Limits of Scalar Multiples of Vector-Valued Functions

# Limits of Scalar Multiples of Vector-Valued Functions

Recall from the Limits of Sums and Difference of Vector-Valued Functions page that if $(S, d_S)$ and $(\mathbb{R}^n, d)$ are metric spaces such that $d$ is the usual (Euclidean) metric on $\mathbb{R}^n$ defined for all $\mathbf{x} = (x_1, x_2, ..., x_n), \mathbf{y} = (y_1, y_2, ..., y_n) \in \mathbb{R}^n$, $A \subseteq S$, $p \in S$ is an accumulation point of $A$, $\mathbf{f} : A \to \mathbb{R}^n$, and $\lim_{x \to p} \mathbf{f}(x) = \mathbf{a}$, $\lim_{x \to p} \mathbf{g}(x) = \mathbf{b}$ then:

(1)
\begin{align} \quad \lim_{x \to p} [\mathbf{f}(x) + \mathbf{g}(x)] = \mathbf{a} + \mathbf{b} \end{align}

If $\lambda$ is a any nonzero scalar, then our next question to pose is if $\lim_{x \to p} \mathbf{f}(x) = \mathbf{a}$ then for any scalar $\lambda$, does the function $\lambda \mathbf{f} : A \to \mathbb{R}^n$ defined below exists?:

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

As we will see in the following theorem, for any scalar $\lambda$ we have that $\lim_{x \to p} \lambda \mathbf{f}(x)$ exists and equals $\lambda \mathbf{a}$.

 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 $A$, $\mathbf{f}, \mathbf{g} : A \to \mathbb{R}^n$, $\lim_{x \to p} \mathbf{f}(x) = \mathbf{a}$ and $\lambda$ is a scalar, then $\lim_{x \to p} \lambda \mathbf{f}(x) = \lambda \mathbf{a}$.
• Proof: Let $\epsilon > 0$ be given.
• First suppose that $\lambda \neq 0$. Since $\lim_{x \to p} \mathbf{f}(x) = \mathbf{a}$ then for $\epsilon_1 = \frac{\epsilon}{\mid \lambda \mid} > 0$ 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:
(3)
\begin{align} \quad d(\mathbf{f}(x), \mathbf{a}) < \epsilon_1 = \frac{\epsilon}{\mid \lambda \mid} \quad (*) \end{align}
• Let $\delta = \delta_1$, so then if $d_S(x, p) < \delta$ then $(*)$ holds and:
(4)
\begin{align} \quad d(\lambda \mathbf{f}(x), \lambda \mathbf{a}) = \| \lambda \mathbf{f}(x) - \lambda \bf{a} \| = \| \lambda [\mathbf{f}(x) - \mathbf{a}] \| = \mid \lambda \mid \| \mathbf{f}(x) - \mathbf{a} \| < \mid \lambda \mid \epsilon_1 = \mid \lambda \mid \frac{\epsilon}{\mid \lambda \mid} = \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 $d(\lambda \mathbf{f}(x), \lambda \mathbf{a}) < \epsilon$, so $\lim_{x \to p} \lambda \mathbf{f}(x) = \lambda \mathbf{a}$. $\blacksquare$