Existence of Limits of Vector-Valued Functions

Existence of Limits of Vector-Valued Functions

Let $(S, d_S)$ be any arbitrary metric space and consider the metric space $(\mathbb{R}^n, d)$ where $d$ is the usual Euclidean metric 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}

If $A \subseteq S$, $p \in S$ is an accumulation point of $A$, and $\mathbf{f} : A \to \mathbb{R}^n$ then the function $\mathbf{f}(x) = (f_1(x), f_2(x), ..., f_n(x))$ (where $f_j : A \to \mathbb{R}$ for each $j \in \{1, 2, ..., n \}$) is a vector-valued function, and additional theorems can be proved as a result of the basic properties of vectors (similarly to how we proved some additional theorems for complex-valued functions).

We begin by proving that the limit of $\mathbf{f}$ as $x \to p$ exists if and only if the limits of the components $f_j$ for $j \in \{1, 2, ..., n \}$ exist as $x \to p$.

 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 $d(\mathbf{x}, \mathbf{y}) = \| \mathbf{x} - \mathbf{y} \|$. Furthermore let $A \subseteq S$, $p \in S$ is an accumulation point of $A$, $\mathbf{f} : A \to \mathbb{R}^n$. Then $\lim_{x \to p} \mathbf{f}(x) = \mathbf{a}$ if and only if $\lim_{x \to p} f_j(x) = a_j$ for all $j \in \{1, 2, ..., n \}$.

Here the function $f_j : A \to \mathbb{R}$ is the $j^{\mathrm{th}}$ component of $\mathbf{f}$, and similarly, $a_j$ is the $j^{\mathrm{th}}$ component of $\mathbf{a}$.

• Proof: $\Rightarrow$ Suppose that $\lim_{x \to p} \mathbf{f}(x) = \mathbf{a}$. Then for all $\epsilon > 0$ there exists a $\delta^* > 0$ such that if $x \in D(\mathbf{f}) \setminus \{ p \} = A \setminus \{ p \}$ and $d_S(x, p) < \delta^*$ then:
(2)
\begin{align} \quad d(\mathbf{f}(x), \mathbf{a}) = \| \mathbf{f}(x) - \mathbf{a} \| < \epsilon \quad (*) \end{align}
• Expanding the inequality above gives us:
(3)
\begin{align} \quad \| \mathbf{f}(x) - \mathbf{a} \| = \sqrt{(f_1(x) - a_1)^2 + (f_2(x) - a_2)^2 + ... + (f_n(x) - a_n)^2} < \epsilon \end{align}
• For each $j \in \{ 1, 2, ..., n \}$ we have that then $\mid f_j(x) - a_j \mid < \| \mathbf{f}(x) - \mathbf{a} \| < \epsilon$. Taking $\delta = \delta^*$ and we see that if $x \in D(f) \setminus \{ p \} = A \setminus \{ p \}$ and $d_S(x, p) < \delta$ then $(*)$ holds and $\mid f_j(x) - a_j \mid < \epsilon$, so $\lim_{x \to p} f_j(x) = a_j$.
• $\Leftarrow$ Let $\epsilon > 0$ be given. Suppose that $\lim_{x \to p} f_j(x) = a_j$ for all $j \in \{1, 2, ..., n \}$. Then for $\epsilon_j = \frac{\epsilon}{n} > 0$ there exists a $\delta_j > 0$ such that if $x \in D(\mathbf{f}) \setminus \{ p \} = A \setminus \{ p \}$ and $d_S(x, p) < \delta_j$ then:
(4)
\begin{align} \quad \mid f_j(x) - a_j \mid < \epsilon_j = \frac{\epsilon}{n} \quad (*) \end{align}
• Let $\delta = \min \{ \delta_1, \delta_2, ..., \delta_n \}$. If $d_S(x, p) < \delta$ then $(*)$ holds for each $j \in \{1, 2, ..., n \}$ and so:
(5)
\begin{align} \quad d(\mathbf{f}(x), \mathbf{a}) = \| \mathbf{f}(x) - \mathbf{a} \| = \sqrt{(f_1(x) - a_1)^2 + (f_2(x) - a_2)^2 + ... + (f_n(x) - a_n)^2} \leq \sum_{j=1}^{n} \mid f_j(x) - a_j \mid < \sum_{j=1}^{n} \epsilon_j = \sum_{j=1}^{n} \frac{\epsilon}{n} = \epsilon \end{align}
• So 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 $d(\mathbf{f}(x), \mathbf{a}) < \epsilon$, so $\lim_{x \to p} \mathbf{f}(x) = \mathbf{a}$. $\blacksquare$