Limits of Dot Products of Vector-Valued Functions

# Limits of Dot Products of Vector-Valued Functions

If $(S, d_s)$ and $(\mathbb{R}^n, d)$ are metric spaces where $d$ is the usual Euclidean metric on $\mathbb{R}^n$, $A \subseteq S$, and $\mathbf{f}, \mathbf{g} : A \to \mathbb{R}^n$, then we can define the dot product of $\mathbf{f}$ and $\mathbf{g}$ for all $x \in A$ by:

(1)
\begin{align} \quad \mathbf{f}(x) \cdot \mathbf{g}(x) = (f_1(x), f_2(x), ..., f_n(x)) \cdot (g_1(x), g_2(x), ..., g_n(x)) = (f_1(x)g_1(x), f_2(x)g_2(x), ..., f_n(x)g_n(x)) \end{align}

If $p \in S$ is an accumulation point of $A$ and if $\lim_{x \to p} \mathbf{f}(x) = \mathbf{a}$ and $\lim_{x \to p} \mathbf{g}(x) = \mathbf{b}$ then what can we say about $\lim_{x \to p} \mathbf{f}(x) \cdot \mathbf{g}(x)$? The following theorem answers this question.

 Theorem 1: Let $(S, d_S)$ and $(\mathbb{R}^n, d)$ be metric spaces where $d$ the usual metric on $\mathbb{R}^n$ defined for all $\mathbf{x} = (x_1, x_2, ..., x_n), \mathbf{y} = (y_1, y_2, ..., y_n) \in \mathbb{R}^n$ by $d(\mathbf{x}, \mathbf{y}) = \| \mathbf{x} - \mathbf{y} \|$. Then if $A \subseteq S$, $p \in S$ is an accumulation point of $A$, $\mathbf{f}, \mathbf{g} : A \to \mathbb{R}^n$, and $\lim_{x \to p} \mathbf{f}(x) = \mathbf{a}$, $\lim_{x \to p} \mathbf{g}(x) = \mathbf{b}$ then $\lim_{x \to p} [\mathbf{f}(x) \cdot \mathbf{g}(x) ] = \mathbf{a} \cdot \mathbf{b}$.

Notice that the function $\mathbf{f}(x) \cdot \mathbf{g}(x)$ maps into $\mathbb{R}$ and NOT into $\mathbb{R}^n$! To prove that $\lim_{x \to p} [\mathbf{f}(x) \cdot \mathbf{g}(x)] = \mathbf{a} \cdot \mathbf{b}$ we must show that for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $x \in D(f) \setminus \{ p \} = A \setminus \{ p \}$ and $d_S(x, p) < \delta$ then $\mid \mathbf{f}(x) \cdot \mathbf{g}(x) - \mathbf{a} \cdot \mathbf{b} \mid < \epsilon$.

• Proof: Let $\epsilon > 0$ be given and assume that $1 > \epsilon$. If $\epsilon \geq 1$ then we can always choose a $\delta$ corresponding to a smaller epsilon to satisfy the definition of the limit of a function.
• Since $\lim_{x \to p} \mathbf{f}(x) = \mathbf{a}$ we have that for $\epsilon_1 = \frac{\epsilon}{4(\| \mathbf{b} \| + 1)} > 0$ (note that $\epsilon_1 < 1$) there exists a $\delta_1 > 0$ such that if $x \in D(\mathbf{f}) \setminus \{ p \} = A \setminus \{ p \}$ and $d_S(x, p) < \delta_1$ then:
(2)
\begin{align} \quad d(\mathbf{f}(x), \mathbf{a}) = \| \mathbf{f}(x) - \mathbf{a} \| < \epsilon_1 = \frac{\epsilon}{4(\| \mathbf{b} \| + 1)} \quad (*) \end{align}
• Similarly, since $\lim_{x \to p} \mathbf{g}(x) = \mathbf{b}$ we have that for $\epsilon_2 = \frac{\epsilon}{4(\| \mathbf{a} \| + 1)} > 0$ (note that $\epsilon_2 < 1$) there exists a $\delta_2 > 0$ such that if $x \in D(\mathbf{g}) \setminus \{ p \} = A \setminus \{ p \}$ and $d_S(x, p) < \delta_2$ then:
(3)
\begin{align} \quad d(\mathbf{g}(x), \mathbf{b}) = \| \mathbf{g}(x) - \mathbf{b} \| < \epsilon_2 = \frac{\epsilon}{4(\| \mathbf{a} \| + 1)} \quad (**) \end{align}
• Now let $\delta = \min \{ \delta_1, \delta_2 \}$. Then if $x \in A \setminus \{ p \}$ and $d_S(x, p) < \delta$ then $d_S(x, p) < \delta_1$ AND $d_S(x, p) < \delta_2$ so $(*)$ and $(**)$ hold and:
(4)
\begin{align} \quad \mid \mathbf{f}(x) \cdot \mathbf{g}(x) - \mathbf{a} \cdot \mathbf{b} \mid = \mid \mathbf{f}(x) \cdot \mathbf{g}(x) - \mathbf{b} \cdot \mathbf{f}(x) + \mathbf{b} \cdot \mathbf{f}(x) - \mathbf{a} \cdot \mathbf{b} \mid & = \mid \mathbf{f}(x) \cdot [\mathbf{g}(x) - \mathbf{b}] + \mathbf{b} \cdot [\mathbf{f}(x) - \mathbf{a}] \mid \\ & \leq \mid \mathbf{f}(x) \cdot [ \mathbf{g}(x) - \mathbf{b} ] \mid + \mid \mathbf{b} \cdot [\mathbf{f}(x) - \mathbf{a}] \mid \end{align}
• Applying the Cauchy-Schwarz inequality yields:
(5)
\begin{align} \quad \mid \mathbf{f}(x) \cdot \mathbf{g}(x) - \mathbf{a} \cdot \mathbf{b} \mid & \leq \| \mathbf{f}(x) \| \| \mathbf{g}(x) - \mathbf{b} \| + \| \mathbf{b} \| \| \mathbf{f}(x) - \mathbf{a} \| \\ \quad & \leq \| [\mathbf{f}(x) - \mathbf{a}] + \| \mathbf{a} \| \| \mathbf{g}(x) - \mathbf{b} \| + \| \mathbf{b} \| \| \mathbf{f}(x) - \mathbf{a} \| \\ \quad & \leq \| \mathbf{f}(x) - \mathbf{a} \| \| \mathbf{g}(x) - \mathbf{b} \| + \| \mathbf{a} \| \| \mathbf{g}(x) - \mathbf{b} \| + \| \mathbf{b} \| \| \mathbf{f}(x) - \mathbf{a} \| \\ \quad & < \epsilon_1 \epsilon_2 + \| \mathbf{a} \| \epsilon_2 + \| b \| \epsilon_1 \\ \quad & < \frac{\epsilon}{4(\| b \| + 1)} \frac{\epsilon}{4(\| \mathbf{a} \| + 1)} + \| \mathbf{a} \| \frac{\epsilon}{4(\| \mathbf{a} \| + 1)} + \| \mathbf{b} \| \frac{\epsilon}{4(\| \mathbf{b} \| + 1)} \\ \quad & < \frac{\epsilon^2}{16} + \frac{\epsilon}{4} + \frac{\epsilon}{4} \\ \quad & < \frac{\epsilon}{4} + \frac{\epsilon}{2} \\ \quad & < \epsilon \end{align}
• Therefore, for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $x \in A \setminus \{ p \}$ and $d_S(x, p) < \delta$ then $\mid \mathbf{f}(x) \cdot \mathbf{g}(x) - \mathbf{a} \cdot \mathbf{b} \mid < \epsilon$, so $\lim_{x \to p} \mathbf{f}(x) \cdot \mathbf{g}(x) = \mathbf{a} \cdot \mathbf{b}$. $\blacksquare$