Uniform Continuity

In this lesson you’ll learn the difference between continuity and uniform continuity, why the distinction is about quantifier order, the theorem that continuity on a compact set is automatically uniform, and how Lipschitz conditions fit in.

Continuity on a set $S$:

$$\forall a \in S \; \forall \varepsilon>0 \; \exists \delta>0 \; \forall x \in S: \; |x-a|<\delta \implies |f(x)-f(a)|<\varepsilon$$

Uniform continuity on $S$:

$$\forall \varepsilon>0 \; \exists\delta>0 \; \forall x,y \in S: \; |x-y|<\delta \implies |f(x)-f(y)|<\varepsilon$$

The symbols are nearly identical. The order of the quantifiers is the entire difference.

In ordinary continuity, $\forall a$ comes before $\exists \delta$, so $\delta$ may depend on the point. In uniform continuity, $\exists\delta$ comes first, so one $\delta$ must serve every point at once.

Uniform continuity implies continuity. The converse is where the interest is.

Take $f(x) = \frac1x$ on $(0,1]$. It is continuous at every point of its domain. But for a fixed $\varepsilon$, the required $\delta$ shrinks as the point approaches 0, because the graph steepens without bound. There is no smallest positive $\delta$ that works everywhere.

Formally: for any candidate $\delta$, you can find two points closer than $\delta$ whose outputs differ by more than $\varepsilon$. Near 0 the function changes so fast that any fixed input window produces an arbitrarily large output change.

$f(x) = x^2$ on $\mathbb{R}$ fails for the same reason at the other end. Take $x = n$ and $y = n + \frac{1}{n}$. Then $|x-y| = \frac1n \to 0$, but

$$|f(y)-f(x)| = \left(n+\tfrac1n\right)^2 - n^2 = 2 + \tfrac{1}{n^2} > 2$$

So $\varepsilon = 1$ has no workable $\delta$. Unbounded slope over an unbounded domain is the obstruction, and squaring is a perfectly nice function otherwise.

If $f$ is continuous on a compact set $K$, then $f$ is uniformly continuous on $K$.

For $\mathbb{R}$ this means: continuous on a closed bounded interval implies uniformly continuous there.

Proof. Let $\varepsilon>0$. For each $a \in K$, continuity gives $\delta_a > 0$ with $|f(x)-f(a)| < \frac{\varepsilon}{2}$ whenever $|x-a|<\delta_a$.

Cover $K$ by the intervals $\left(a - \frac{\delta_a}{2}, a + \frac{\delta_a}{2}\right)$. By compactness, finitely many suffice, centred at $a_1,\dots,a_n$. Set

$$\delta = \min\left(\frac{\delta_{a_1}}{2},\dots,\frac{\delta_{a_n}}{2}\right) > 0$$

positive because it is a minimum of finitely many positive numbers.

Now take any $x, y \in K$ with $|x-y|<\delta$. The point $x$ lies in some $\left(a_i - \frac{\delta_{a_i}}{2}, a_i + \frac{\delta_{a_i}}{2}\right)$, and then $|y - a_i| \le |y-x| + |x-a_i| < \delta + \frac{\delta_{a_i}}{2} \le \delta_{a_i}$. So both $x$ and $y$ are within $\delta_{a_i}$ of $a_i$, giving

$$|f(x)-f(y)| \le |f(x)-f(a_i)| + |f(a_i)-f(y)| < \frac{\varepsilon}{2}+\frac{\varepsilon}{2} = \varepsilon$$

∎

The whole proof is the finiteness trick. Infinitely many $\delta_a$ have an infimum that may be 0; finitely many have a positive minimum. Compactness is what makes the collection finite. This is the argument previewed two lessons ago, now cashed in.

$f$ is Lipschitz on $S$ with constant $K$ if

$$|f(x)-f(y)| \le K|x-y| \quad \text{for all } x,y \in S$$

Lipschitz implies uniformly continuous: take $\delta = \frac{\varepsilon}{K}$, which depends only on $\varepsilon$.

The chain of strength:

$$\text{Lipschitz} \implies \text{uniformly continuous} \implies \text{continuous}$$

Both implications are strict. $\sqrt x$ on $[0,1]$ is uniformly continuous (continuous on a compact set) but not Lipschitz, because its slope $\frac{1}{2\sqrt x}$ blows up at 0. And $\frac1x$ on $(0,1]$ is continuous but not uniformly continuous.

A useful sufficient condition: if $f'$ is bounded by $K$ on an interval, then $f$ is Lipschitz with constant $K$, by the mean value theorem. That is usually the easiest way to establish uniform continuity in practice.

Uniform continuity is exactly what you need to make an approximation valid across a whole domain rather than point by point. Two places it is indispensable:

• Riemann integrability. Proving every continuous function on $[a,b]$ is integrable requires making the oscillation on every subinterval small at once, with a single mesh size. That is uniform continuity.
• Extending functions and swapping limits. Uniform continuity lets a function defined on a dense subset extend continuously to the whole space, which is how many constructions in analysis get off the ground.

Example 1: Show $f(x) = 3x+2$ is uniformly continuous on $\mathbb{R}$.

Solution. Given $\varepsilon>0$, take $\delta = \frac{\varepsilon}{3}$. For any $x,y$ with $|x-y|<\delta$,

$$|f(x)-f(y)| = 3|x-y| < 3\delta = \varepsilon$$

∎

$\delta$ depends only on $\varepsilon$. Every linear function is Lipschitz with $K = |m|$, hence uniformly continuous on all of $\mathbb{R}$.

Example 2: Show $f(x)=x^2$ is uniformly continuous on $[0,5]$ but not on $\mathbb{R}$.

Solution. On $[0,5]$. $|x^2-y^2| = |x-y||x+y| \le 10|x-y|$ since $x,y \le 5$. Take $\delta = \frac{\varepsilon}{10}$. ✓ (Lipschitz with $K=10$.)

On $\mathbb{R}$. Take $x_n = n$, $y_n = n+\frac1n$. Then $|x_n-y_n| = \frac1n \to 0$ but

$$|f(y_n)-f(x_n)| = 2 + \frac{1}{n^2} > 2$$

So with $\varepsilon=1$, no $\delta$ works. Not uniformly continuous on $\mathbb{R}$. ∎

Note the same function is uniformly continuous on every bounded interval and fails globally. Uniform continuity is a property of the function together with its domain.

Example 3: Show $\frac1x$ is not uniformly continuous on $(0,1]$.

Solution. Let $\varepsilon=1$ and take any $\delta>0$. Choose $n$ with $\frac1n < \delta$ and set

$$x_n = \frac{1}{n}, \qquad y_n = \frac{1}{n+1}$$

Then $|x_n-y_n| = \frac{1}{n(n+1)} < \frac1n < \delta$, while

$$|f(x_n)-f(y_n)| = |n - (n+1)| = 1 \ge \varepsilon$$

So no $\delta$ works. Not uniformly continuous. ∎

The construction pattern is worth remembering: to disprove uniformity, produce two sequences with $|x_n-y_n|\to0$ and $|f(x_n)-f(y_n)|$ bounded away from 0.

Example 4: $\sqrt x$ is uniformly continuous but not Lipschitz on $[0,1]$.

Solution. Uniformly continuous. Continuous on the compact $[0,1]$, so the theorem applies. ✓

Not Lipschitz. Suppose $|\sqrt x - \sqrt y| \le K|x-y|$ for all $x,y$. Take $y=0$:

$$\sqrt x \le Kx \implies K \ge \frac{1}{\sqrt x}$$

which is unbounded as $x \to 0^+$. No finite $K$ exists. ∎

The middle implication is strict, and this is the standard example.

Example 5: A bounded derivative gives Lipschitz.

Show $\sin x$ is uniformly continuous on $\mathbb{R}$.

Solution. $|\sin'x| = |\cos x| \le 1$, so by the mean value theorem, for any $x<y$ there is $c$ with

$$|\sin y - \sin x| = |\cos c||y-x| \le |y-x|$$

Lipschitz with $K=1$, hence uniformly continuous on $\mathbb{R}$ with $\delta = \varepsilon$. ∎

Contrast $x^2$, whose derivative $2x$ is unbounded on $\mathbb{R}$. A bounded derivative is the practical test.

Example 6: Continuous on an unbounded domain, and still uniform.

Is $f(x) = \frac{1}{1+x^2}$ uniformly continuous on $\mathbb{R}$?

Solution. Yes. Its derivative is $\frac{-2x}{(1+x^2)^2}$, which is bounded on $\mathbb{R}$ (it tends to 0 at both ends and is continuous, so it attains a finite maximum modulus).

Bounded derivative gives Lipschitz gives uniform continuity. ✓

So compactness is sufficient, not necessary. Plenty of functions on unbounded domains are uniformly continuous; they just have to flatten out rather than steepen.

Example 7: Quantifier order made concrete.

For $f(x)=x^2$ on $[0,5]$ and $\varepsilon = 0.1$, compare the $\delta$ needed at $x=0$ and at $x=5$.

Solution. Near 0: $|x^2 - 0| < 0.1$ needs $|x| < 0.316$, so $\delta \approx 0.316$ is fine.

Near 5: $|x^2-25| = |x-5||x+5| \approx 10|x-5|$, so we need $|x-5| < 0.01$.

Different points need very different $\delta$. Uniform continuity does not deny that; it only requires that some single $\delta$ works for all, and here $\delta = 0.01$ does. Uniformity fails only when the required $\delta$ has infimum 0, which cannot happen on a compact set.

Numerical integration and step size. Choosing one step size $h$ for a whole interval assumes a uniform error bound, which requires uniform continuity of the integrand (or of a derivative). Adaptive quadrature exists precisely because the required step varies when uniformity is weak, refining where the function changes fast.

ODE solvers and existence theory. The Picard-Lindelöf theorem requires the right-hand side to be Lipschitz in the state variable. Without it, solutions can fail to be unique - $y' = \sqrt y$ with $y(0)=0$ has more than one solution, and $\sqrt{\ }$ failing Lipschitz at 0 is exactly why.

Machine learning robustness. The Lipschitz constant of a network bounds how much the output can move when the input is perturbed, so it directly bounds vulnerability to adversarial examples. Spectral normalization is a technique for controlling it deliberately.

Signal sampling. Reconstructing a signal from samples with a uniform error guarantee needs uniform continuity or a bandwidth limit. A signal with unbounded rate of change cannot be reconstructed uniformly from any fixed sample rate.

Control system gain. A Lipschitz constant is a worst-case gain: the largest output change per unit input change. Stability margins are computed from exactly this quantity.

Floating-point error propagation. Whether a fixed input precision guarantees a fixed output precision across the whole input range is a uniform continuity question. Functions that steepen, like division near zero, lose precision catastrophically, which is why numerical code special-cases them.