Continuity, uniform continuity, Lipschitz

In one sentence

Continuity is a promise made separately at each point; uniform continuity is one promise for the whole domain; Lipschitz puts a number on it — and the exam lives in the two gaps.

Why the exam asks it

Because the same three-word chain generates an unlimited supply of true-looking false statements, and because the answer usually turns on the domain rather than the formula. A candidate who has memorised "$\sqrt{x}$ is uniformly continuous" without the interval will get it wrong the moment the paper changes the interval, which it does.

The idea

Fix $f$ on a set $E$.

Continuous at $a$: for every $\varepsilon > 0$ there is a $\delta > 0$ with $|f(x) - f(a)| < \varepsilon$ whenever $|x - a| < \delta$. The $\delta$ may depend on both $\varepsilon$ and $a$.

Uniformly continuous on $E$: for every $\varepsilon > 0$ there is a $\delta > 0$ that works for every pair of points at distance less than $\delta$. One $\delta$, whole domain.

Lipschitz: there is a constant $L$ with $|f(x) - f(y)| \le L|x - y|$ for all $x, y \in E$. Then $\delta = \varepsilon/L$ works everywhere, so Lipschitz is uniform continuity with an explicit rate.

Both implications run one way. Both converses fail, and each failure has a standard witness you should know by name.

$\sqrt{x}$ is uniformly continuous and not Lipschitz

On $[0,1]$, $\sqrt x$ is continuous on a compact set, so it is uniformly continuous — that is Heine–Cantor, and it is the single most useful theorem in this topic. But

so no constant $L$ can bound the ratio. The graph goes vertical at the origin, and Lipschitz is precisely a bound on steepness.

1/x is continuous and not uniformly continuous

On $(0,1)$, $1/x$ is continuous at every point. It is not uniformly continuous: near the origin, two points at distance $\delta$ have images arbitrarily far apart. Heine–Cantor does not apply, because $(0,1)$ is not compact — and that missing hypothesis is the whole reason.

$\sin(x^{2})$ is bounded and continuous and not uniformly continuous

This one is worth attention, because it kills the plausible rescue "surely a bounded continuous function on $\mathbb{R}$ is uniformly continuous".

Take $x_n = \sqrt{2\pi n}$ and $y_n = \sqrt{2\pi n + \pi/2}$. Then $|x_n - y_n| \to 0$, because consecutive square roots crowd together. But $f(x_n) = 0$ and $f(y_n) = 1$ always, so the outputs stay a full unit apart. The domain is unbounded, the oscillation accelerates, and no $\delta$ survives.

Where intuition breaks

"Continuous on a bounded interval, so bounded." $1/x$ on $(0,1)$. The theorem needs compact — closed and bounded. This is the single most common hypothesis dropped in the topic, and it is easy to miss because "bounded interval" sounds like enough.

"Uniformly continuous, so Lipschitz." $\sqrt x$ on $[0,1]$. Tempting because both feel like statements about a uniform rate, and they are — but uniform continuity constrains only that a rate exists, not that it is linear.

"Bounded and continuous on $\mathbb{R}$, so uniformly continuous." $\sin(x^2)$. Boundedness controls the size of the output; uniform continuity controls how fast it can change. Those are different, and an unbounded domain leaves room for the second to fail while the first holds.

"A product of uniformly continuous functions is uniformly continuous." Take $f(x) = g(x) = x$ on $\mathbb{R}$: each is Lipschitz with $L = 1$, and the product $x^2$ is not uniformly continuous. Sums behave; products do not, because the growth of one amplifies the oscillation of the other.

"Continuous and injective, so monotone." This one is true on an interval — and false in general, which is exactly how it is set. On a disconnected domain, $f(x) = x$ on $[0,1]$ and $f(x) = 3 - x$ on $[2,3]$ is continuous and injective and not monotone. A converse assumed in disguise: the theorem carries a connectedness hypothesis that the option quietly removes.

"$f$ is uniformly continuous on $(0,1)$, so it extends continuously to $[0,1]$." Also true — and worth knowing as a true statement, because it appears among false ones. A uniformly continuous function on a bounded interval maps Cauchy sequences to Cauchy sequences, so the limits at the endpoints exist. That is why $\sin(1/x)$, which has no limit at $0$, cannot be uniformly continuous on $(0,1)$.

The exam's angle

Read the domain before the formula. Most options in this topic are the same function on different sets, and the set decides.

1. Is the domain compact? If yes, continuous $\Rightarrow$ uniformly continuous, and the question is nearly over. Heine–Cantor does most of the work in this topic.
2. If not compact, hunt for the failure. Two sequences with $|x_n - y_n| \to 0$ and $|f(x_n) - f(y_n)|$ bounded away from zero. This is the standard disproof and it is worth practising until it is automatic.
3. For Lipschitz, look at the derivative. Bounded $f'$ on an interval gives Lipschitz, by MVT. Unbounded derivative near a point — $\sqrt x$ at $0$ — kills it.
4. Check whether the claim is about a product. Products are where uniform continuity fails most often, and where candidates expect it to survive.

The night before

• Lipschitz $\Rightarrow$ uniformly continuous $\Rightarrow$ continuous. Neither converse holds.
• Heine–Cantor: continuous on a compact set $\Rightarrow$ uniformly continuous. Reach for this first.
• $\sqrt x$ on $[0,1]$: uniformly continuous, not Lipschitz.
• $1/x$ on $(0,1)$: continuous, not uniformly, not bounded. Domain is not compact.
• $\sin(x^2)$ on $\mathbb{R}$: bounded, continuous, not uniformly continuous.
• $x \cdot x$: product of Lipschitz functions that is not uniformly continuous on $\mathbb{R}$.
• Bounded derivative $\Rightarrow$ Lipschitz, via the mean value theorem.
• Uniformly continuous on a bounded interval $\Rightarrow$ extends continuously to the closure.