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The identity is uniformly continuous on $\mathbb{R}$ and its square is not, so uniform continuity is not preserved by products

Statement refuted

Refuted claim: if $f, g : A \to \mathbb{R}$ are uniformly continuous on $A \subseteq \mathbb{R}$ (Uniform continuity of $f : A \to \mathbb{R}$ : one $\delta$ serving every pair of points of $A$ ) then so is their pointwise product $fg$ .

The witness is the simplest possible one: $A := \mathbb{R}$ and $f = g = \mathrm{id}$ , the identity. The identity is uniformly continuous on $\mathbb{R}$ — one may take $\delta := \varepsilon$ — and its product with itself is $x \mapsto x^{2}$ , which is not uniformly continuous on $\mathbb{R}$ ( $x \mapsto x^2$ is continuous on $\mathbb{R}$ and not uniformly continuous, the pairs $k+1$ and $k+1+1/(k+1)$ defeating every $\delta$ ).

The contrast with continuity is the point. Products do preserve continuity, with no hypothesis at all (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function). The reason the proof there does not transfer is visible in the estimate: bounding $|f(x)g(x) - f(x')g(x')|$ requires a bound on the values of $f$ and $g$ near the two points, and for continuity that bound is local, supplied afresh at each point, whereas uniform continuity would need one bound valid on the whole domain. On an unbounded domain no such bound need exist.

Facts & Assumptions

Given: The domain $A := \mathbb{R}$ , the identity $\mathrm{id} : \mathbb{R} \to \mathbb{R}$ , $\mathrm{id}(x) = x$ , and its square $x \mapsto x^{2}$ (Integer powers $a^m$ ).

[L1]

Uniform continuity on $A$ : for every real $\varepsilon > 0$ there is a real $\delta > 0$ such that all $x, x' \in A$ with $|x - x'| < \delta$ satisfy $|h(x) - h(x')| < \varepsilon$ (Uniform continuity of $f : A \to \mathbb{R}$ : one $\delta$ serving every pair of points of $A$ ).

[L2]

$x \mapsto x^{2}$ is continuous on $\mathbb{R}$ and is not uniformly continuous on $\mathbb{R}$ ( $x \mapsto x^2$ is continuous on $\mathbb{R}$ and not uniformly continuous, the pairs $k+1$ and $k+1+1/(k+1)$ defeating every $\delta$ , Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point).

[L3]

The pointwise product of $\mathrm{id}$ with itself is $x \mapsto x \cdot x = x^{2}$ , and products of continuous functions are continuous (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, Integer powers $a^m$ ).

[L4]

Absolute value and order: $|u| \ge 0$ and $|u| = 0$ exactly when $u = 0$ ; the order is total (Basic properties of the absolute value, Ordered field).

Counterexample

technique · direct

1.1

The identity is uniformly continuous on $\mathbb{R}$ : given a real $\varepsilon > 0$ , take $\delta := \varepsilon$ ; then all $x, x' \in \mathbb{R}$ with $|x - x'| < \delta$ satisfy $|\mathrm{id}(x) - \mathrm{id}(x')| = |x - x'| < \varepsilon$ .

L1L4

1.2

The pointwise product of the identity with itself is the function $x \mapsto x^{2}$ .

1.3

By [L2] the function $x \mapsto x^{2}$ is not uniformly continuous on $\mathbb{R}$ .

2.1

So $f := \mathrm{id}$ and $g := \mathrm{id}$ are uniformly continuous on $A = \mathbb{R}$ while $fg$ is not: the refuted claim is false.

step 1.1step 1.2step 1.3L1∎

Remarks

Where the implication survives. On a bounded domain the product of two uniformly continuous functions is uniformly continuous, because each factor is then bounded: a uniformly continuous function on a bounded set is bounded, and the estimate $|fg - f'g'| \le |f|\,|g - g'| + |g'|\,|f - f'|$ closes. Nothing on this page asserts that, and it is not used anywhere here; the witness above shows only that the unrestricted claim fails.
Sums and scalar multiples are safe. If $f$ and $g$ are uniformly continuous on $A$ then so are $f + g$ and $\alpha f$ , by the same $\varepsilon/2$ and $\varepsilon/|\alpha|$ arguments that work for continuity, with the $\delta$ 's inherited uniformly. It is only the product, and the quotient, that need a bound on values.
A second witness on a bounded domain. $x \mapsto 1/x$ is not uniformly continuous on $(0,1)$ ( $x \mapsto 1/x$ is continuous on $(0,1)$ and not uniformly continuous there, the pairs $1/(k+2)$ and $1/(k+3)$ defeating every $\delta$ ) although the identity is; that failure is a quotient rather than a product, and it shows that boundedness of the domain does not rescue every algebraic operation.

Depends on

Used by

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 78 results over 18 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Uniform continuity (Wikipedia) (standard reference, not scraped)
J. Lebl, Basic Analysis I, §3.4 (standard reference, not scraped)
Rutgers Mathematics 300, Fall 2017 examination solutions (standard reference, not scraped)

The identity is uniformly continuous on R\mathbb{R}R and its square is not, so uniform continuity is not preserved by products