Continuity of Combinations of Complex-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 complex-valued functions from an arbitrary metric space $(S, d_S)$ to $(\mathbb{C}, d)$ where $d$ is the usual Euclidean metric on $\mathbb{C}$.

Theorem 1: Let $(S, d_S)$ and $(\mathbb{C}, d)$ be metric spaces where $d$ is the usual Euclidean metric on $\mathbb{C}$ defined for all $x = a + bi, y = c + di \in \mathbb{C}$ by $d(x, y) = \mid x - y \mid$. If $f, g : S \to \mathbb{C}$, $p \in S$, and both $f$ and $g$ are continuous at $p$ then: a) $f + g$ is continuous at $p$. b) $f - g$ 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.

• Since $f$ 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(f(x), f(p)) = \mid f(x) - f(p) \mid < \epsilon_1 = \frac{\epsilon}{2} \quad (*) \end{align}

• Similarly since $g$ 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(g(x), g(p)) = \mid g(x) - g(p) \mid < \epsilon_2 = \frac{\epsilon}{2} \quad (**) \end{align}

• Let $\delta = \min \{ \delta_1, \delta_2 \}$. Then if $d_S(x, p) < \delta$ then both $(*)$ and $(**)$ hold and:

(4)

\begin{align} \quad \: d((f + g)(x), (f + g)(p)) = \mid f(x) + g(x) - f(p) - g(p) \mid = \mid [f(x) - f(p)] + [g(x) - g(p)] \mid \leq \mid f(x) - f(p) \mid + \mid g(x) - g(p) \mid & < \epsilon_1 + \epsilon_2 \\ & < \frac{\epsilon}{2} + \frac{\epsilon}{2} \\ & < \epsilon \end{align}

• Therefore $f + g$ is continuous at $p$. $\blacksquare$

Theorem 2: Let $(S, d_S)$ and $(\mathbb{C}, d)$ be metric spaces where $d$ is the usual Euclidean metric on $\mathbb{C}$ defined for all $x, y \in \mathbb{C}$ by $d(x, y) = \mid x - y \mid$. If $f, g : S \to \mathbb{C}$, $p \in S$, $c$ is a constant, and both $f$ and $g$ are continuous at $p$ then: a) $cf$ is continuous at $p$. b) $fg$ is continuous at $p$.