Continuity of Combinations of Vector-Valued Functions

# Continuity of Combinations of Vector-Valued Functions

Recall from the Continuity of Functions on Metric Spaces that if $(S, d_S)$ and $(T, d_T)$ are metric spaces, $f : S \to T$, and $p \in S$ then we say that $f$ is continuous at $p$ if for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $d_S(x, p) < \delta$ then $d_T(f(x), f(p)) < \epsilon$.

We noted that if $p$ is an isolated point of $S$ then $f$ is always continuous at $p$ since for sufficiently small $\delta$ the we have that the only point $x$ of distance less than $\delta$ from $p$ will be itself, and $d_T(f(p), f(p)) = 0 < \epsilon$.

If instead $p$ is an accumulation point of $S$ then $f$ being continuous at $p$ implies that:

(1)
\begin{align} \quad \lim_{x \to p} f(x) = f(p) \end{align}

We will now consider the case of continuity of various combinations (sums, differences, etc…) of vector-valued functions from an arbitrary metric space $(S, d_S)$ to $(\mathbb{R}^n, d)$ where $d$ is the usual Euclidean metric on $\mathbb{R}^n$.

 Theorem 1: Let $(S, d_S)$ and $(\mathbb{R}^n, d)$ be metric spaces 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 $d(\mathbf{x}, \mathbf{y}) = \| \mathbf{x} - \mathbf{y} \|$. If $\mathbf{f}, \mathbf{g} : S \to \mathbb{R}^n$, $p \in S$, and both $\mathbf{f}$ and $\mathbf{g}$ are continuous at $p$ then: a) $\mathbf{f} + \mathbf{g}$ is continuous at $p$. b) $\mathbf{f} - \mathbf{g}$ is continuous at $p$.

We will only prove (a) since the proof of (b) is similar and these type of continuity proofs resemble the proofs regarding limits very closely.

• Proof of a) Let $\epsilon > 0$ be given.
• If $p \in S$ is an isolated point of $S$ then $\mathbf{f} + \mathbf{g}$ is trivially continuous at $p$.
• Suppose instead that $p$ is an accumulation point of $S$. Since $\mathbf{f}$ is continuous at $p$ we have that for $\epsilon_1 = \frac{\epsilon}{2} > 0$ there exists a $\delta_1 > 0$ such that if $d_S(x, p) < \delta_1$ then:
(2)
\begin{align} \quad d(\mathbf{f}(x), \mathbf{f}(p)) = \| \mathbf{f}(x) - \mathbf{f}(p) \| < \epsilon_1 = \frac{\epsilon}{2} \quad (*) \end{align}
• Similarly since $\mathbf{g}$ is continuous at $p$ we have that for $\epsilon_2 = \frac{\epsilon}{2} > 0$ there exists a $\delta_2 > 0$ such that if $d_S(x, p) < \delta_2$ then:
(3)
\begin{align} \quad d(\mathbf{g}(x), \mathbf{g}(p)) = \| \mathbf{g}(x) - \mathbf{g}(p) \| < \epsilon_2 = \frac{\epsilon}{2} \quad (**) \end{align}
• Let $\delta = \min \{ \delta_1, \delta_2 \}$. Then both $(*)$ and $(**)$ hold so:
(4)
\begin{align} \quad d(\mathbf{f}(x) + \mathbf{g}(x), \mathbf{f}(p) + \mathbf{g}(p)) = \| [\mathbf{f}(x) + \mathbf{g}(x)] - [\mathbf{f}(p) + \mathbf{g}(p)] \| = \| [\mathbf{f}(x) - \mathbf{f}(p)] + [\mathbf{g}(x) - \mathbf{g}(p)] \| & \leq \| \mathbf{f}(x) - \mathbf{f}(p) \| + \| \mathbf{g}(x) - \mathbf{g}(p) \| \\ & < \epsilon_1 + \epsilon_2 \\ & < \frac{\epsilon}{2} + \frac{\epsilon}{2} \\ & < \epsilon \end{align}
• Therefore $\mathbf{f} + \mathbf{g}$ is continuous at $p$. $\blacksquare$
 Theorem 2: Let $(S, d_S)$ and $(\mathbb{R}^n, d)$ be metric spaces 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 $d(\mathbf{x}, \mathbf{y}) = \| \mathbf{x} - \mathbf{y} \|$. If $\mathbf{f}, \mathbf{g} : S \to \mathbb{R}^n$, $p \in S$, $\lambda$ is a scalar, and both $\mathbf{f}$ and $\mathbf{g}$ are continuous at $p$ then: a) $\lambda \mathbf{f}$ is continuous at $p$. b) $\mathbf{f} \cdot \mathbf{g}$ is continuous at $p$. c) $\| \mathbf{f} \|$ is continuous at $p$.

Note that (a) and (c) are functions mapping into the reals and not into $\mathbb{R}^n$!