Sequences, Series, and Special Functions 589*6. 00
(a) Let f(z) = L CnZn for lzl < R, and let A(r) = maxRef(z), where
n=O JzJ=r
0 < r < R. Prove that for n ;:::: 1 the inequality
lcnlrn :::; max {4A(r), O} - 2 Re f(O)
holds.
(b) Use the preceding inequality to show that if f(z) 1s an entire
function, and if
A(r):::; Crk
for r ;:::: r 0 , where C and k ;:::: 0 are constants, then f(z) is a
polynomial of degree not greater than k.8.14 Hadamard's Three-Circles Theorem
As still another application of the maximum modulus principle, we have
the following theorem due to J. Hadamard [16].Theorem 8.40 Let f be analytic in the ring ri < lzl < ra and continuous
on the boundaries lzl = ri and lzl = ra. Suppose that r1 < r2 < ra and
let Mk= maXJzJ=rk IJ(z)I (k = 1,2,3). Then
Mlog(r3/r1) < Mlog(r3/r2) Mlog(r2/ri) (8.14-1)
2 - i a
where the real logarithms are taken to any suitable base greater than 1.
Proof Let F(z) = zo: f(z), where a is a real constant to be determined.
Then F(z) is in general a multiple-valued function but its branches are
locally analytic in the ring and continuous on the bounding circles. Since
IF(z)I is single-valued on the closed ring ri :::; lzl :::; ra, the maximum of
IF(z)I will be attained on the boundary, that is,
IF(z)I:::; max(rf M1,r~Ma)
and it follows that on lzl = r2 we have
IJ(z)I :::; max(rfr2o: Mi, r~, r2o: Ma)Now choosing a so that rf Mi = rf Ma, i.e.,
log Ma - log Mi log( Ma/ Mi)
O!=-------=-log ra - log ri log( ra / ri)
(8.14-2) gives(8.14-2)