Continuous Extension to a Closed Endpoint

Adding a Value at a Missing Endpoint

The previous tutorial showed how a uniformly continuous function on a dense subset of a compact set can be extended to the whole set. For a single missing endpoint, there is a more direct criterion: the function must already be continuous on its domain, and its values must approach a finite limit at the missing endpoint. Uniform continuity is not required for this endpoint result.

The continuity hypothesis on the original domain is important. A limit at the missing endpoint controls behavior only near that endpoint; it says nothing about a discontinuity at an interior point. We will state and prove the criterion with both requirements explicit, then use examples to distinguish the roles they play.

The Endpoint Extension Criterion

Definition: Let $a<b$, and let $f:(a,b]\to\mathbb{R}$. We say that $f$ has the finite right-hand limit $A$ at $a$ if, for every $\varepsilon>0$, there exists $\delta>0$ such that $$ x\in(a,b],\quad 0<x-a<\delta \quad\Longrightarrow\quad |f(x)-A|<\varepsilon. $$

The point $a$ is not in the domain of $f$, so this definition compares nearby values of $f$ with a proposed number $A$, rather than with $f(a)$. The number $A$, if it exists, is unique. Indeed, if both $A$ and $B$ satisfy the definition, choose points $x\in(a,b]$ sufficiently close to $a$ that $|f(x)-A|<\varepsilon/2$ and $|f(x)-B|<\varepsilon/2$. Then

|A-B|\leq |A-f(x)|+|f(x)-B|<\varepsilon.

Since this holds for every $\varepsilon>0$, $A=B$. The interval contains points arbitrarily close to $a$, so such an $x$ can be chosen.

Theorem (Endpoint Extension Criterion): Let $a<b$, and let $f:(a,b]\to\mathbb{R}$. There exists a continuous function $F:[a,b]\to\mathbb{R}$ satisfying $F(x)=f(x)$ for every $x\in(a,b]$ if and only if $f$ is continuous on $(a,b]$ and has a finite right-hand limit at $a$. If that limit is $A$, the extension is given by $F(a)=A$ and $F(x)=f(x)$ for $x\in(a,b]$.

Proof. First suppose that such a continuous extension $F$ exists. Since $f$ is the restriction of $F$ to $(a,b]$, it is continuous there. Set $A=F(a)$. Given $\varepsilon>0$, continuity of $F$ at $a$ gives $\delta>0$ such that $x\in[a,b]$ and $|x-a|<\delta$ imply $|F(x)-F(a)|<\varepsilon$. For $x\in(a,b]$ with $0<x-a<\delta$, we have $f(x)=F(x)$, and consequently

|f(x)-A|=|F(x)-F(a)|<\varepsilon.

Thus $f$ has the finite right-hand limit $A$ at $a$.

Conversely, suppose that $f$ is continuous on $(a,b]$ and that $\lim_{x\to a^+}f(x)=A$ for some $A\in\mathbb{R}$. Define $F:[a,b]\to\mathbb{R}$ by

F(a)=A,\qquad F(x)=f(x)\quad\text{for }x\in(a,b].

We check continuity at every point of $[a,b]$. At $a$, the definition of the right-hand limit says that for every $\varepsilon>0$, some $\delta>0$ satisfies

x\in[a,b],\quad |x-a|<\delta \quad\Longrightarrow\quad |F(x)-F(a)|<\varepsilon.

For $x=a$, the difference is zero; for $x>a$, the implication is exactly the assumed limit condition. Hence $F$ is continuous at $a$. At every $x\in(a,b]$, continuity follows from the continuity of $f$ on its domain and the equality $F=f$ there. In particular, continuity at $b$ is understood relative to $[a,b]$, just as continuity of $f$ at $b$ is relative to $(a,b]$. Therefore $F$ is continuous on $[a,b]$, as required. $\square$

Theorem (Uniqueness of the Endpoint Extension): If $F,G:[a,b]\to\mathbb{R}$ are continuous and agree with the same function $f:(a,b]\to\mathbb{R}$ on $(a,b]$, then $F=G$ on $[a,b]$.

Proof. The functions already agree at every point of $(a,b]$. It remains to compare their values at $a$. Choose $x_n=a+(b-a)/(n+1)$ for each positive integer $n$. Then $x_n\in(a,b]$ and $x_n\to a$. Since $F$ and $G$ are continuous at $a$, $F(x_n)\to F(a)$ and $G(x_n)\to G(a)$. But $F(x_n)=f(x_n)=G(x_n)$ for every $n$, so uniqueness of limits gives $F(a)=G(a)$. Thus the functions agree throughout $[a,b]$. $\square$

The criterion separates two questions. Continuity on $(a,b]$ ensures that no already-defined point prevents a continuous extension. The finite right-hand limit determines whether a value can be assigned at the missing endpoint so that continuity holds there. If both conditions are satisfied, the endpoint value is forced; there is no choice left in the extension.

Worked Examples

Worked Example: Extending a Square-Root Function

Let $f:(a,b]\to\mathbb{R}$ be defined by $f(x)=\sqrt{x-a}$. This function is continuous on $(a,b]$. For every $x\in(a,b]$,

|f(x)-0|=\sqrt{x-a}.

Given $\varepsilon>0$, choose $\delta=\varepsilon^2$. If $0<x-a<\delta$, then

|f(x)-0|=\sqrt{x-a}<\sqrt{\delta}=\varepsilon.

Thus the right-hand limit at $a$ is $0$, and the criterion gives the continuous extension $F(a)=0$, $F(x)=\sqrt{x-a}$ for $x>a$. Equivalently, $F(x)=\sqrt{x-a}$ on all of $[a,b]$. The endpoint value is not an arbitrary convenient assignment: continuity forces it to be $0$.

Worked Example: An Oscillating Factor with a Vanishing Amplitude

Define $f:(a,b]\to\mathbb{R}$ by

f(x)=(x-a)\sin\!\left(\frac{1}{x-a}\right).

The function is continuous at every $x\in(a,b]$, since $x-a$ is nonzero there and the expression is formed from continuous functions. The oscillating sine factor does not itself settle to a limit near $a$, but its magnitude is at most $1$. Therefore

|f(x)-0| =(x-a)\left|\sin\!\left(\frac{1}{x-a}\right)\right| \leq x-a.

Given $\varepsilon>0$, take $\delta=\varepsilon$. Whenever $0<x-a<\delta$, the displayed inequality yields $|f(x)|<\varepsilon$. Hence $f(x)\to0$ as $x\to a^+$, and the unique continuous extension assigns $F(a)=0$. This example shows that the values of $f$ need not themselves be monotone or visibly settle down; a bound forcing their distance from one number to vanish is enough.

Worked Example: Oscillation Prevents an Endpoint Extension

Now let $f(x)=\sin(1/(x-a))$ on $(a,b]$. For all sufficiently large positive integers $n$, the points

x_n=a+\frac{1}{\pi/2+2\pi n}, \qquad y_n=a+\frac{1}{3\pi/2+2\pi n}

belong to $(a,b]$, because both positive increments tend to $0$. Both sequences approach $a$. Direct substitution gives

f(x_n)=\sin(\pi/2+2\pi n)=1, \qquad f(y_n)=\sin(3\pi/2+2\pi n)=-1.

If $f$ had a finite right-hand limit $A$, the values along both sequences would tend to $A$, by the sequential criterion for limits. But the first sequence of values is constantly $1$, while the second is constantly $-1$. Their limits differ, so no such $A$ exists. The endpoint extension criterion rules out a continuous extension to $[a,b]$.

Worked Example: An Endpoint Limit Cannot Repair an Interior Discontinuity

Choose $c\in(a,b)$ and define $f:(a,b]\to\mathbb{R}$ by $f(c)=1$ and $f(x)=0$ for $x\ne c$. For $0<x-a<c-a$, we have $x\ne c$, so $f(x)=0$. In particular, $f$ has right-hand limit $0$ at $a$.

Nevertheless, $f$ is not continuous at $c$. For example, the points $z_n=c+(b-c)/(n+1)$ lie in $(a,b]$, are different from $c$, and tend to $c$. Thus $f(z_n)=0$ for every $n$, whereas $f(c)=1$. If a continuous $F:[a,b]\to\mathbb{R}$ agreed with $f$ on $(a,b]$, then it would be discontinuous at $c$, which is impossible. The finite endpoint limit is present, but the required continuity of $f$ on its domain is not.

Using the Criterion Carefully

A reliable way to test for a continuous extension is to check the two conditions separately. First, verify continuity at every point where the function is already defined, including the included endpoint $b$. Second, determine whether the function has a finite right-hand limit at the missing endpoint $a$. If both checks succeed, set the new endpoint value equal to that limit. If either fails, a continuous extension agreeing with the original function cannot exist.

Check the existing domain.
Confirm that $f$ is continuous on $(a,b]$; a limit at $a$ cannot fix a discontinuity at an interior point.

Test the endpoint behavior.
Find a finite number $A$ such that values of $f(x)$ approach $A$ as $x$ approaches $a$ from the right.

Assign the forced value.
If both conditions hold, define $F(a)=A$ and retain $F(x)=f(x)$ for $x>a$.

This result is local to the missing endpoint in an important sense: the limit condition controls continuity at $a$, while the original continuity assumption handles every other point. It is also distinct from the Uniform Extension to a Compact Closure theorem. That theorem starts with uniform continuity on a dense set and constructs values throughout the closure; here, ordinary continuity on the half-open interval plus one finite endpoint limit is enough to fill in the single missing endpoint.

The same reasoning applies at a missing right endpoint. For a function on $[a,b)$, a continuous extension to $[a,b]$ exists exactly when the function is continuous on $[a,b)$ and has a finite left-hand limit at $b$. One can obtain this version by reversing the direction of approach in the definition and proof.

Check Your Understanding

Use the endpoint extension criterion and its proof to answer the following questions.

What two hypotheses on $f:(a,b]\to\mathbb{R}$ are needed for a continuous extension to $[a,b]$?
If a continuous extension exists, how does continuity at $a$ identify the right-hand limit of $f$?
Why does the function $(x-a)\sin(1/(x-a))$ have a finite right-hand limit even though its sine factor oscillates?
How do two sequences approaching $a$ show that $\sin(1/(x-a))$ has no finite right-hand limit?
Why does a finite right-hand limit fail to guarantee a continuous extension when the original function has an interior discontinuity?
Why can there be at most one continuous extension to $[a,b]$ that agrees with $f$ on $(a,b]$?