Limits of Norms of Vector-Valued Functions

Limits of Norms of Vector-Valued Functions

Let $(S, d_S)$ and $(\mathbb{R}^n, d)$ be metric spaces 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}

We have already seen a bunch of theorems regarding the limits of these vector-valued functions, and we will look at one more for the time being regarding the norm of $\mathbf{f}(x)$.

 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 $A$, $\mathbf{f} : A \to \mathbb{R}^n$, and $\lim_{x \to p} \mathbf{f}(x) = \mathbf{a}$ then $\lim_{x \to p} \| \mathbf{f}(x) \| = \| \mathbf{a} \|$.

Note that the function $\| \mathbf{f}(x) \|$ outputs real numbers!

• Let $\epsilon > 0$ be given.
• First assume that $\mathbf{a} \neq \mathbf{0}$. Since $\lim_{x \to p} \mathbf{f}(x) = \mathbf{a}$ then for $\epsilon > 0$ 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 \end{align}
• Let $\delta = \delta^*$ so if $x \in D(\mathbf{f}) \setminus \{ p \} = A \setminus \{ p \}$ and $d_S(x, p) < \delta$ then $(*)$ holds and:
(3)
\begin{align} \quad \mid \| \mathbf{f}(x) \| - \| \mathbf{a} \| \mid \leq \| \mathbf{f}(x) - \mathbf{a} \| < \epsilon \end{align}
• Therefore $\lim_{x \to p} \| \mathbf{f}(x) \| = \| \mathbf{a} \|$. $\blacksquare$