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We remark here that the proof presented here connects the eigenvalue problem of L with that of L~x o / : H\ —> H\ between the same space H\. 1 is obvious. 2 Let L : Hi —» H be a linear completely continuous field, and L be a sectorial operator. If L is symmetric, then L has a complete eigenvalue sequence {Afc} C R1, and all eigenvectors {

Let (ai, • • • , am) G Km with |a| 2 = 1 satisfy / m \ Fk+i ^ Q i e i , A \i=i / m = max Fk+i / P u t t i n g ( x i , - - - , x m ) = {ait,--- |x| - 1 ,amt) ±(1 - XX^t2 + Fk+i \ ^^e^A \fc=i . 23) we get (JTaiei,x\tk+1. Without loss of generality, let Fk+i(Y^Lx a^i, Ao) > 0. Then, for A > Ao we take tkx~l = (AAo1 - l)/(fe + l)F fc+1 f f ; aiei, A j . 23). e. C. Clark. 13) from (u, A) = (0,A0). 23) is as follows. We take (ai, • • • , am) £ Rm such that F k+\ / m Y]onei,\ \~T^ \ / / m = minm max Fk+\ I Y]xiei,X K'-cK ieK'-, |a:|=i \ ^ \ I .

3 Let f3j•, (j = 1, • • • ,n) be the eigenvalues of annxn matrix M. Then we can take the eigenvectors {£j £ Rn \ 1 < j < n} of M and eigenvectors {£,• € Rn | 1 < j < n} of M* such that &,l) = sij. The Fredholm-Jordan theorem can be generalized to general linear completely continuous fields. For this purpose, consider a linear completely continuous field L — —A-\- B : H\ —• H. 2, we assume that L has a compact inverse given by L-^Ii-A + B)-1 :H->H, where I : Hi —* H is the inclusion mapping. Therefore, the eigenvalue problem of L can be equivalently written in the following form (L'1 - Pk)m^k = 0 , (3k = X^\ for some m > 1.