When the Limit Breaks

fn(x)=xnf_n(x)=x^n converges at every point of [0,1][0,1], yet sup⁡∣fn−f∣=1↛0\sup|f_n-f|=1\not\to 0: not uniformly. Watch the limit break at x=1x=1.

fn(x)=xn→f(x)={0x<11x=1f_n(x)=x^n\to f(x)=\begin{cases}0&x<1\\1&x=1\end{cases}
sup⁡[0,1]∣fn−f∣=1↛0\sup_{[0,1]}|f_n-f|=1\not\to 0

Part of these learning paths

Examples & Exercises

Find the pointwise limit of fn(x)=xnf_n(x) = x^n on [0,1][0,1]: prove it equals 00 on [0,1)[0,1) and 11 at x=1x = 1.

f(x)={00≤x<11x=1f(x) = \begin{cases} 0 & 0 \leq x < 1 \\ 1 & x = 1 \end{cases}
n=8n = 8 — the curve hugs 00, then shoots up at x=1x = 1.
Show proof
  1. At x=1x = 1, fn(1)=1n=1f_n(1) = 1^n = 1 for every nn, so the limit is 11.
  2. At x=0x = 0 the limit is trivially 00. For 0<x<10 < x < 1, write t=1x>1t = \frac1x > 1 as t=1+ht = 1 + h with h>0h > 0; Bernoulli's inequality gives tn≥1+nht^n \geq 1 + nh. Hence 0≤xn=1tn≤11+nh→00 \leq x^n = \frac{1}{t^n} \leq \frac{1}{1+nh} \to 0, so xn→0x^n \to 0 by the squeeze theorem.
  3. The pointwise limit is therefore 00 on [0,1)[0,1) and 11 at x=1x = 1 — discontinuous at x=1x = 1. A first warning that pointwise limits of continuous functions need not be continuous.

Prove sup⁡[0,1]∣fn−f∣=1\sup_{[0,1]} |f_n - f| = 1 for every nn, and conclude the convergence is not uniform.

sup⁡[0,1]∣fn−f∣=1↛0\sup_{[0,1]} |f_n - f| = 1 \not\to 0
n=8n = 8, ε=0.2\varepsilon = 0.2 — the chart reads ≈0.98\approx 0.98 on its grid; the exact supremum is 11 (never attained).
Show proof
  1. On [0,1)[0,1), ∣fn−f∣=xn|f_n - f| = x^n, and it is 00 at x=1x = 1, so sup⁡[0,1]∣fn−f∣=sup⁡[0,1)xn\sup_{[0,1]} |f_n - f| = \sup_{[0,1)} x^n.
  2. Since xn<1x^n < 1 for x<1x < 1, the supremum is at most 11. But xn→1x^n \to 1 as x→1−x \to 1^{-} (continuity of the power function), so no number below 11 is an upper bound. The supremum is exactly 11.
  3. Uniform convergence would require sup⁡∣fn−f∣→0\sup |f_n - f| \to 0, but here it equals 11 for every nn. Hence (xn)(x^n) does not converge uniformly.

Prove that a uniformly convergent sequence of continuous functions has a continuous limit — the ε/3\varepsilon/3 argument.

∣f(x)−f(x0)∣≤ε3+ε3+ε3=ε|f(x) - f(x_0)| \leq \tfrac{\varepsilon}{3} + \tfrac{\varepsilon}{3} + \tfrac{\varepsilon}{3} = \varepsilon
n=30n = 30 — for xnx^n, no NN works for all xx at once. Uniform convergence is exactly that simultaneity.
Show proof
  1. Let each fnf_n be continuous and fn→ff_n \to f uniformly. Fix x0x_0 and ε>0\varepsilon > 0.
  2. By uniform convergence there is NN with sup⁡∣fN−f∣<ε3\sup |f_N - f| < \frac{\varepsilon}{3}. Fix this NN.
  3. Since fNf_N is continuous at x0x_0, there is δ>0\delta > 0 with ∣fN(x)−fN(x0)∣<ε3|f_N(x) - f_N(x_0)| < \frac{\varepsilon}{3} whenever ∣x−x0∣<δ|x - x_0| < \delta.
  4. For ∣x−x0∣<δ|x - x_0| < \delta, the triangle inequality gives ∣f(x)−f(x0)∣≤∣f(x)−fN(x)∣+∣fN(x)−fN(x0)∣+∣fN(x0)−f(x0)∣<ε3+ε3+ε3=ε|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)| < \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon. So ff is continuous at x0x_0, and x0x_0 was arbitrary.

3. Continuity