Pointwise Limit Versus Integral Limit for a Density Approximation
An analyst approximates a positive density on [0,1] by the sequence f_n(x)=n x (1-x^2)^n for n=1,2,... . The plan is to treat the pointwise limit as the true density and to estimate total mass by lim_{n→∞} ∫_0^1 f_n(x) dx. A reviewer objects that pointwise convergence is not enough to justify exchanging limit and integral. The analyst notes that each f_n is continuous, nonnegative, and vanishes at both endpoints, so the integrals seem well behaved. The team needs three judgments: the pointwise limit on [0,1], whether convergence is uniform, and whether the limit of the integrals equals the integral of the pointwise limit. A numerical check at n=100 suggests the maximum of f_n is not shrinking, which is a warning sign. No dominated convergence theorem may be invoked without checking its hypothesis.
1. Find the pointwise limit f(x)=lim f_n(x) for x in [0,1]. Justify the endpoint cases.
Read the worked answer
At x=0, f_n(0)=0 for every n, so f(0)=0. For 0<x≤1, let r=1-x^2. Then 0<r<1 and f_n(x)=n x r^n. For fixed x, r^n decays geometrically, and n r^n→0, so f_n(x)→0. Hence the pointwise limit is f(x)=0 on all of [0,1].
2. Is the convergence uniform on [0,1]? Compute or bound sup f_n.
Read the worked answer
Differentiate: f_n'(x)=n(1-x^2)^{n-1}[1-(2n+1)x^2]. The maximum on [0,1] occurs at x=1/√(2n+1). Then sup f_n = f_n(1/√(2n+1)) = n/√(2n+1)·(2n/(2n+1))^n. As n→∞, (2n/(2n+1))^n→e^{-1/2} and n/√(2n+1)√(n/2), so sup f_n√(n/(2e))→∞. Since the sup norm does not go to 0, convergence to 0 is not uniform.
3. Compute lim ∫ f_n and compare with ∫ f. Does the exchange of limit and integral fail?
Read the worked answer
Compute ∫_0^1 f_n(x) dx. Let u=1-x^2, du=-2x dx. Then ∫_0^1 n x(1-x^2)^n dx = n·(1/2)∫_0^1 u^n du = n/[2(n+1)]→1/2. The integral of the pointwise limit is ∫_0^1 0 dx=0. Since 1/2≠0, lim ∫ f_n ≠ ∫ lim f_n. The exchange fails; dominated convergence cannot apply because there is no integrable g with |f_n(x)|≤g(x) for all n and x, as the sup norms grow without bound.