Asymptotic distribution of eigenvalues of Hill's equation with ...

6 T. Nagasawa and H. Ohrui3. Proof of Theorem 1.3In this section λ is λ 2n−1 or λ 2n , and put√λ − 2n = d.The functions y i satisfyy i ′′ +(2n) 2 y i = −(4nd + d 2 + Q(x))y i .Taking their initial conditions into account, we havesin 2nx(3.1)y 1 =cos2nx + Ky 1 , y 2 = + Ky 2 ,2nwhere K is a bounded operator from L ∞ (0,π) into itself defined by(Ky)(x) =− 1 ∫ x{sin 2n(x − x 1 )}(4nd + d 2 + Q(x 1 ))y(x 1 ) dx 1 .2n 0Since d = o(1) as n →∞,itiseasytosee‖K‖ L∞ →L ∞ ≦ 1 {π|4nd + d 2 }| + ‖Q‖ L 1 = o(1).2nFor n so large that ‖K‖ L∞ →L ∞‖y 1 ‖ L ∞ ≦The derivative of y 2 isy ′ 2 =cos2nx −< 1, we have11 −‖K‖ L∞ →L ∞ , ‖y 2 ‖ L ∞ ≦∫ x0Therefore the condition Δ(λ) =2isequivalentto∫ π012n(1 −‖K‖ L∞ →L ∞).{cos 2n(x − x 1 )} (4nd + d 2 + Q(x 1 ))y 2 (x 1 ) dx 1 .U n (x)(4nd + d 2 + Q(x)) dx =0,wheresin 2nxU n (x) =2ny 1(x) − (cos 2nx) y 2 (x).Putting (3.1) into this twice, we havesin 2nxU n (x) =2n(Ky 1)(x) − (cos 2nx)(Ky 2 )(x)= − 1 ∫ x{sin 24n 2 2n(x − x 1 ) } (4nd + d 2 + Q(x 1 )) dx 1+0sin 2nx2n(K2 y 1 )(x) − (cos 2nx)(K 2 y 2 )(x).