Uniform convergence and the limit of an integral
During a revision session on sequences of functions, a student considers f_n(x) = n x(1 − x²)^n on [0,1], n ≥ 1. The student computes the pointwise limit: at x=0, f_n(0)=0; for 0<x≤1, the factor (1−x²)^n decays exponentially while n grows linearly, so f_n(x)→0; at x=1, f_n(1)=0. Hence f_n→0 pointwise. The student then evaluates ∫₀¹ f_n(x) dx = n/[2(n+1)], so the integral tends to 1/2. Since lim ∫ = 1/2 but ∫ lim = 0, the student is puzzled but concludes that the limit-integral interchange must still hold because each f_n is continuous and the pointwise limit is continuous. A classmate claims that continuity plus pointwise convergence is enough. They ask whether the convergence could be uniform despite the integral mismatch, and whether the maximum value of f_n tends to 0.
1. Compute the maximum value of f_n on [0,1] and use it to decide whether f_n→0 uniformly.
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On (0,1), f_n′(x)= n(1−x²)^{n−1}[1−(2n+1)x²]. The only interior critical point is x_n=1/√(2n+1). Since f_n(0)=f_n(1)=0 and f_n>0 inside, this is the maximum. Thus ||f_n||∞=f_n(x_n)= n/√(2n+1) · (2n/(2n+1))^n. As n→∞, n/√(2n+1) ~ √(n/2) and (2n/(2n+1))^n→e^{−1/2}, so ||f_n||∞→∞. A sequence converging uniformly to 0 must have sup norm tending to 0; here it does not. Hence f_n does not converge uniformly to 0 on [0,1], even though the pointwise limit is 0.
2. Explain why the equality lim ∫ f_n = ∫ lim f_n fails here, and identify which hypothesis is missing.
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The equality requires more than pointwise convergence plus continuity. Here the graph develops a spike of height about √(n/(2e)) near x_n=1/√(2n+1), whose width is of order 1/√n. The area under the spike stays 1/2: ∫₀¹ f_n = n/[2(n+1)]→1/2, while the pointwise limit has integral 0. The mass escapes to a moving concentration point near 0. Uniform convergence is missing; dominated convergence also cannot be applied with an integrable dominating function because sup_n f_n(x) grows like 1/(e x) as x→0+, whose integral diverges. So continuity and pointwise convergence alone are insufficient.
3. State a sufficient condition that would justify the interchange and check whether it holds.
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A sufficient condition is uniform convergence on [0,1]: if f_n→f uniformly and each f_n is integrable, then ∫ f_n→∫ f. It fails here because ||f_n||_∞→∞. Another sufficient condition is domination by an integrable g with |f_n(x)|≤g(x); that also fails, since the spike height grows without bound and the envelope near 0 is not integrable. Hence the student's interchange is unjustified.