One-sided continuity.

Definition 5.4.1   The function is left continuous (resp. right) at $x_0$ if it has a left limit (resp. right limit) at $x_0$ and if this limit is equal to $f(x_0)$ .

For example, the integer part function is right continuous but not left continuous at 1 (v.s. figure  fig integer part 2).

Proposition 5.4.2   Let $f$ be a function defined over a neighborhood of the real number $x_0$ . The function $f$ is continuous at $x_0$ if, and only if, the two conditions hold:

1. the function $f$ is left-continuous at $x_0$ ;
2. the function $f$ is right-continuous at $x_0$ .

Example 5.4.3   Take $f(x)=\vert x\vert$ . We have:

(i)

If $x \leq 0$ , then $f(x)=-x$ . It follows that $\underset{x \underset{<}{\rightarrow} 0}{\lim} f(x)=0=f(0)$ ; thus $f$ is left continuous at 0.

(ii)

If $x \geq 0$ , then $f(x)=x$ . It follows that $\underset{x \underset{>}{\rightarrow} 0}{\lim} f(x)=0=f(0)$ ; thus $f$ is right continuous at 0.

Thus, $f$ is continuous at 0; see Figure 4.

Figure 4: The absolute value function.

Example 5.4.4   Let $f$ be given as follows:

$$\begin{cases}f(x)=2x^2-3x+5, x \leq 0 f(x)=\frac {2x+5}{x+1} \end{cases}.$$

Let us compute the one-sided limits of $f$ at 0:

$$\underset{x \underset{<}{\rightarrow} 0}{\lim} f(x) = \underset{x \underset{<}{\rightarrow} 0}{\lim} (2x^2-3x+5)=5$$

$$\underset{x \underset{>}{\rightarrow} 0}{\lim} f(x) = \underset{x \underset{>}{\rightarrow} 0}{\lim} \frac {2x+5}{x+1}=5.$$

As $f(0)=5$ , this shows that $f$ is left-continuous at 0 and right-continuous at 0, thus is continuous at 0.

Example 5.4.5   Consider the "integer part" function, i.e. $f(x)=[x]$ . Let $n \in \mathbb{Z}$ . Then the following holds:

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

As we know that $f(n)=n$ , we conclude that $f$ is right-continuous at $n$ but not left-continuous at $n$ .