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One delta serving the whole set; Lipschitz functions, the closed bounded interval theorem, and why the integral will need it.
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By the end of this lesson you will be able to state uniform continuity with its quantifiers in the right order, decide whether a function has it on a given set, deny it correctly by building a pair of points from an arbitrary delta, and use the theorem that continuity on a closed bounded interval is automatically uniform. You will also be able to say which later arguments need the uniform version and which do not.
Continuity at a point, and the extreme value theorem's habit of using Bolzano-Weierstrass on a closed bounded interval. This lesson moves one quantifier and gets a strictly stronger property — and then proves that on such an interval the move is free.
Continuity is uniform on a set when one delta serves every point of it at once. A function is Lipschitz with constant $K$ on a set when the change in its values is always at most $K$ times the change in the input there, which gives uniform continuity immediately.
Continuous on a set $S$: for every $c \in S$ and every $\varepsilon > 0$ there is $\delta > 0$ such that $|f(x) - f(c)| < \varepsilon$ whenever $x \in S$ and $|x - c| < \delta$. The delta may depend on $c$.
Uniformly continuous on $S$: for every $\varepsilon > 0$ there is $\delta > 0$ such that $|f(x) - f(y)| < \varepsilon$ whenever $x, y \in S$ and $|x - y| < \delta$. The delta may not depend on the point.
Write the quantifiers out and the difference is visible: $\forall c\, \forall \varepsilon\, \exists \delta$ against $\forall \varepsilon\, \exists \delta\, \forall c$. Moving an existential quantifier to the left always strengthens a statement, and here the strengthening has content.
Uniform implies continuous, never the reverse in general. $x^2$ on $\mathbb{R}$ and $1/x$ on $(0,1]$ are continuous at every point and uniformly continuous on neither: in each case the delta that works at a point shrinks to nothing as the point moves.
The theorem. A continuous function on a closed bounded interval is uniformly continuous there.
Proof. Suppose not: some $\varepsilon_0$ has no delta, so for each $n$ there are $x_n, y_n$ with $|x_n - y_n| < 1/n$ and $|f(x_n) - f(y_n)| \ge \varepsilon_0$. By Bolzano-Weierstrass a subsequence of $(x_n)$ converges to some $p$ in the interval; the matching $y$ terms converge to $p$ too. Continuity at $p$ sends both image sequences to $f(p)$, so their difference tends to $0$ — contradicting $\varepsilon_0$.
Another way: picture
Slide a rectangle of width $\delta$ and height $\varepsilon$ along the graph. Uniform continuity says one rectangle shape works everywhere: wherever you put it, the graph enters through the left edge and leaves through the right rather than through the top or bottom. Where the graph is arbitrarily steep, no fixed shape survives.
Another way: steps
To decide uniform continuity on a set $S$:
$f$ is Lipschitz on $S$ with constant $K$ when $|f(x) - f(y)| \le K|x - y|$ for all $x, y \in S$. Then $\delta = \varepsilon/K$ works everywhere at once, so Lipschitz implies uniformly continuous — immediately, with no theorem needed.
For a differentiable function, a bound $|f'| \le K$ on an interval gives exactly this, by the mean value theorem of the next unit. That is why $\sin x$ is uniformly continuous on all of $\mathbb{R}$ despite the unbounded domain: its derivative never exceeds $1$ in size.
The converse fails. $\sqrt{x}$ on $[0,1]$ is uniformly continuous, by the theorem, and is not Lipschitz: near $0$ the slope is unbounded. So Lipschitz is a convenient sufficient condition and not the property itself.
| Situation | Continuity enough? | Uniform continuity needed? |
|---|---|---|
| the limit of $f(x_n)$ for a convergent $(x_n)$ | yes | no |
| the image of a Cauchy sequence being Cauchy | no | yes |
| extending $f$ from the rationals to the reals | no | yes |
| one partition fine enough across a whole interval | no | yes |
| the intermediate value theorem | yes | no |
The fourth row is the one the next unit turns on. Showing a continuous function integrable means making the gap between the upper and lower sums small everywhere at once, and a partition fine enough near one point but not another does not do it. Uniform continuity on a closed bounded interval is what makes that argument three lines long.
Producing a delta for each point and calling it uniform. That is ordinary continuity restated. The delta must not depend on the point, and checking that means taking an infimum over all points and asking whether it is positive.
Thinking a bounded function is uniformly continuous. $\sin(1/x)$ on $(0,1]$ is bounded, continuous and not uniformly continuous. Boundedness of the values has nothing to do with it; boundedness of the steepness has everything.
Thinking an unbounded domain rules it out. $\sin x$ and every Lipschitz function on $\mathbb{R}$ are uniformly continuous. The theorem is sufficient, not necessary.
Choosing the points before the delta when denying it. The denial has to defeat every delta, so the points are built from the delta rather than fixed in advance.
Is $f(x) = 3x + 1$ uniformly continuous on $\mathbb{R}$? Compute $|f(x) - f(y)| = 3|x - y|$.
Look for a Lipschitz constant first.
So $K = 3$, and $\delta = \varepsilon/3$ works for every pair of points.
One delta, no point mentioned.
Yes, uniformly continuous — and the unbounded domain was no obstacle.
The theorem was not needed.
Is $f(x) = 1/x$ uniformly continuous on $(0, 1]$? Fix $\varepsilon_0 = 1$ and let $\delta > 0$ be arbitrary.
One epsilon, then an arbitrary delta.
Take $x = 1/n$ and $y = 1/(n+1)$; they are closer than $\delta$ once $n$ is large, since their gap is below $1/n^2$.
The points are built from the delta.
But $|f(x) - f(y)| = 1$ for every $n$. So no delta works, and it is not uniformly continuous — although it is continuous at every point of the domain.
A gap that never closes.
On $[0,1]$ it is, by the theorem: continuous on a closed bounded interval.
Split the domain at a convenient point.
On $[1, \infty)$ the derivative is at most $1/2$, so it is Lipschitz there with constant $1/2$.
The other piece, by the shortcut.
Taking the smaller of the two deltas handles a pair straddling $1$ as well, so it is uniformly continuous on the whole half-line — despite not being Lipschitz near $0$, where the slope is unbounded. Splitting a domain into a compact piece and a piece with bounded slope is the standard way to settle these.
Match each function and set to the reason it is, or is not, uniformly continuous there.
| uniformly continuous, by the theorem about closed bounded intervals | not uniformly continuous: arbitrarily steep far out | not uniformly continuous: arbitrarily steep near a missing end | uniformly continuous, because its slope is bounded | |
|---|---|---|---|---|
| $x^2$ on the closed interval from $0$ to $3$ | ||||
| $x^2$ on the whole real line | ||||
| $1/x$ on the interval from $0$ to $1$, left end excluded | ||||
| $\sin x$ on the whole real line |
On the closed interval from $0$ to $3$, the function $x^2$ satisfies $|x^2 - y^2| \le K|x - y|$. What is the smallest such $K$?
Answer:
Put the steps of the proof that $x^2$ is not uniformly continuous on the whole line into order.
Number the steps in order (write the number in the box):
Here is an argument that $x^2$ is uniformly continuous on the whole line. Mark the one step that does not follow.
This task has no paper form; do it on a device.
Continuity on which kind of set upgrades automatically to uniform continuity?
Build the proof that a uniformly continuous function sends a Cauchy sequence to a Cauchy sequence.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Match each function and set to the reason it is, or is not, uniformly continuous there.
| uniformly continuous, by the theorem about closed bounded intervals | not uniformly continuous: arbitrarily steep far out | not uniformly continuous: arbitrarily steep near a missing end | uniformly continuous, because its slope is bounded | |
|---|---|---|---|---|
| $x^2$ on the closed interval from $0$ to $8$ | ||||
| $x^2$ on the whole real line | ||||
| $1/x$ on the interval from $0$ to $1$, left end excluded | ||||
| $\sin x$ on the whole real line |
You can tell uniform continuity from continuity, find a Lipschitz constant, deny uniform continuity on an unbounded or open domain, and quote the theorem that makes it free on a closed bounded interval. Say in your own words which quantifier moves. Next: the derivative, from the definition.
10. Your turn: is $\sqrt{x}$ uniformly continuous on $[0, \infty)$?, step 3