Continuity on an Interval

Continuity Throughout a Domain

In “One-Sided Continuity,” continuity was examined at a particular point, with separate conditions for approaching from the left or the right. We now consider continuity across an entire interval. This means checking the appropriate pointwise condition at every point in the domain. The endpoint cases matter: an endpoint may have domain points approaching from only one side, while an interior point has points approaching from both sides.

Definition (Continuity on an Interval): Let $I\subseteq\mathbb{R}$ be an interval and $f:I\to\mathbb{R}$. We say that $f$ is continuous on $I$ if $f$ is continuous at every point $a\in I$, using the epsilon-delta definition relative to the domain $I$. Thus, for every $a\in I$ and every $\varepsilon>0$, there is a $\delta>0$ such that $$ x\in I,\quad |x-a|<\delta \quad\Longrightarrow\quad |f(x)-f(a)|<\varepsilon. $$ At a left endpoint of a nondegenerate interval, this condition is right-continuity; at a right endpoint, it is left-continuity. At an interior point, it requires continuity from both sides.

The definition only considers inputs in $I$. For instance, if $I=[a,b]$, continuity at $a$ concerns points of $[a,b]$ to the right of $a$; it does not require the function to be defined or continuous to the left of $a$. The Two-Sided Characterization of Continuity from the previous tutorial applies at interior points, and the endpoint characterization there applies at the ends of the interval. If $I$ consists of a single point, every function on $I$ is continuous: there are no other domain points to check, and $|f(a)-f(a)|=0<\varepsilon$.

Continuity Is a Local Property

Although continuity on $I$ asks about every point, the test at any one point only uses inputs sufficiently close to it. This makes continuity local: restricting a continuous function to a smaller part of its domain preserves continuity, and checking continuity on suitable neighborhoods is enough to establish continuity throughout the domain.

Theorem (Locality of Continuity): Let $I\subseteq\mathbb{R}$ and $f:I\to\mathbb{R}$. The function $f$ is continuous on $I$ if and only if, for every $a\in I$, there is an $r_a>0$ such that the restriction of $f$ to $I\cap(a-r_a,a+r_a)$ is continuous at every point of that restricted domain.

Proof. Suppose first that $f$ is continuous on $I$. Fix $a\in I$, and choose any $r_a>0$. Let $J=I\cap(a-r_a,a+r_a)$. If $y\in J$, then $y\in I$, so $f$ is continuous at $y$. The epsilon-delta condition for continuity at $y$, when restricted to inputs $x\in J$, still holds. Therefore the restriction of $f$ to $J$ is continuous at every point of $J$.

Conversely, suppose the stated neighborhood condition holds for every $a\in I$. Fix $a\in I$, and choose $r_a>0$ as in the condition. The point $a$ belongs to $J=I\cap(a-r_a,a+r_a)$, and the restriction of $f$ to $J$ is continuous at $a$. Given $\varepsilon>0$, there is a $\delta>0$ such that $|f(x)-f(a)|<\varepsilon$ whenever $x\in J$ and $|x-a|<\delta$. Replace $\delta$, if necessary, by $\min(\delta,r_a)$. If $x\in I$ and $|x-a|<\delta$, then $x\in J$, so the same inequality holds. Thus $f$ is continuous at $a$. Since $a$ was arbitrary, $f$ is continuous on $I$. $\square$

The point of this result is not that the neighborhoods must all have the same size. They may vary from point to point. Rather, continuity at each point can be checked using a neighborhood around that point, without examining the function everywhere at once. This is useful when a function is given by different formulas on different parts of an interval: away from the points where the formulas change, each formula can be checked locally.

Worked Example: A Polynomial on an Unbounded Interval

Consider $f:(-2,\infty)\to\mathbb{R}$ given by $f(x)=x^3-2x+4$. The Polynomial Continuity Theorem establishes that this polynomial is continuous at every real number. In particular, for any $a\in(-2,\infty)$ and any $\varepsilon>0$, there is a $\delta>0$ such that

|x-a|<\delta \quad\Longrightarrow\quad |f(x)-f(a)|<\varepsilon

for real $x$. The same implication holds when $x$ is restricted to the domain $(-2,\infty)$. Therefore $f$ is continuous on $(-2,\infty)$. The left endpoint $-2$ is not part of this interval, so there is no continuity condition to check at $-2$. There is also no finite right endpoint. The definition asks for continuity at each point actually belonging to the domain.

Joining Two Continuous Pieces

A common use of local continuity is to establish continuity for a piecewise-defined function. If two formulas apply on opposite sides of a point, continuity away from that point can often be checked within each piece. At the joining point, the values and the one-sided behavior must fit together. The following result makes that requirement precise on a closed interval.

Theorem (Pasting at a Shared Endpoint): Let $a<c<b$. Suppose $f_1:[a,c]\to\mathbb{R}$ and $f_2:[c,b]\to\mathbb{R}$ are continuous on their respective intervals and $f_1(c)=f_2(c)$. Define $f:[a,b]\to\mathbb{R}$ by $f(x)=f_1(x)$ for $a\leq x\leq c$, and $f(x)=f_2(x)$ for $c\leq x\leq b$. Then $f$ is continuous on $[a,b]$.

Proof. The condition $f_1(c)=f_2(c)$ ensures that the definition assigns a single value at $c$. Consider first a point $x\in[a,c)$. At $x$, the function $f$ agrees with $f_1$. If $x>a$, choose a positive radius smaller than $c-x$ and $x-a$. Within that radius, every point of $[a,b]$ lies in $[a,c]$, where $f=f_1$; continuity of $f_1$ at $x$ therefore gives continuity of $f$ at $x$. If $x=a$, choose a radius smaller than $c-a$; the relative neighborhood of $a$ in $[a,b]$ then lies in $[a,c]$, and right-continuity of $f_1$ at $a$ gives continuity of $f$ there. The same reasoning for $x\in(c,b]$, using $f_2$ and left-continuity when $x=b$, gives continuity at every such $x$.

It remains to check $c$. The equality $f(c)=f_1(c)=f_2(c)$ and continuity of $f_1$ at the right endpoint $c$ show that $f$ is left-continuous at $c$. Continuity of $f_2$ at the left endpoint $c$, with the same assigned value, shows that $f$ is right-continuous there. Since $c$ is an interior point of $[a,b]$, the Two-Sided Characterization of Continuity gives continuity at $c$. Hence $f$ is continuous at every point of $[a,b]$. $\square$

Worked Example: Joining Two Polynomial Formulas

Define $f:[-2,3]\to\mathbb{R}$ by

f(x)= \begin{cases} x^2, & -2\leq x\leq 1,\\ 2x-1, & 1\leq x\leq 3. \end{cases}

The first formula is a polynomial, so it is continuous on $[-2,1]$; the second is a polynomial, so it is continuous on $[1,3]$. At the shared endpoint, the formulas agree:

1^2=1, \qquad 2(1)-1=2-1=1.

The Pasting at a Shared Endpoint Theorem therefore shows that $f$ is continuous on $[-2,3]$. For example, its values on either side are $f(0)=0^2=0$ and $f(2)=2(2)-1=3$, while both formulas assign $f(1)=1$. Agreement at the joining point is essential: without it, the hypotheses of the theorem would fail, and continuity at that point would need a separate check.

Continuity on an Interval Is Not Uniform Continuity

Continuity on an interval means continuity at each point separately. The radius $\delta$ in the definition may depend on both the point $a$ and the tolerance $\varepsilon$. Uniform continuity is stronger: for each $\varepsilon>0$, one radius must work for every point of the interval. A function may therefore be continuous throughout an interval without being uniformly continuous there.

Worked Example: Continuous but Not Uniformly Continuous

Let $f:(0,\infty)\to\mathbb{R}$ be $f(x)=1/x$. The Continuity of Rational Functions Theorem shows that $f$ is continuous at every $x>0$, because its denominator is nonzero throughout the domain. Thus $f$ is continuous on the interval $(0,\infty)$.

To show that $f$ is not uniformly continuous, take the sequences $x_n=1/n$ and $y_n=1/(n+1)$, for positive integers $n$. Both sequences lie in the domain, and

|x_n-y_n| = \left|\frac{1}{n}-\frac{1}{n+1}\right| = \frac{1}{n(n+1)} \longrightarrow 0.

However,

|f(x_n)-f(y_n)| = \left|n-(n+1)\right| =1

for every $n$. If $f$ were uniformly continuous, inputs whose distance tends to zero would have output differences tending to zero, by the epsilon-delta definition with one $\delta$ valid everywhere. These sequences contradict that requirement. Thus $f$ is continuous on $(0,\infty)$ but not uniformly continuous there. The issue is that points can get arbitrarily close to the excluded endpoint $0$, where the function grows without bound.

What to Check Across an Interval

To establish continuity on an interval, first identify the points where the formula changes and the endpoints of the domain. At ordinary interior points, check continuity using the applicable pointwise theorem or definition. At a domain endpoint, check the one-sided condition pointing into the interval. At each joining point, check that the assigned value agrees with the behavior on both sides; the Pasting at a Shared Endpoint Theorem applies when the pieces are continuous on their respective intervals and agree at their shared endpoint.

A frequent pitfall is to check each formula on its own region and conclude that the entire piecewise function is continuous. That conclusion can fail at a boundary between regions. Continuity on the full interval includes the joining point, where the left-hand and right-hand behavior must both approach the function’s assigned value. A different pitfall is to treat continuity throughout an interval as if it automatically supplied a single $\delta$ for all points. It does not; that stronger conclusion is uniform continuity.

Check Your Understanding

Use the definitions and results in this tutorial to answer the following questions.

What does it mean for a function to be continuous on an interval?
Which one-sided condition is relevant at the left endpoint of a nondegenerate interval?
How does the Locality of Continuity theorem reduce a global continuity question to checks near individual points?
Why must the two formulas agree at the shared endpoint in the Pasting at a Shared Endpoint Theorem?
What do the sequences in the reciprocal-function example show about uniform continuity?
Why is continuity on an interval not, by itself, enough to conclude uniform continuity?

What You'll Learn