Sequences of Functions

Jiří Lebl · Basic Analysis I, Ch. 7 · CC BY-SA 4.0

Pointwise convergence is not enough: the limit of continuous functions can be discontinuous, and you can't always interchange limits with integrals or derivatives. Uniform convergence is the fix. It preserves continuity, and the Weierstrass M-test detects it for series.

Pointwise vs uniform convergence

A sequence f_n converges pointwise to f if for each x, f_n(x) → f(x). It converges uniformly if sup_x |f_n(x) - f(x)| → 0. Uniform convergence means the entire graph of f_n gets close to f, not just individual points. The classic counterexample: f_n(x) = x^n on [0, 1] converges pointwise to a discontinuous function but not uniformly.

Scheme

; Pointwise but NOT uniform convergence
; f_n(x) = x^n on [0, 1]
; Pointwise limit: f(x) = 0 for x in [0,1), f(1) = 1
; Not uniform: sup |f_n(x) - f(x)| does not -> 0

(define (f-n x n) (expt x n))

(display "Pointwise: f_n(0.5) as n grows:") (newline)
(for-each (lambda (n)
  (display "  f_") (display n) (display "(0.5) = ")
  (display (f-n 0.5 n)) (newline))
  (list 1 5 10 50 100))

(display "Pointwise: f_n(0.99) as n grows:") (newline)
(for-each (lambda (n)
  (display "  f_") (display n) (display "(0.99) = ")
  (display (f-n 0.99 n)) (newline))
  (list 1 10 100 1000))

; sup |f_n(x) - f(x)| on [0,1) is always close to 1
; (take x = (1-1/n)^(1/n) which is close to 1/e)
(display "sup error on [0,1): always near ")
(display (/ 1.0 (exp 1)))
(display " (not -> 0)")

Uniform convergence preserves continuity

If each f_n is continuous and f_n → f uniformly, then f is continuous. This fails for pointwise convergence: the x^n example gives a discontinuous limit. Uniform convergence also allows interchanging limits with integrals: ∫ lim f_n = lim ∫ f_n.

Scheme

; Uniform convergence: f_n(x) = sin(x)/n -> 0 uniformly
; Each f_n is continuous, limit f(x)=0 is continuous
; sup |f_n(x) - 0| = 1/n -> 0

(define (f-n x n) (/ (sin x) n))

(display "sup |f_n(x)| = 1/n:") (newline)
(for-each (lambda (n)
  (display "  n=") (display n)
  (display ": sup = ") (display (/ 1.0 n)) (newline))
  (list 1 5 10 100 1000))

; Interchange of limit and integral works:
; integral of sin(x)/n from 0 to pi = 2/n -> 0
; integral of limit (0) from 0 to pi = 0
(display "Integral of f_n on [0,pi]: 2/n -> 0") (newline)
(display "Integral of limit on [0,pi]: 0") (newline)
(display "They match because convergence is uniform")

Weierstrass M-test

If |f_n(x)| ≤ M_n for all x, and Σ M_n converges, then Σ f_n converges uniformly (and absolutely). The M_n are constants that bound each term independently of x.

Scheme

; Weierstrass M-test
; sum sin(nx)/n^2 converges uniformly on all of R
; |sin(nx)/n^2| <= 1/n^2 = M_n
; sum 1/n^2 = pi^2/6 converges

(define (partial-series x n-terms)
  (let loop ((n 1) (s 0.0))
    (if (> n n-terms) s
      (loop (+ n 1) (+ s (/ (sin (* n x)) (* n n)))))))

(display "sum sin(nx)/n^2 at various x:") (newline)
(for-each (lambda (x)
  (display "  x=") (display x)
  (display ": S_100 = ")
  (display (partial-series x 100)) (newline))
  (list 0.5 1.0 2.0 3.0))

(display "Bound: sum 1/n^2 = pi^2/6 = ")
(display (/ (* 3.141592653589793 3.141592653589793) 6.0)) (newline)
(display "Convergence is uniform because the bound is x-independent")

Notation reference

Symbol	Scheme	Meaning
f_n → f pointwise	f_n(x) → f(x) each x	Pointwise convergence
f_n ⇒ f	sup\|f_n - f\| → 0	Uniform convergence
M_n	bound on \|f_n\|	Weierstrass M-test bound
Σ M_n < ∞	convergent bound series	Implies uniform convergence

Neighbors

Ch. 2: Sequences — sequences of numbers vs sequences of functions
Ch. 3: Series — power series are the main application
Ch. 4: Continuous Functions — uniform convergence preserves continuity
🎰 Probability Ch.9 — CLT as convergence in distribution: a limit theorem for random functions
🤖 ML Ch.7 — Gaussian processes are distributions over functions, where convergence questions arise
🎛 Control Ch.7 — Fourier series: trigonometric function sequences and their convergence behavior

Translation notes

The x^n example is the canonical counterexample in every analysis course. Our numerical evaluation shows pointwise convergence (each x^n → 0 for x < 1), but the lack of uniform convergence is a theoretical fact about the supremum, not something a finite computation proves.

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