Two-layer iteration sche1nes 661refers to self-adjoint operators, since any degree of the operator A also is a
self-adjoint operator: Am= (Am)*,
Let { A 8 , ~s} be eigenvalues an cl orthononnal eigenfunctions of the op-
erator A:A~s =As ~s, S = 1, 2, ... , N, 0 < A 1 < A 2 <···<AN,
where N is the dimension of the space H and A 1 = mins A 8 = fl, AN
maxs As = f 2 • By definition,111eaning that A~ is one of the eigenvalues of the operator Ak. This serves
as a basis for the representations
and, on the saine grounds, A(P 11 (A)) = P 11 (A(A)).
Thus, the eigenvalues of the polynornia.l Pn(A) are equal to the poly-
nomial Pn(A) of the eigenvalues A = A(A) of the operator A. vVith the
relation (P 11 (A))* = P 11 (A) established, we find that( 21) II Pn(A) II < rnax IPn(x)I.
/1 .:Sx.:S12Because of this, the proble1n of searching for mm II P 11 (A) II can be
Tl )72)··· ~Tn
reduced to the well-known 1ninimax problem for the polynomial Pn ( x) in
question. By interchanging the variables by the rule
(22)the segment [fl>f 2 ] carries into the segment [-1, 1], so that P 11 (x) = P 11 (t),
t E [-1, 1], and P 11 (0) = 1.
With the detailed fonns in mind, a revised statement of the problem
consists of finding a polyno111ia.l with n1inin1al deviations from zero on the
segment [-1, 1] such that 1nax IPn (t) I is n1inimal under the additional
-l<t<l
condition of nonnalization P 11 (t 0 ) = 1, where the point t 0 corresponds to
the point x = 0. Frmn fonnula (22) it follows for x = 0 that
(23) t o-- - f2 +fl
f2 - fl