For the equation x + q(t)x = 0, let x(t) be a solution with consecutive zeroes at t = a and t = b. A simple inequality is proven that relates not only a and b to the integral of q (t) but also any point c e (a, 6) where \x(t)\ is maximized. As a corollary, it is shown that if the above equation is oscillatory and if q (t) £ L [0, °»), 1 < p < °°, then the distance between consecutive zeroes must become unbounded. Consider the following second order linear differential equation: continuous on

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... e appropriate t interval. Let q u) = max[^(z), 0]. Pertaining to (1), the following theorem of Hartman [3, p. 345] is known. Theorem 1. Let q{t) be real-valued and continuous for a .< t < b. If x{t) is a solution of (1) with two zeroes in [a, b\, then (2) Cb(t-a)(b-t)q+(t)dt>{b-a). J a Since (b -a)2/4 > it -a)(b -t) for t £ {a, b), equation (2) =» that (3) {b-a)2 fh q + (t)dt>(b-a), 4 J a or (4) f%+(^>-. J a b-a Thus Theorem 1 has as a corollary the following condition of Lyapunov. Again, see Hartman [3, p, 345]. Corollary 1. A necessary condition for any solution x(t) of (1) to have two zeroes in [a, b] is that (a (t) dt > 4/(b -a). The lemma that we would like to present is the following. Lemma L Let x{t) be a solution of (1), where x(a) =* x{b) = 0, and Received by the editors July 2, 1974.