Passing a limit through an integral: the moving peak
While revising sequences of functions, a candidate writes this solution for f_n(x) = n x (1 − x²)^n on [0, 1]: 'For each fixed x in (0, 1], (1 − x²)^n decays exponentially, so f_n(x) → 0; the pointwise limit is the zero function, hence ∫₀¹ f_n(x) dx → 0.' A numerical check contradicts the conclusion: the substitution u = 1 − x² gives ∫₀¹ f_n = n/(2(n + 1)), so the integrals are 0.4545 for n = 10, 0.4950 for n = 100, and they tend to 1/2. The candidate also notices that f_n(1/√(2n+1)) ≈ 4.3 when n = 100, so the functions are not small everywhere on the interval. Two further statements appear on the page: 'the convergence is uniform on [0, 1]' and 'the interchange of limit and integral is licensed by the dominated convergence theorem'. The candidate wants to know which single step is illegitimate, since every pointwise assertion written down is true.
1. Find the pointwise limit and the exact value of lim ∫₀¹ f_n, then identify precisely which inference in the candidate's argument fails and why the numerical values are not a computational error.
Read the worked answer
Pointwise limit: fix x ∈ (0, 1]; then 0 < 1 − x² < 1, so (1 − x²)^n → 0 and f_n(x) = n x (1 − x²)^n → 0 (the power beats the factor n). At x = 0, f_n(0) = 0. Hence f_n → 0 pointwise on [0, 1].
Integrals: with u = 1 − x², du = −2x dx, so ∫₀¹ n x (1 − x²)^n dx = (n/2)∫₀¹ u^n du = n/(2(n + 1)) → 1/2.
The failed inference is 'pointwise convergence implies convergence of the integrals'. Pointwise convergence constrains each fixed x separately and says nothing about the height of f_n near its moving peak x_n = 1/√(2n+1), which drifts towards 0 as n grows. To move a limit inside an integral one needs sup |f_n| → 0 (uniform convergence) or an integrable dominating function; both fail here. The numbers 0.4545 and 0.4950 are correct, so there is no arithmetic slip - the error is the passage to the limit.
2. Decide whether f_n → 0 uniformly on [0, 1], and whether it does on [a, 1] for a fixed a with 0 < a < 1.
Read the worked answer
Differentiate: f_n′(x) = n(1 − x²)^{n−1}[(1 − x²) − 2n x²], which vanishes exactly at x_n = 1/√(2n+1). There f_n(x_n) = n/√(2n+1) · (2n/(2n+1))^n ≈ √(n/2)·e^{−1/2} → ∞.
Since the supremum on [0, 1] tends to infinity, sup|f_n − 0| does not tend to 0 and the convergence is not uniform. (This also explains why the claim 'uniform on [0, 1]' on the candidate's page is false.)
On [a, 1] with a > 0 we have x ≥ a, so 1 − x² ≤ 1 − a² < 1, giving f_n(x) ≤ n(1 − a²)^n → 0 because an exponential decay beats any polynomial factor. Thus f_n → 0 uniformly on [a, 1]. Consistency check: ∫_a^1 f_n → 0 agrees with the limit of the integrals, because essentially all of the mass of f_n lies in the shrinking interval [0, a].
3. Is there an integrable g on (0, 1) with f_n(x) ≤ g(x) for every n and every x? What does the answer say about the invoked dominated convergence theorem?
Read the worked answer
No such g exists. Fix a small x > 0 and pick an integer n with n x² ∈ [1/2, 2]. Then (1 − x²)^n ≥ (1 − x²)^{2/x²} → e^{−2} ≥ c > 0 for a constant c, so f_n(x) = n x (1 − x²)^n ≥ (1/(2x))·c = c/(2x). Any dominating g must therefore satisfy g(x) ≥ c/(2x) for small x, and ∫₀^ε dx/x = ∞, so g cannot be integrable. Equivalently, sup_n f_n(x) is of order 1/(e x) as x ↓ 0 (maximise n x e^{−n x²} over n, at n = 1/x²).
So the dominated convergence theorem is simply unavailable: its hypothesis is violated, not its conclusion. The example is a counterexample to the unchecked interchange of limit and integral, not to the theorem. It also shows why a sequential check at n = 10, 100, 1000 can never settle such a question - the defect sits at a scale that moves with n.