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Uniform Continuity & Function Sequences

Useful preparation: Suprema, Epsilon Bounds & Compactness

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Meaning & notation

Uniform convergence sets one accuracy deadline for the whole domain. Uniform continuity similarly chooses one input tolerance that protects output accuracy everywhere.

Uniform continuity: For every $\displaystyle \epsilon > 0$ there exists $\displaystyle \delta > 0$ such that all domain points x,y with $\displaystyle \left|x - y\right| < \delta$ satisfy $\displaystyle \left|f\, \left(x\right) - f\, \left(y\right)\right| < \epsilon$ . The choice of $\displaystyle \delta$ cannot depend on x or y.
Pointwise versus uniform convergence: Pointwise convergence chooses N after both x and $\displaystyle \epsilon$ . Uniform convergence chooses N after $\displaystyle \epsilon$ alone, and that N works for every x and every $\displaystyle n \ge N$ .
Uniform error: The uniform error $\displaystyle E_{n}$ is the extended supremum of $\displaystyle \left|f_{n}\, \left(x\right) - f\, \left(x\right)\right|$ over the entire domain; it may be $\displaystyle +\infty$ when those errors are unbounded. Uniform convergence is equivalent to $\displaystyle E_{n}$ tending to zero.

Conditions before calculation

Continuity on a compact domain implies uniform continuity. Continuity alone on an arbitrary domain does not supply this global guarantee.
A uniform limit of continuous functions on a fixed domain is continuous. On a finite closed interval, uniform convergence of Riemann-integrable functions also permits passage of the limit through the integral.
Uniform convergence alone does not justify differentiating limits. A sufficient condition on a finite closed interval is continuous differentiability, uniform convergence of the derivatives, and convergence of function values at one fixed point.

Why the method works

1. Put the deadline before the location

Uniformity is precisely this order of choices. In the pointwise version, the universal choice of x occurs before the existence of N.

\forall\varepsilon>0\ \exists N\ \forall x\in D\ \forall n\ge N:\ |f_n(x)-f(x)|<\varepsilon

2. Seek a bound without x

An estimate depending only on n gives a single deadline. An estimate still containing x may establish only pointwise convergence unless that dependence is bounded uniformly.

E_n=\sup_{x\in D}|f_n(x)-f(x)|

3. Understand the continuity transfer

Approximate f at two nearby points by one fixed continuous $\displaystyle f_{n}$ . Two approximation errors and one continuity error each below $\displaystyle \frac{\epsilon}{3}$ make the total below $\displaystyle \epsilon$ .

Worked examples

EXAMPLE 1

Show $\displaystyle f\, \left(x\right) = x^{2}$ is uniformly continuous on [−2,2], and explain why the same claim fails on R.

Factor the difference. Within the bounded interval, $\displaystyle \left|x + y\right| \le 4$ .
$|x^2-y^2|\le4|x-y|$
Choose $\displaystyle \delta = \frac{\epsilon}{4}$ . It works for every pair in the interval, independently of their locations.
On R choose $\displaystyle x_{n} = n$ and $\displaystyle y_{n} = n + \frac{1}{n}$ . Their distance tends to zero, but their squared outputs stay more than 2 apart.
$y_n^2-x_n^2=2+1/n^2$

Conclusion: The compact restriction is uniformly continuous; the function on all of R is not.

EXAMPLE 2

For $\displaystyle f_{n}\, \left(x\right) = 2\, x + \frac{3}{n + 1}$ on R, prove uniform convergence and find the first N making every later error below $\displaystyle \frac{1}{10}$ .

Subtract the candidate limit $\displaystyle f\, \left(x\right) = 2\, x$ . The error is independent of x.
$E_n=\frac3{n+1}$
The strict inequality $\displaystyle \frac{3}{n + 1} < \frac{1}{10}$ is equivalent to $\displaystyle n > 29$ , so choose $\displaystyle N = 30$ .
For $\displaystyle n \ge 30$ the denominator only increases. For any smaller tolerance, choosing N sufficiently large repeats the same argument.

Conclusion: Convergence to 2x is uniform on R, and the requested smallest threshold is 30.

Pitfalls & counterexamples

Tempting claim: Continuous functions converging at every point must converge uniformly.

Correction: The deadline may still depend on the point.

Test the claim: On [0,1], $\displaystyle x^{n}$ tends to zero below 1 and to 1 at 1. The limit is discontinuous, so the convergence cannot be uniform.

Check your understanding

Do $\displaystyle f_{n}\, \left(x\right) = \frac{x}{n}$ converge uniformly to zero on [−3,3]? What about on R?

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Compare your reasoning

Yes on [−3,3]; no on R.

The first supremum error is $\displaystyle \frac{3}{n}$ . On R the error is unbounded for each fixed n.

Connect this idea

Uniform bounds explain when approximations preserve continuity or permit integration. Differentiation needs stronger control.

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Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.