This page is intended to be a part of the Real Analysis section of Math Online. Similar topics can also be found in the Calculus section of the site.

Properties of Continuous Functions

We will now look at a whole bunch of theorems regarding the continuity of functions at a point $c$ in their domain.

Theorem 1: Let $f : A \to \mathbb{R}$ and $g : A \to \mathbb{R}$ be continuous functions at $c \in A$. Then the function $f + g$ is continuous at $c \in A$.

• Proof: Let $f$ and $g$ be continuous functions at $c \in A$.
• Since $f$ is continuous at $c$ then for $\epsilon_1 = \frac{\epsilon}{2} > 0$ there exists a $\delta_1 > 0$ such that if $x \in A$ and $\mid x - c \mid < \delta_1$ then $\mid f(x) - f(c) \mid < \epsilon_1 = \frac{\epsilon}{2}$.

• Similarly since $g$ is continuous at $c$ then for $\epsilon_2 = \frac{\epsilon}{2} > 0$ there exists a $\delta_2 > 0$ such that if $x \in A$ and $\mid x - c \mid < \delta_2$ then $\mid g(x) - g(c) \mid < \epsilon_2 = \frac{\epsilon}{2}$.

• Now we want to show that $\forall \epsilon > 0$ then $\exists \delta > 0$ such that if $\mid x - c \mid < \delta$ then $\mid (f(x) + g(x)) - (f(c) + g(c)) \mid < \epsilon$. Let $\delta = \mathrm{min} \{\delta_1, \delta_2 \}$. Then:

(1)

\begin{align} \quad \quad \mid (f(x) + g(x)) - (f(c) + g(c)) \mid = \mid (f(x) - f(c)) + (g(x) - g(c)) \mid ≤ \mid f(x) - f(c) \mid + \mid g(x) - g(c) \mid < \epsilon_1 + \epsilon_2 = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}

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

A similar proof can be constructed to show that $f - g$ is also continuous at $c \in A$.

Theorem 2: Let $f : A \to \mathbb{R}$ be a continuous function at $c \in A$ and let $k \in \mathbb{R}$. Then the functions $kf$ is continuous at $c \in A$.

• Proof: Let $f$ be continuous at $c \in A$ and let $k \in \mathbb{R}$.

• First consider the case where $k = 0$. Then $kf$ will be the constant function of zero which we've established that constant functions are continuous for all $\mathbb{R}$.

• Now suppose that $k \neq 0$. Since $f$ is continuous at $c \in A$ then for $\epsilon_0 = \frac{\epsilon}{\mid k \mid} > 0$ there exists a $\delta_0 > 0$ such that if $x \in A$ and $\mid x - c \mid < \delta_0$ then $\mid f(x) - f(c) \mid < \epsilon_0$.

• We want to show that $\forall \epsilon > 0$ $\exists \delta > 0$ such that if $x \in A$ and $\mid x - c \mid < \delta$ then $\mid kf(x) - kf(c) \mid < \epsilon$.

• Let $\delta = \delta_0$ and so:

(2)

\begin{align} \quad \mid kf(x) - kf(c) \mid = \mid k(f(x) - f(c)) \mid = \mid k \mid \mid f(x) - f(c) \mid < \mid k \mid \epsilon_1 = \mid k \mid \frac{\epsilon}{\mid k \mid} = \epsilon \end{align}

• Therefore $kf$ is continuous at $c$. $\blacksquare$

Theorems 3, 4, 5, and 6 below deal with the product and quotient of two continuous functions and the absolutely value and square root of a continuous function. All of these results we've seen with regards to sequences and limits already, and their proofs are similar.

Theorem 3: Let $f : A \to \mathbb{R}$ and $g : A \to \mathbb{R}$ be continuous functions at $c \in A$. Then the function $fg$ is continuous at $c \in A$.

Theorem 4: Let $f : A \to \mathbb{R}$ and $g : A \to \mathbb{R}$ be continuous functions at $c \in A$. Then the function $\frac{f}{g}$ is continuous at $c \in A$ provided $g(c) \neq 0$.

Theorem 5: Let $f : A \to \mathbb{R}$ be a continuous function at $c \in A$. Then the function $\mid f \mid$ is continuous at $c \in A$.

• Proof: Let $f$ be a continuous function at $c \in A$.

• Since $f$ is continuous at $c$ then $\forall \epsilon > 0$ $\exists \delta > 0$ such that if $x \in A$ and $\mid x - c \mid < \delta$ then $\mid f(x) - f(c) \mid < \epsilon$.

• We want to show that $\mid f \mid$ is also continuous at $c$, that is $\forall \epsilon > 0$ $\exists \delta > 0$ such that if $x \in A$ and $\mid x - c \mid < \delta$ then $\mid \mid f(x) \mid - \mid f(c) \mid \mid < \epsilon$. Now notice from one of the corollaries to The Triangle Inequality that:

(3)

\begin{align} \quad \mid \mid f(x) \mid - \mid f(c) \mid \mid ≤ \mid f(x) - f(c) \mid < \epsilon \end{align}

• Therefore $\mid f \mid$ is continuous at $c$. $\blacksquare$

Theorem 6: Let $f : A \to \mathbb{R}$ be a continuous function at $c \in A$ such that $f(x) ≥ 0$ at $c \in A$. Then the function $\sqrt{f}$ is continuous at $c \in A$.

We should note that theorems 1-6 can be generalized such that if $f$ and $g$ are continuous on all of $A$ or even all of $B \subseteq A$, then all of the previous results hold.