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A(X,q) = YI(r) + y ' (r) is known as the discriminant of Hill's equation and depends on the parameters X and q. It is well known that for all X and q for which (2) A(X, q) = 2 (1) has solutions of period 7r. For all X and q for which (3) A, q) = -2 (1) has solutions of period 27r. For proofs of these facts and other pertinent background material see Hochstadt [1], Magnus and Winkler [2 ]. It is also known that the functions X(q) satisfying (2) or (3) are analytic functions of q for small q, see Hochstadt [3]. If we stipulate that f(t) be an even function of t and X(q) satisfies (2), then in general periodic solutions of (1) will be either even or odd. If even, they can be characterized by the boundary conditions (4) y' (0) = y (i/2) 0 and if odd, by

Let A A be a compact selfadjoint operator acting on a Hilbert space H H . P P denotes a one dimensional projection also acting on H H . It is shown that the eigenvalues of A A and A + t P ( t > 0 ) A + tP(t > 0) interlace on the real axis. A converse of this result is also proved.

According to Fink and St. Mary [2] the proof of (2) given in [1] is incorrect, and therefore the inequality is, as yet, undecided. For n = 2 and P2 0, (2) is known to be correct and also the best possible. This case was first proved by Lyapunov and is generally referred to as Lyapunov's theorem. A recent proof of this result may be found in Hochstadt [3]. In [2] a similar technique is used to prove (2) for n= 2, where P2 iS merely integrable on [a, b]. In fact, one can extract from [2] the inequality

It is shown that regular ordinary differential operators have a semibounded spectrum. The proof requires fewer prerequisites than other proofs found in the literature and also yields estimates on the lower bound of the spectrum.

Let Δ ( λ ) \Delta (\lambda ) be the discriminant of a Hill’s equation with a π \pi -periodic potential q ( x ) q(x) . It is shown that if 2 + Δ ( λ ) 2 + \Delta (\lambda ) has only double zeros then q ( x ) q(x) is necessarily π / 2 \pi /2 -periodic.

THEOREM. Let p (t) 0 0 be a non-negative, integrable, periodic function of period T, and X the smallest eigenvalue of the boundary value problem y" + Xp(t)y = 0, y(O) + y(T) = 0, y'(0) + Y'(T) =0. If TfTp(t)dt_4 then X>1. PROOF. From the boundary conditions it is clear that the solution y cannot be of constant sign. If X is the smallest eigenvalue y must vanish precisely once in the half-open interval (0, T]. Also Y"f 0 for y > 0 Y"t 0 for y 0. Since y" >0 the function has nonpositive curvature