Uniform Convergence and Its Consequences

Part 2, Chapter 2: Single-Variable Real Analysis

Learning objectives

  • Define uniform convergence and contrast precisely with pointwise convergence
  • Apply the sup-norm criterion sup⁡x∣fn(x)−f(x)∣→0\sup_x|f_n(x)-f(x)|\to 0
  • State and use the theorems: uniform limit of continuous is continuous; lim⁡∫=∫lim⁡\lim\int=\int\lim under uniform convergence
  • Apply the Weierstrass M-test to verify uniform convergence of series

Uniform convergence is the right notion for analysis. It fixes everything pointwise convergence broke: the limit of continuous functions is continuous, the limit of integrals is the integral of the limit, and (with one extra hypothesis) the limit of derivatives is the derivative of the limit. The price for these clean theorems is a stronger hypothesis, the same NN must work for ALL xx simultaneously. Once you internalise the geometric picture (the entire graph of fnf_nn trapped in an epsilon\epsilon-tube around the graph of ff), uniform convergence becomes intuitive, and the failure modes of pointwise convergence become easy to diagnose.

The definition

A sequence fn\{f_n\}n converges uniformly to ff on SS if for every epsilon>0\epsilon>0 there exists NN (depending only on epsilon\epsilon, NOT on xx) such that for all n>Nn>N and ALL xinSx\in S, ∣fn(x)−f(x)∣<epsilon|f_n(x)-f(x)|<\epsilonn(x)−f(x)∣<epsilon. The single most useful equivalent formulation is the sup-norm criterion: fntoff_n\to fntof uniformly on SS iff supxinS∣fn(x)−f(x)∣to0\sup_{x\in S}|f_n(x)-f(x)|\to 0f(x)∣to0 as ntoinftyn\to\infty.

What uniform convergence buys you

Three landmark theorems hold under uniform convergence: (i) if each fnf_nn is continuous and fntoff_n\to fntof uniformly on a set, then ff is continuous, the epsilontext−delta\epsilon\text{-}\delta proof is a clean triple inequality. (ii) If each fnf_nn is Riemann integrable on [a,b] and fntoff_n\to fntof uniformly, then ff is integrable and intabfntointabf\int_a^b f_n\to\int_a^b fabfntointabf, the limit slides INSIDE the integral. (iii) If each fn′f_n'n′ is continuous, fn′togf_n'\to gn′tog uniformly, and fnf_nn converges at one point, then f=limfnf=\lim f_nn is differentiable and f′=gf'=g.

Example contrasting pointwise vs uniform

For fn(x)=x/nf_n(x)=x/nn(x)=x/n on [0,1]: \sup_{[0,1]}|x/n|=1/n\to 0, so convergence is UNIFORM. For fn(x)=x/nf_n(x)=x/nn(x)=x/n on mathbbR\mathbb{R}: supmathbbR∣x/n∣=+infty\sup_{\mathbb{R}}|x/n|=+\inftymathbbR∣x/n∣=+infty, so convergence is NOT uniform, only pointwise. The example illustrates the dependence on the domain: uniform convergence is a global property, and changing the domain can change the verdict.

Plot fn(x)=xnf_n(x)=x^nn(x)=xn on [0,1] for n=2,5,10,50n=2, 5, 10, 50. The graphs hug the xx-axis on most of [0,1] but jump steeply to 1 near x=1x=1. The sup-norm sup∣fn(x)−f(x)∣\sup|f_n(x)-f(x)|n(x)−f(x)∣ stays near 1/21/2 for every nn (peak located at xn=2−1/nx_n=2^{-1/n}n=2−1/n), so the convergence is NOT uniform on [0,1]. But on [0,1/2]: supxn=(1/2)nto0\sup x^n=(1/2)^n\to 0, so convergence IS uniform there. The grapher visualises the gap shrinking on [0,1/2] and never shrinking on [0,1].

The Weierstrass M-test

The Weierstrass M-test is the standard tool for uniform convergence of series: if ∣fn(x)∣leqMn|f_n(x)|\leq M_nn for all xinSx\in S and sumMn\sum M_nn converges, then sumfn\sum f_nn converges uniformly (and absolutely) on SS. Example: sumnsin(nx)/n2\sum_n \sin(nx)/n^2nsin(nx)/n2 on mathbbR\mathbb{R} has ∣sin(nx)/n2∣leq1/n2|\sin(nx)/n^2|\leq 1/n^2 and sum1/n2=pi2/6\sum 1/n^2=\pi^2/6 converges, so the series converges uniformly, immediately implying the sum is a continuous function of xx.

Where this shows up
  • Series approximation & Taylor / Fourier series: a Taylor series converges uniformly on compact sets inside its radius of convergence, that is why you can differentiate or integrate it term-by-term. The same idea makes Fourier series usable for solving PDEs (heat equation, wave equation) by separation of variables.
  • Numerical computation & error guarantees: when a numerical method approximates a function uniformly, you can quote a SINGLE error bound valid everywhere in the domain. Pointwise convergence would force you to quote a different error at every input, useless for guarantees.
  • Stone-Weierstrass theorem & modern analysis: the Stone-Weierstrass theorem says every continuous function on a compact set can be uniformly approximated by polynomials. This underlies neural-network universal-approximation theorems, polynomial regression error bounds, and the theory of orthogonal polynomials (Legendre, Chebyshev, Hermite).

Pause and think: Why does uniform convergence preserve continuity but pointwise convergence does not? Sketch a triple-inequality argument: ∣f(x)−f(x0)∣leq∣f(x)−fN(x)∣+∣fN(x)−fN(x0)∣+∣fN(x0)−f(x0)∣|f(x)-f(x_0)|\leq|f(x)-f_N(x)|+|f_N(x)-f_N(x_0)|+|f_N(x_0)-f(x_0)|N(x)∣+∣fN(x)−fN(x0)∣+∣fN(x_0)−f(x_0)∣. Which term needs uniformity?

Try it

  • Predict first: does fn(x)=x2/nf_n(x)=x^2/nn(x)=x2/n converge uniformly on [0,5]? On mathbbR\mathbb{R}? Compute \sup_{[0,5]}|x^2/n| and supmathbbR∣x2/n∣\sup_{\mathbb{R}}|x^2/n|mathbbR∣x2/n∣ to decide.
  • Show by the M-test that \sum_{n=1}^\infty \cos(nx)/n^3 converges uniformly on mathbbR\mathbb{R}. Conclude that the sum is a continuous function.
  • Construct a sequence fntoff_n\to fntof uniformly with each fn′f_n'n′ existing but fn′f_n'n′ NOT converging to f′f' uniformly (uniform convergence does NOT in general imply uniform convergence of derivatives). Hint: try fn(x)=sin(nx)/sqrtnf_n(x)=\sin(nx)/\sqrt{n}n(x)=sin(nx)/sqrtn.
  • True or false: if fntoff_n\to fntof uniformly on [a,b] and each fnf_nn is Riemann integrable, then intabfntointabf\int_a^b f_n\to\int_a^b fabfntointabf. Justify briefly using the epsilon(b−a)\epsilon(b-a) bound.

    The series-summer above is ideal for visualising the M-test: pick a series like sumsin(nx)/n2\sum \sin(nx)/n^2 and watch partial sums sN(x)s_N(x)N(x) get within sumn>N1/n2\sum_{n>N}1/n^2n>N1/n2 of the full sum uniformly in xx. The tail bound is what the M-test exploits.

    A trap to watch for

    Uniform convergence does NOT automatically give uniform convergence of derivatives. The example fn(x)=sin(nx)/sqrtnf_n(x)=\sin(nx)/\sqrt nn(x)=sin(nx)/sqrtn shows fnto0f_n\to 0nto0 uniformly (since ∣fn∣leq1/sqrtn|f_n|\leq 1/\sqrt nn∣leq1/sqrtn), but fn′(x)=sqrtncos(nx)f_n'(x)=\sqrt n\cos(nx)n′(x)=sqrtncos(nx) has ∣fn′(0)∣=sqrtntoinfty|f_n'(0)|=\sqrt n\to\inftyn′(0)∣=sqrtntoinfty. To swap lim\lim and d/dxd/dx you need the SEPARATE hypothesis that fn′f_n'n′ converges uniformly (and fnf_nn converges at one point).

    What you now know

    You can apply the sup-norm criterion, distinguish uniform from pointwise convergence by example, use the Weierstrass M-test, and quote the three landmark uniform-convergence theorems (continuity, integration, differentiation). This concludes Garrity ch. 2 on real analysis, the next chapter generalises everything to several variables, where pointwise vs uniform vs L^p convergence becomes even richer.

    Mark section complete →

    References

    • Garrity, T. (2002). All the Mathematics You Missed. Cambridge University Press, ch. 2.
    • Rudin, W. (1976). Principles of Mathematical Analysis (3rd ed.). McGraw-Hill, ch. 7.
    • Abbott, S. (2015). Understanding Analysis (2nd ed.). Springer, ch. 6.
    • Apostol, T. M. (1974). Mathematical Analysis (2nd ed.). Addison-Wesley, ch. 9.
    • Bartle, R. G., Sherbert, D. R. (2011). Introduction to Real Analysis (4th ed.). Wiley, ch. 8.

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