The algebra of continuous functions.

Theorem 5.2.1   If $ f$ and $ g$ are two continuous functions at $ x_0$ , then $ f+g$ is continuous at $ x_0$ .

\begin{figure}\mbox{\subfigure{\epsfig{file=error.eps, height=1cm}} \qquad
Be careful! The converse is not true!}
\end{figure}
For example, take $ f$ and $ g$ as follows:

\begin{displaymath}\begin{cases}f(x)=-1 \text{ if } x \leq 0  f(x)=1 \text{ if...
...1 \text{ if } x \leq 0  g(x)=-1 \text{ if } x > 0 \end{cases}\end{displaymath}    

Figure 1: The graph of the absolute value function.
\begin{figure}\mbox{\epsfig{file=AbsoluteValue.eps,height=4cm}}\end{figure}

The functions $ f$ and $ g$ are disontinuous at 0, but $ f+g$ is continuous at 0 ($ f+g$ is constant on $ \mathbb{R}$ ).

Theorem 5.2.2   If $ f$ and $ g$ are two continuous functions at $ x_0$ , then $ fg$ is continuous at $ x_0$ .

\begin{figure}\mbox{\subfigure{\epsfig{file=error.eps, height=1cm}} \qquad
Be careful! The converse is not true!}
\end{figure}
For example, take $ f$ and $ g$ as follows:

\begin{displaymath}\begin{cases}f(x)=-1 \text{ if } x \leq 0  f(x)=1 \text{ if...
...1 \text{ if } x \leq 0  g(x)=-1 \text{ if } x > 0 \end{cases}\end{displaymath}    

The functions $ f$ and $ g$ are discontinuous at 0, but $ fg$ is continuous at 0 ($ fg$ is constant on $ \mathbb{R}$ ).

Using the fact that a constant function is continuous on its domain, Theorems 2.1 and 2.2 imply that the set $ \mathcal{C}^0{I}$ of all the continuous functions on the interval $ I$ is a real vector space.

Theorem 5.2.3   If $ f$ and $ g$ are two continuous functions at $ x_0$ , and if $ g(x_0) \neq 0$ , then $ \frac fg$ is continuous at $ x_0$ .

\begin{figure}\mbox{\subfigure{\epsfig{file=error.eps, height=1cm}} \qquad
Be careful! The converse is not true!}
\end{figure}

The same functions as above yield a counter-example to the converse: the function $ f/g$ is constant and is equal to $ -1$ over $ \mathbb{R}$ , thus it is continuous at every point.

Theorem 5.2.4   If $ f$ is continuous at $ x_0$ and if $ g$ is continuous at $ y_0=f(x_0)$ , then $ gof$ is continuous at $ x_0$ .

\begin{figure}\mbox{\subfigure{\epsfig{file=error.eps, height=1cm}} \qquad
Be careful! The converse is not true!}
\end{figure}

For example, take a function $ f$ defined over some open interval $ I$ , with at least one discontinuity at, say, $ x_0$ ; now take a function $ g$ , constant over $ f(I)$ . Then $ g \circ f$ is constant over $ I$ , thus is continuous at every point.

Noah Dana-Picard 2007-12-28