*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:

- 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:

- 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:

- 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.