Continuity on an interval.

Definition 5.5.1   Let $ f$ be a function defined on an open interval $ I$ . The function $ f$ is continuous on $ I$ if it is continuous at every point of $ I$ .

Example 5.5.2       

Example 5.5.3   Let $ f$ be defined as follows:

\begin{displaymath}\begin{cases}f(x)=x^2+mx, x \leq 1 f(x)=x^3-2x+1, x> 1 \end{cases} ,\; \text{where } m \text{ is a real parameter}.\end{displaymath}    

On the interval $ (-\infty, 1]$ , $ f$ is the restriction of a polynomial function, thus it is continuous. In particular it is left continuous at $ 1$ .

On the interval $ (1,+\infty)$ , $ f$ is the restriction of a polynomial function, thus it is continuous. Let us compute the left limit of $ f$ at 1:

$\displaystyle \underset{x \underset{>}{\rightarrow} 1}{\lim} f(x)= \underset{x \underset{>}{\rightarrow} 1}{\lim} (x^3-2x+1)=0.$    

The function $ f$ is right-continuous at 1 if, and only if $ f(1)=0$ , i.e. if, and only if, $ 1+m=0$ .

In conclusion, $ f$ is continuous over the whole of $ \mathbb{R}$ whenever $ m=-1$ ; otherwise, it is continuous at every point of $ \mathbb{R}-\{ 1 \}$ and has a discontinuity at $ 1$ .

Example 5.5.4   Let $ f: \; x \mapsto x[x]$ . We wish to study the continuity of $ f$ . We have:

$\displaystyle \forall n \in \mathbb{Z}, \; x \in [n,n+1) \Longrightarrow [x]=n \Longrightarrow f(x)=nx.$    

Part of the graph of $ f$ is displayed on Figure 5.

Figure: The graph of $ x \mapsto x[x]$ .
\begin{figure}\centering
\mbox{\epsfig{file=xFloorx.eps,height=4.5cm}}\end{figure}

On each interval of the form $ [n,n+1)$ , the function $ f$ coincides with the restriction of a linear function, thus it is continuous on each interval of this form. In particular $ f$ is right-continuous at every integer $ n$ . Let us compute the left limit of $ f$ at an integer $ n$ :

$\displaystyle \underset{x \underset{<}{\rightarrow} n}{\lim} f(x) =\underset{x \underset{<}{\rightarrow} n}{\lim} f(x) (n-1)x = n(n-1).$    

At the point $ n$ , the value of $ f$ is given by $ f(n)=n[n]=n^2$ . Thus the following results hold:

Proposition 5.5.5   If $ f$ is continuous on the interval $ I$ , then $ f(I)$ is an interval.

Example 5.5.6 (see Figure 5)   Take $ f(x)=x^2$ . Then $ f((-1,2))=[0,4)$ .

Figure 6: The image of an interval by a continuous function.
\begin{figure}\mbox{\epsfig{file=ParabInterv1.eps,height=4cm}}\end{figure}

Proposition 5.5.7   If $ f$ is continuous on the closed interval $ I$ , then $ f(I)$ is a closed interval.

Example 5.5.8 (see Figure parab interv 1)   . Take $ f(x)=x^2$ . Then $ f([-1,2])=[0,4]$ .

The importance of the continuity hypothesis appears with the function whose graph is displayed on Figure 5(b): there is a point of discontinuity at $ 1$ . Take the interval $ I=(1/2,3/2)$ ; then $ g ([1/2,3/2]) = [3/2,5/2] \cup [13/4,4]$ and this is not an interval.

An important corollary of this proposition will come in the next subsection (i.e Thm intermediate value 1 and Thm 6.3).

Noah Dana-Picard 2007-12-28