarrow 0} \frac{\abs{f(h, k)}}{\abs{(h, k)}} &\leq \lim\limits_{(h, k) \rightarrow 0} \frac{M(m + n)\abs{(h, k)}^2}{\abs{(h, k)}} \\ &= \lim\limits_{(h, k) \rightarrow 0} M(m + n)\abs{(h, k)} \\ &= 0. \end{align*}\]

2

Let $(a, b) \in \mathbb{R}^n \times \mathbb{R}^m$.

Let $A: \mathbb{R}^n \times \mathbb{R}^m \rightarrow \mathbb{R}^p$ be defined such that $A(x, y) = f(x, b) + f(a, y)$. Then $A$ is linear since $\forall (x_1, y_1), (x_2, y_2) \in \mathbb{R}^n \times \mathbb{R}^m, \forall t \in \mathbb{R},$

\[\begin{align*} A(c(x_1, y_1) + (x_2, y_2)) &= A((cx_1 + x_2, cy_1 + y_2)) \\ &= f(cx_1 + x_2, b) + f(a, cy_1 + y_2) \\ &= cf(x_1, b) + f(x_2, b) + cf(a, y_1) + f(a, y_2) \\ &= c(f(x_1, b) + f(a, y_1)) + f(x_2, b) + f(a, y_2) \\ &= cA(x_1, y_1) + A(x_2, y_2). \end{align*}\] \[\begin{align*} \lim\limits_{(h, k) \rightarrow 0} \frac{\abs{f((a, b) + (h, k)) - f(a, b) - A(h, k)}}{\abs{(h, k)}} &= \lim\limits_{(h, k) \rightarrow 0} \frac{\abs{f(a, b) + f(h, b) + f(a, k) + f(h, k) - f(a, b) - A(h, k)}}{\abs{(h, k)}} \\ &= \lim\limits_{(h, k) \rightarrow 0} \frac{\abs{f(h, b) + f(a, k) + f(h, k) - A(h, k)}}{\abs{(h, k)}} \\ &= \lim\limits_{(h, k) \rightarrow 0} \frac{\abs{f(h, b) + f(a, k) + f(h, k) - f(h, b) - f(a, k)}}{\abs{(h, k)}} \\ &= \lim\limits_{(h, k) \rightarrow 0} \frac{\abs{f(h, k)}}{\abs{(h, k)}} \\ &= 0. \end{align*}\]

Therefore, $f$ is differentiable at $(a, b)$ and $Df(a, b)(x, y) = f(x, b) + f(a, y)$.

3

Let $s: \mathbb{R}^2 \rightarrow \mathbb{R}$ be defined such that $s(x, y) = xy$.

$as(x, y) = a(xy) = (ax)y = s(ax, y)$.
$as(x, y) = a(xy) = x(ay) = s(x, ay)$.
$s(x_1 + x_2, y) = (x_1 + x_2)y = x_1y + x_2y = s(x_1, y) + s(x_2, y)$.
$s(x, y_1 + y_2) = x(y_1 + y_2) = xy_1 + xy_2 = s(x, y_1) + s(x, y_2)$.

Therefore, $s$ is bilinear. From (b), we know that $Ds(a, b)(x, y) = s(a, y) + s(x, b) = ay + xb$.