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The family $\sin (n x) \sin (n y)$ is uniformly bounded but not equicontinuous

Example

Let $K = [0, π]^{2}$ with the Euclidean metric. For every natural $n \geq 1$ , define

$f_{n} : K ⟶ R, f_{n} (x, y) = \sin (n x) \sin (n y) .$

Every $f_{n}$ is continuous, and the family $F = {f_{n} : n \geq 1}$ is uniformly bounded by $1$ . It is not equicontinuous at the origin, and hence is not equicontinuous on $K$ .

Facts & Assumptions

Given: The compact square $K = [0, π]^{2}$ and the family $F$ in the Example.

[L1]

For a nonempty compact metric space $K$ , a family $F \subseteq C (K, R)$ is equicontinuous at $a$ when every $ε > 0$ admits one $δ > 0$ that works for every $f \in F$ and every $x \in K$ ; it is uniformly bounded when one $M \geq 0$ bounds $∣ f (x) ∣$ for all $f$ and $x$ (Equicontinuity, pointwise boundedness, and uniform boundedness for families in $C (K, R)$ ).

[L2]

For every real $t$ , $\sin (- t) = - \sin t$ and $∣ \sin t ∣ \leq 1$ (Parity and the Pythagorean identity for sine and cosine).

[L3]

For all reals $u, v$ , $∣ \sin u - \sin v ∣ \leq ∣ u - v ∣$ (Sine and cosine are $1$ -Lipschitz on $R$ ).

[L4]

$\sin (π / 2) = 1$ (Quarter-turn values and shifts by pi/2 and pi).

[L5]

For every real $ε > 0$ there is a natural $n \geq 1$ with $1 / n < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L6]

The number $π$ is positive (Pi as twice the smallest positive zero of cosine).

[L7]

Every closed box in $R^{2}$ is compact in the Euclidean metric (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

[L8]

For $(u, v) \in R^{2}$ , $∥ (u, v) ∥_{2} = \sqrt{u^{2} + v^{2}}$ (The $p$ -norms $∥ x ∥_{p}$ for rational $p \geq 1$ , and $∥ x ∥_{\infty}$ ).

Verification

technique · direct

1.1givenL6L7

The set $K$ is a nonempty closed box in $R^{2}$ , so it is compact.

1.2givenL2L3L8algebra

For fixed $n \geq 1$ and points $(x, y), (u, v) \in K$ , the sine bound and Lipschitz estimate give $∣ f_{n} (x, y) - f_{n} (u, v) ∣ \leq n ∣ x - u ∣ + n ∣ y - v ∣ \leq 2 n ∥ (x - u, y - v) ∥_{2}$ ; hence $f_{n}$ is continuous on $K$ .

1.3givenL1L2algebra

For every $n \geq 1$ and $(x, y) \in K$ , $∣ f_{n} (x, y) ∣ \leq 1$ , so $F$ is uniformly bounded.

1.4givenL2L4L6algebra

For every $n \geq 1$ , $f_{n} (0, 0) = 0$ , while at $p_{n} = (π / (2 n), π / (2 n)) \in K$ one has $f_{n} (p_{n}) = 1$ .

2.1step 1.4L5L6L8choosealgebra

Let $δ > 0$ . Applying [L5] to $δ / π > 0$ gives some natural $n \geq 1$ with $π / n < δ$ , and then $d_{2} (p_{n}, (0, 0)) = π / (\sqrt{2} n) < π / n < δ$ .

3.1step 1.1step 1.2step 1.4step 2.1L1∎

Taking $ε = 1 / 2$ , steps 1.4 and 2.1 show that every $δ > 0$ admits an $f_{n} \in F$ and a point $p_{n} \in K$ within $δ$ of the origin for which $∣ f_{n} (p_{n}) - f_{n} (0, 0) ∣ = 1 > ε$ . Thus the family is not equicontinuous at the origin.

Remarks

Uniform boundedness controls the range of every function in the family. Equicontinuity asks for a common spatial scale, and the oscillation scale $1 / n$ prevents such a scale at the origin.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

56 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. Lebl, Basic Analysis II, sections 8.3 and 11.4 (standard reference, not scraped)
University of Toronto MAT237, section 2.1 Differentiation of real-valued functions (standard reference, not scraped)

The family sin⁡(nx)sin⁡(ny)sin(nx)sin(ny) is uniformly bounded but not equicontinuous