Pointwise versus uniform convergence of a quadrature sequence
On [0,1] define f_n(x)=n x (1-x²)^n for n≥1. A numerical analyst uses ∫_0^1 f_n(x) dx as a quadrature weight and observes that the computed values approach 0.5, while pointwise values appear to vanish. A student argues that since f_n(x)→0 for every x, the convergence is uniform and therefore the limit of the integrals must be the integral of the limit, namely 0. The analyst suspects the inference is wrong and wants a rigorous diagnosis: locate the maximum of f_n, decide whether sup_x |f_n(x)| tends to 0, compute the integral exactly, and explain which theorem fails. A second question concerns a possible subsequence on which uniform convergence does hold on the subinterval [δ,1] for fixed δ>0, and why that does not rescue the full interval.
1. Find the pointwise limit of f_n on [0,1] and determine whether the convergence is uniform.
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Pointwise, f_n(0)=0 for all n. For 0<x≤1, write f_n(x)=n x exp(n log(1-x²)); since log(1-x²)<0, the exponential decays faster than n grows, so f_n(x)→0. Thus the pointwise limit is f≡0. For the supremum, differentiate: f_n'(x)=n(1-x²)^{n-1}[1-(2n+1)x²], so the maximum on [0,1] occurs at x_n=1/√(2n+1). The maximum value is M_n=n/√(2n+1) · (2n/(2n+1))^n. Since (2n/(2n+1))^n→e^{-1/2} and n/√(2n+1)~√(n/2)→∞, we get M_n→∞. Hence sup|f_n-0| does not tend to 0, and the convergence is not uniform.
2. Compute ∫_0^1 f_n(x) dx and reconcile the result with the pointwise limit.
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Substitute u=1-x², du=-2x dx. Then ∫_0^1 n x(1-x²)^n dx = n/2 ∫_0^1 u^n du = n/(2(n+1)) → 1/2. Therefore lim ∫ f_n = 1/2, while ∫ lim f_n = ∫ 0 = 0. The interchange of limit and integral fails because the convergence is not uniform. The sequence develops a spike of height about √n and width about 1/√n near 0, so the mass escapes pointwise convergence. Dominated convergence also does not apply, since sup f_n→∞.
3. Explain the behaviour on [δ,1] for fixed δ>0 and why it is consistent with failure on [0,1].
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On [δ,1], 1-x² ≤ 1-δ², so 0≤f_n(x)≤n(1-δ²)^n. The bound is independent of x and tends to 0, so f_n→0 uniformly on [δ,1]. This local uniform convergence is consistent with the full-interval failure: the mass producing the integral 1/2 is concentrated in a boundary layer near 0. Uniform convergence on every compact subinterval does not imply uniform convergence on the whole interval.