Limits of Sums and Differences of Vector-Valued Functions

# Limits of Sums and Difference of Vector-Valued Functions

We recently looked at some nice properties of complex-valued functions and we will now focus our attention on vector-valued functions. Let $(S, d_S)$ be any metric space and consider the metric space $(\mathbb{R}^n, d)$ where $d$ is the usual Euclidean 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:

(1)
\begin{align} \quad d(\mathbf{x}, \mathbf{y}) = \| \mathbf{x} - \mathbf{y} \| = \sqrt{(x_1 - y_1)^2 + (x_2 - y_2)^2 + ... + (x_n - y_n)^2} \end{align}

Now consider a function $\mathbf{f} : A \to \mathbb{R}^n$. The function $\mathbf{f}$ is actually an $n$-component vector where $\mathbf{f}(x) = (f_1(x), f_2(x), ..., f_n(x))$ where $f_j : A \to \mathbb{R}$. So for each $x \in A$ we defined $\mathbf{f}(x)$ by:

(2)
\begin{align} \quad \mathbf{f}(x) = (f_1(x), f_2(x), ..., f_n(x)) \end{align}

If $\mathbf{g} : A \to \mathbb{R}^n$ is another function, then with vector addition we have that:

(3)
\begin{align} \quad (\mathbf{f} + \mathbf{g})(x) = (f_1(x) + g_1(x), f_2(x) + g_2(x), ..., f_n(x) + g_n(x)) = (f_1(x), f_2(x), ..., f_n(x)) + (g_1(x), g_2(x), ..., g_n(x)) = \mathbf{f}(x) + \mathbf{g}(x) \end{align}

We will now see that if $\lim_{x \to p} \mathbf{f}(x) = \mathbf{a}$ and $\lim_{x \to p} \mathbf{g}(x) = \mathbf{b}$ then $\lim_{x \to p} [\mathbf{f}(x) + \mathbf{g}(x)] = \mathbf{a} + \mathbf{b}$.

 Theorem 1: Let $(S, d_S)$ and $(\mathbb{R}^n, d)$ be metric spaces where $d$ the usual Euclidean 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 $A$, $\mathbf{f}, \mathbf{g} : A \to \mathbb{R}^n$, $\lim_{x \to p} \mathbf{f} (x) = \mathbf{a}$ and $\lim_{x \to p} \mathbf{g} (x) = \mathbf{b}$ then $\lim_{x \to p} [\mathbf{f}(x) + \mathbf{g}(x)] = \mathbf{a} + \mathbf{b}$.
• Proof: Let $\epsilon > 0$ be given.
• Since $\lim_{x \to p} \mathbf{f}(x) = \mathbf{a}$ we have that for $\epsilon_1 = \frac{\epsilon}{2} > 0$ there exists a $\delta_1 > 0$ such that if $x \in D(f) \setminus \{ p \} = A \setminus \{ p \}$ and $d_S(x, p) < \delta_1$ then:
(4)
• Similarly, since $\lim_{x \to p} \mathbf{g}(x) = \mathbf{b}$ we have that for $\epsilon_2 = \frac{\epsilon}{2} > 0$ there exists a $\delta_2 > 0$ such that if $x \in D(g) \setminus \{ p \} = A \setminus \{ p \}$ and $d_S(x, p) < \delta_2$ then:
(5)
• 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 $(**)$ will hold and so:
(6)
\begin{align} \quad d(\mathbf{f}(x) + \mathbf{g}(x), \mathbf{a} + \mathbf{b}) = \| \mathbf{f}(x) + \mathbf{g}(x) - (\mathbf{a} + \mathbf{b}) \| = \| (\mathbf{f}(x) - \mathbf{a}) + (\mathbf{g}(x) - \mathbf{b}) \| \leq \| \mathbf{f}(x) - \mathbf{a} \| + \| \mathbf{g}(x) - \mathbf{b} \| &= d(\mathbf{f}(x), \mathbf{a}) + d(\mathbf{g}(x), \mathbf{b}) \\ & < \epsilon_1 + \epsilon_2 \\ & < \frac{\epsilon}{2} + \frac{\epsilon}{2} \\ & < \epsilon \end{align}
• So for all $\epsilon > 0$ there exists a $\delta = \min \{ \delta_1, \delta_2 \} > 0$ such that if $x \in D(f + g) \setminus \{ p \} = A \setminus \{ p \}$ and $d_S(x, p) < \delta$ then $d(\mathbf{f}(x) + \mathbf{g}(x), \mathbf{a} + \mathbf{b}) < \epsilon$, so $\lim_{x \to p} [\mathbf{f}(x) + \mathbf{g}(x)] = \mathbf{a} + \mathbf{b}$. $\blacksquare$

We also have a similar result for the difference of two vector-valued functions.

 Theorem 2: 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 $A$, $f, g : A \to \mathbb{R}^n$, $\lim_{x \to p} \mathbf{f} (x) = \mathbf{a}$ and $\lim_{x \to p} \mathbf{g} (x) = \mathbf{b}$ then $\lim_{x \to p} [\mathbf{f}(x) - \mathbf{g}(x)] = \mathbf{a} - \mathbf{b}$.
• Proof: Let $\epsilon > 0$ be given.
• Since $\lim_{x \to p} \mathbf{f}(x) = \mathbf{a}$ we have that for $\epsilon_1 = \frac{\epsilon}{2} > 0$ there exists a $\delta_1 > 0$ such that if $x \in D(f) \setminus \{ p \} = A \setminus \{ p \}$ and $d_S(x, p) < \delta_1$ then:
(7)
• Similarly, since $\lim_{x \to p} \mathbf{g}(x) = \mathbf{b}$ we have that for $\epsilon_2 = \frac{\epsilon}{2} > 0$ there exists a $\delta_2 > 0$ such that if $x \in D(g) \setminus \{ p \} = A \setminus \{ p \}$ and $d_S(x, p) < \delta_2$ then:
• 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 $(**)$ will hold and so:
• So for all $\epsilon > 0$ there exists a $\delta = \min \{ \delta_1, \delta_2 \} > 0$ such that if $x \in D(f - g) \setminus \{ p \} = A \setminus \{ p \}$ and $d_S(x, p) < \delta$ then $d(\mathbf{f}(x) - \mathbf{g}(x), \mathbf{a} - \mathbf{b}) < \epsilon$, so $\lim_{x \to p} [\mathbf{f}(x) - \mathbf{g}(x)] = \mathbf{a} - \mathbf{b}$. $\blacksquare$