Proof. We may suppose \(y > 0\) on \((t_1, t_2)\), so \(y'(t_1) > 0\) and \(y'(t_2) < 0\) (they cannot vanish, by uniqueness). Suppose \(z > 0\) on \((t_1, t_2)\); then \(z(t_1), z(t_2) \geq 0\). Let \(W = yz' - y'z\). Then \(W' = yz'' - y''z = (q_1 - q_2)yz \leq 0\) on \((t_1, t_2)\), and \(W' < 0\) on some subinterval, so \(W(t_2) < W(t_1)\). But \(W(t_1) = -y'(t_1)z(t_1) \leq 0\) and \(W(t_2) = -y'(t_2)z(t_2) \geq 0\), so \(W(t_2) \geq W(t_1)\), a contradiction. □

Proof. (i) Apply the argument of the theorem with \(q_1 = q_2\): \(W\) is now constant and non-zero, and the same sign argument shows that \(z\) must vanish between consecutive zeros of \(y\); it cannot vanish twice, or \(y\) would vanish between those zeros by symmetry. (ii) Compare with \(z'' + m^2z = 0\), whose solutions \(\sin m(t - c)\) have zeros at spacing \(\pi /m\): if a solution of \(y'' + qy = 0\) had consecutive zeros more than \(\pi /m\) apart, a translate of \(\sin m(t - c)\) would have two zeros strictly between them, contradicting the theorem with the roles reversed. The bound \(\pi /M\) is similar. □