A Note on Search Trees

Proof. Proceed by contradiction. Suppose that f has more than one root in (0,∞). Let r1 and r2, with r1 < r2, be two consecutive such roots (note that we can always find two consecutive roots r1 and r2 because f (r) > 0 for any root r). From the hypothesis, we have f (r1) > 0 and f (r2) > 0. Since f ′ is continuous at r1, there exists an 0 < 2 < (r2− r1) such that f (x) > 0 in the interval (r1, r1 + 2). This shows that f is strictly increasing on (r1, r1 + 2), and by the choice of r1 and r2, it follows that f(x) > 0 for all x ∈ (r1, r2). Similarly, since f (r2) > 0, and by continuity of f ′ at r2, there exists 0 < δ < (r2 − r1) such that f (x) > 0 in (r2 − δ, r2 + δ), and hence, f is strictly increasing in the interval (r2 − δ, r2 + δ). Let a ∈ (r2 − δ, r2 + δ), then f(a) < f(r2)= 0. But a ∈ (r1, r2), and f(x) > 0 for all x ∈ (r1, r2), a contradiction. This completes the proof.