Uniform continuity — difference from pointwise: Question bank

Before the drills, a one-line refresher of the two players you're comparing (built fully in the parent note):

Recall The one structural difference (open if you need it)

Pointwise: $\forall x_{0} \forall ε \exists δ \forall x$ — the $\exists δ$ comes after the point $x_{0}$ , so $δ = δ (ε, x_{0})$ may peek at the point. Uniform: $\forall ε \exists δ \forall x, y$ — the $\exists δ$ comes before the points, so one $δ = δ (ε)$ must serve everybody. Quantifier order is the entire story.

True or false — justify

The rule for these: a bare "true"/"false" scores nothing. The reason is the answer.

x^{2}

is uniformly continuous on every bounded interval

[0, M]

True — on

[0, M]

the amplifier satisfies

∣ x + y ∣ \leq 2 M

, so

δ = ε / (2 M)

works for all pairs; boundedness of the slope-factor is exactly what saves it.

x^{2}

is uniformly continuous on

[0, \infty)

False — the factor

∣ x + y ∣

∣ x^{2} - y^{2} ∣ = ∣ x + y ∣∣ x - y ∣

is unbounded, so the required gap

ε /∣ x + y ∣ \to 0

and no single

δ > 0

survives.

f

is continuous on all of

R

, then it is uniformly continuous on

R

False — continuity only promises a local

δ

at each point; over an unbounded domain the infimum of those

δ

's can be

0

. Counterexample:

x^{2}

R

f

is uniformly continuous on

A

, then it is (pointwise) continuous on

A

True — uniform continuity is strictly stronger; the single

δ

that works for all pairs in particular works at each fixed point, so it implies pointwise continuity.

Every bounded domain forces continuity to upgrade to uniform continuity.

False — bounded is not enough; you need bounded and closed (compact). On the bounded-but-open

(0, 1)

1/ x

is continuous yet not uniformly continuous.

A uniformly continuous function must have a bounded derivative.

False —

x

[0, \infty)

has

f^{'} (x) = 1/ (2 x) \to \infty

near

0

yet is uniformly continuous via

∣ x - y ∣ \leq ∣ x - y ∣

Lipschitz continuity implies uniform continuity.

True — from

∣ f (x) - f (y) ∣ \leq L ∣ x - y ∣

, choose

δ = ε / L

; the

δ

contains no

x

, which is precisely uniformity.

Uniform continuity implies Lipschitz continuity.

False —

x

is uniformly continuous but not Lipschitz; near

0

its slope is unbounded so no single

L

bounds all difference-quotients.

A continuous function on the compact set

[a, b]

is always uniformly continuous there.

True — this is Heine–Cantor: cover

[a, b]

by local

δ

-balls, extract a finite subcover, take the minimum

δ > 0

The sum of two uniformly continuous functions on

A

is uniformly continuous on

A

True — if

∣ f (x) - f (y) ∣ < ε /2

and

∣ g (x) - g (y) ∣ < ε /2

for

∣ x - y ∣ < δ

(use the smaller of the two

δ

's), the triangle inequality gives

∣ (f + g) (x) - (f + g) (y) ∣ < ε

The product of two uniformly continuous functions on an unbounded domain is uniformly continuous.

False —

f (x) = g (x) = x

are (Lipschitz, hence) uniformly continuous on

R

, but

f g = x^{2}

is not uniformly continuous on

R

; unbounded functions can amplify each other.

A uniformly continuous function sends every Cauchy sequence to a Cauchy sequence.

True — this is a defining strength: given

ε

, its single

δ

turns "terms eventually within

δ

" into "images eventually within

ε

." Merely pointwise-continuous

f

can fail this (e.g.

1/ x

(0, 1)

sin x

is uniformly continuous on

R

True —

∣ sin^{'} ∣ = ∣ cos ∣ \leq 1

, so it's Lipschitz with

L = 1

and

δ = ε

works everywhere despite the domain being unbounded.

Spot the error

Each item is a plausible-looking argument with a hidden flaw. Name the flaw.

(0, 1)

is bounded, so continuity on it automatically gives uniform continuity."

The error is dropping closed: Heine–Cantor needs closed and bounded (compact).

(0, 1)

is open, and

1/ x

on it is continuous but not uniformly continuous.

f^{'}

is unbounded on

[0, \infty)

for

x

, therefore

x

is not uniformly continuous."

The error is treating bounded-derivative as necessary. It is only sufficient (via MVE).

x

is a genuine uniformly-continuous function with unbounded derivative.

"To disprove uniform continuity of

x^{2}

, I fixed

δ

and

x

and showed

∣ x^{2} - y^{2} ∣ > ε

for one

y

inside

δ

; done."

No error in structure — but the flaw students commit is fixing

δ

last: you must let the adversary give any

δ

first, then produce

x, y

. The parent's choice

x = 1/ δ, y = 1/ δ + δ /2

does exactly this.

f (x) = x

and

g (x) = x

are both uniformly continuous on

R

, so

f \cdot g

is too."

The error is assuming products of uniformly-continuous functions stay uniformly continuous. They don't in general on unbounded domains:

x \cdot x = x^{2}

fails.

"I picked

δ = ε /∣ x + y ∣

for

x^{2}

, which depends only on the points, so

x^{2}

is uniformly continuous."

The error: a valid uniform

δ

must depend on

ε

only. A

δ

containing

x, y

is exactly the pointwise (non-uniform) situation; here it even

\to 0

∣ x + y ∣ \to \infty

1/ x

is Lipschitz on

(0, 1)

because it's differentiable there."

Differentiable

\neq =

bounded derivative. Here

f^{'} (x) = - 1/ x^{2} \to - \infty

x \to 0^{+}

, so no single

L

bounds the slopes;

1/ x

is not Lipschitz and not uniformly continuous on

(0, 1)

"The minimum of the infinitely many local

δ

's gives a positive uniform

δ

The error is taking min over an infinite family — the infimum may be

0

. Compactness is what reduces it to a finite subcover, where the minimum is genuinely positive.

Why questions

Why is the swap of

\exists δ

and

\forall x_{0}

the entire difference between the two notions?

Because it changes what

δ

is allowed to see: after the point (

δ (ε, x_{0})

) it can adapt locally; before the points (

δ (ε)

) it is frozen and must survive the worst pair in the domain.

Why does factoring

∣ x^{2} - y^{2} ∣ = ∣ x + y ∣∣ x - y ∣

immediately reveal the failure on

R

It separates the input gap

∣ x - y ∣

from the amplifier

∣ x + y ∣

; when the amplifier is unbounded, no fixed input gap keeps the output below

ε

everywhere.

Why does compactness (not just boundedness) rescue continuity into uniform continuity?

Compactness lets an infinite open cover of local

δ

-balls be reduced to a finite subcover, so the minimum

δ

is a positive number (a Lebesgue number); mere boundedness gives no such finite reduction.

Why is Lipschitz sufficient but not necessary for uniform continuity?

Lipschitz gives the clean

δ = ε / L

, but uniform continuity only needs some modulus tying gaps to

ε

;

x

's modulus

∣ x - y ∣

works even though no linear bound

L ∣ x - y ∣

does.

Why does uniform continuity preserve Cauchy sequences while pointwise continuity may not?

A Cauchy sequence's terms eventually lie within any

δ

; a single global

δ

then forces the images within

ε

. Without a global

δ

(only local ones near a bad boundary point), the images can spread out — e.g.

1/ x

on the Cauchy sequence

1/ n

(0, 1)

Why can a function be uniformly continuous on an unbounded domain like

[0, \infty)

at all?

Because uniformity concerns how fast outputs can separate, not domain size; if the increments are controlled (e.g. bounded slope for

sin x

, or concave flattening for

x

), one

δ

suffices no matter how far out you go.

Edge cases

Is a constant function

f (x) = c

uniformly continuous on any set?

Yes, degenerately —

∣ f (x) - f (y) ∣ = 0 < ε

for every pair, so any

δ

works; it is Lipschitz with

L = 0

f (x) = 1/ x

uniformly continuous on

[1, \infty)

Yes — on

[1, \infty)

we have

∣ f^{'} (x) ∣ = 1/ x^{2} \leq 1

, so it's Lipschitz with

L = 1

; the trouble at

0

has been excluded by starting at

1

x

still uniformly continuous if we include the endpoint

0

Yes —

δ = ε^{2}

from

∣ x - y ∣ \leq ∣ x - y ∣

holds for all

x, y \geq 0

, including pairs involving

0

; the endpoint causes no issue.

Does uniform continuity on

(a, b)

let

f

extend continuously to the closed

[a, b]

Yes — uniform continuity maps the Cauchy sequences approaching

a

and

b

to Cauchy (hence convergent) image sequences, giving well-defined limit values at the endpoints.

What happens to "min of local

δ

's" as an interval approaches non-compactness, say

[0, M]

with

M \to \infty

for

x^{2}

The working

δ = ε / (2 M)

shrinks to

0

; in the limit

[0, \infty)

there is no positive

δ

left — a concrete picture of compactness being lost.

x^{2}

uniformly continuous on the unbounded but "thin" set

{0} \cup [10, 10.5]

Yes — the set is closed and bounded (compact), so Heine–Cantor applies directly regardless of the gap; the amplifier

∣ x + y ∣

is bounded by

21

there.

Connections

Continuity (pointwise) — the weaker notion every trap here contrasts against.
Heine–Cantor Theorem — the compact-domain upgrade tested in several items.
Compactness / Lebesgue number lemma — why "finite subcover, take the min" works.
Lipschitz continuity — the sufficient-not-necessary test.
Mean Value Theorem — bridges bounded derivative to Lipschitz.
Cauchy sequences — the preservation property that separates uniform from pointwise.