When does a family of continuous functions hide a uniformly convergent subsequence? Uniform boundedness plus equicontinuity lets the diagonal argument extract one. Contrast with xn, where it fails.
Let K be a compact metric space and F⊂C(K) relatively compact in the sup norm. Prove F is equicontinuous using a finite ε/3-net (hint: each net member is uniformly continuous on the compact K).
∣f(x)−f(y)∣≤∥f−gj∥∞+∣gj(x)−gj(y)∣+∥gj−f∥∞<ε
6 members: sup errors 31→41→51→61→71→81=0.125 — a 23-Lipschitz sequence with ∣ak∣≤43 converging uniformly.Show proof
Relatively compact implies totally bounded: for any ε>0 there is a finite ε/3-net {g1,…,gm} of the closure F. That is, every f∈F has some gj with ∥f−gj∥∞<ε/3. The gj need not lie in F — continuity is all we need.
Each gj is continuous on the compact metric space K, hence uniformly continuous (Heine–Cantor): there is δj>0 with ∣gj(x)−gj(y)∣<ε/3 whenever d(x,y)<δj. Set δ=minjδj>0 — this δ does not depend on f.
Now take f∈F and x,y with d(x,y)<δ. Picking gj with ∥f−gj∥∞<ε/3, the triangle inequality gives ∣f(x)−f(y)∣≤∣f(x)−gj(x)∣+∣gj(x)−gj(y)∣+∣gj(y)−f(y)∣<ε/3+ε/3+ε/3=ε. Since δ works for every f∈F, the family F is equicontinuous.
In addition, relative compactness gives uniform boundedness (a bounded set has bounded closure): M=supf∈F∥f∥∞<∞. Conversely, on compact K, pointwise bounded + equicontinuous implies uniformly bounded: cover K by a finite δ-net {y1,…,yr} for the equicontinuity modulus at ε=1; each x is δ-close to some yi, so ∣f(x)∣≤∣f(yi)∣+1≤maxisupf∣f(yi)∣+1.
Let K be a compact metric space, {xk} a countable dense subset of K, and (fn) a sequence in C(K) with supn∥fn∥∞≤M. Construct, by the diagonal argument, a subsequence converging at every xk.
n(1)⊃n(2)⊃⋯(nested index sequences),mj=nj(j)(diagonal)
6 members: amplitudes ak=21+k+2(−1)k oscillate toward 21 through 61,43,103,32,145,85 — the way diagonal extraction fishes out such an oscillating sequence.Show proof
A compact metric space is separable (take the union of finite 1/r-nets). Now look at (fn). For each k, the scalar sequence (fn(xk))n is trapped in [−M,M], so by Bolzano–Weierstrass it has a convergent subsequence.
Build nested index sequences: choose n1(1)<n2(1)<⋯ so that fnj(1)(x1) converges, and inductively choose n(k) as a subsequence of n(k−1) with fnj(k)(xk) convergent. Fixing k, every later stage refines n(k), so convergence at x1,…,xk is preserved.
Set the diagonal mj=nj(j). Fixing k, the tail (mj)j≥k is a subsequence of (nj(k)) (nesting makes mj=nj(j) a strictly increasing pick from the terms of n(k)). Hence fmj(xk) converges. Since k was arbitrary, (fmj) converges at every xk — pointwise convergence on the dense set.
Note: only pointwise boundedness was used here. Upgrading this pointwise convergence to uniform convergence on all of K is the job of equicontinuity, carried out by the ε/3 argument in the next problem.
Let (fn)⊂C(K) be equicontinuous and pointwise bounded (K compact metric), converging pointwise on a countable dense set {xk}. Prove by the ε/3 argument that the convergence is uniform on all of K, and deduce as a corollary that (xn) has no uniformly convergent subsequence on [0,1].
6 members: ∥fak−f∗∥∞=k+21→0 — equicontinuity spreading convergence from the dense set to the whole interval.Show proof
Given ε>0, equicontinuity yields δ>0: d(x,y)<δ implies ∣fn(x)−fn(y)∣<ε/3 for every n. Since K is compact, finitely many points xk1,…,xkr from the dense set {xk} form a δ-net of K.
At the finitely many net points, pointwise convergence becomes "uniform": choose N so large that j,ℓ≥N gives ∣fmj(xki)−fmℓ(xki)∣<ε/3 for every i (a finite intersection of tails).
For arbitrary x∈K, pick xki with d(x,xki)<δ. By the triangle inequality, ∣fmj(x)−fmℓ(x)∣≤∣fmj(x)−fmj(xki)∣+∣fmj(xki)−fmℓ(xki)∣+∣fmℓ(xki)−fmℓ(x)∣<ε/3+ε/3+ε/3=ε. Hence (fmj) is uniformly Cauchy, so it converges uniformly by completeness of C(K).
This completes (⟸): pointwise bounded + equicontinuous implies uniformly bounded (as noted in the first problem), and every sequence has a uniformly convergent subsequence, so by sequential compactness in metric spaces the set is relatively compact.
Corollary: (xn) has no uniformly convergent subsequence on [0,1]. Its pointwise limit is discontinuous — 0 on [0,1), 1 at x=1. If some subsequence converged uniformly, its limit would be continuous (a uniform limit of continuous functions) and would have to equal the pointwise limit — contradiction. So {xn} is not equicontinuous at x=1: as the scene's contrast chart shows, pairwise sup distances between distinct members stay at least 0.47 apart.