Proof.
We set
(15)F(z)=∑i=1p1∥f(z)-a[i]∥;
then
(16)12π∫02πlog+F(reiθ)dθ ≤m(r,0,f(k))+12π∫02πlog+{F(reiθ)∥f(k)(reiθ)∥}dθ.
By [6], we have
(17)12π∫02πlog+F(reiθ)dθ≥∑i=1pm(r,a[i])-log+2qδ.
From (16) and (17), we can get
(18)∑i=1pm(r,a[i],f) ≤m(r,0,f(k))+12π∫02πlog+{F(reiθ)∥f(k)(reiθ)∥}dθ +log+2qδ.
Hence, we can get from the above inequality and Lemma 4 that
(19)∑i=1pm(r,a[i],f)≤m(r,0,f(k))+S(r,f).
It follows from Theorem A that
(20)T(r,f(k))=m(r,0,f(k))+N(r,0,f(k))+V(r,0,f(k))+O(1).
Thus from (19) and (20) we deduce
(21)∑i=1pm(r,a[i],f)≤T(r,f(k))-N(r,0,f(k))-V(r,0,f(k))+S(r,f).
By Theorem A, we have
(22)pT(r,f)≤T(r,f(k))+∑i=1p[N(r,a[i],f)+V(r,a[i],f)]-N(r,0,f(k))-V(r,0,f(k))+S(r,f).
Now it follows from Theorems A and B and Lemma 4 that
(23)qT(r,f(k)) ≤∑j=1q{N(r,b[j],f(k))+V(r,b[j],f(k))} +N(r,0,f(k))+V(r,0,f(k))+N(r,f(k)) -(N(r,0,f(k+1))+2N(r,f(k))-N(r,f(k+1))) +S(r,f(k)) =∑j=1q{N(r,b[j],f(k))+V(r,b[j],f(k))} +N(r,0,f(k))+V(r,0,f(k))+N(r,f(k+1)) -N(r,f(k))+N(r,0,f(k+1))+S(r,f) ≤∑j=1q{N(r,b[j],f(k))+V(r,b[j],f(k))} +N(r,0,f(k))+V(r,0,f(k))+N¯(r,f) -N(r,0,f(k+1))+S(r,f).
It follows from (22) and (23) that
(24)pqT(r,f)≤N¯(r,f)+(q-1){∑i=1pN(r,a[i],f)-N(r,0,f(k))} +{∑j=1q(r,0,f(k+1))∑i=1pN(r,a[i],f) +∑j=1qN(r,b[j],f(k))-N(r,0,f(k+1))} +q∑i=1pV(r,a[i],f)+∑j=1qV(r,b[j],f(k))+S(r,f).
A zero of f-a of order j>k is a zero of f(k+1) of order j-(k+1) and a zero of f(k)-b of order m is a zero of f(k+1) of order m-1. Moreover, zeros of f-a of order >k are zeros of f(k) and so are not zeros of f(k)-b since b≠0. Hence
(25)∑i=1pN(r,a[i],f)+∑j=1qN(r,b[j],f(k))-N(r,0,f(k+1)) ≤∑i=1pNk+1(r,a[i],f)+∑j=1qN¯(r,b[j],f(k)),∑i=1pN(r,a[i],f)-N(r,0,f(k))≤∑i=1pNk(r,a[i],f).
Substituting (25) to (24), we obtain
(26)pqT(r,f)≤N¯(r,f)+(q-1)∑i=1pNk(r,a[i],f)+∑i=1pNk+1(r,a[i],f)+∑j=1qN¯(r,b[j],f(k))+q∑i=1pV(r,a[i],f)+∑j=1qV(r,b[j],f(k))+S(r,f),
since
(27)Nk(r,a[i],f) ≤kN¯(r,a[i],f) ≤kmi+1{miN¯mi(r,a[i],f)+N(r,a[i],f)} ≤kmi+1{miN¯mi(r,a[i],f)+T(r,f)}+O(1),(28)Nk+1(r,a[i],f) ≤(k+1)N¯(r,a[i],f) ≤k+1mi+1{miN¯mi(r,a[i],f)+N(r,a[i],f)} ≤k+1mi+1{miN¯mi(r,a[i],f)+T(r,f)}+O(1).
Similarly, we can get
(29)N¯(r,b[j],f(k)) ≤1nj+1{njN¯nj(r,b[j],f(k))+T(r,f(k))}+O(1),N¯(r,f)≤1l+1{lN¯l(r,f)+T(r,f)}.
By Lemma 4, we can get
(30)T(r,f(k))=m(r,f(k))+N(r,f(k))≤m(r,f)+N(r,f(k)) +12π∫02πlog+∥f(k)(reiθ)∥∥f(reiθ)∥dθ≤m(r,f)+N(r,f)+kN¯(r,f)+S(r,f)≤T(r,f)+kN¯(r,f)+S(r,f).
Substituting (27)–(30) into (26), we obtain
(31)pqT(r,f) ≤N¯(r,f)+(q-1) ×∑i=1pkmi+1{miN¯mi(r,a[i],f)+T(r,f)} +∑i=1pk+1mi+1{miN¯mi(r,a[i],f)+T(r,f)} +∑j=1q1nj+1{njN¯nj(r,b[j],f(k))+T(r,f(k))} +q∑i=1pV(r,a[i],f)+∑j=1qV(r,b[j],f(k))+S(r,f) ≤(1+∑j=1qknj+1)N¯(r,f)+(q-1) ×∑i=1pkmimi+1N¯mi(r,a[i],f) +∑i=1pk+1mi+1miN¯mi(r,a[i],f) +∑j=1qnjnj+1N¯nj(r,b[j],f(k)) +(q-1)∑i=1pkmi+1T(r,f) +∑i=1pk+1mi+1T(r,f)+∑j=1q1nj+1njT(r,f) +q∑i=1pV(r,a[i],f)+∑j=1qV(r,b[j],f(k))+S(r,f) ≤(1+∑j=1qknj+1)ll+1N¯l(r,f)+(kq+1) ×∑i=1pmimi+1N¯mi(r,a[i],f) +∑j=1qnjnj+1N¯nj(r,b[j],f(k)) +q∑i=1pV(r,a[i],f)+∑j=1qV(r,b[j],f(k)) +(∑i=1pkq+1mi+1+∑j=1q1nj+1+1l+1(1+k∑j=1q1nj+1)) ×T(r,f)+S(r,f).
Since mi, nj, k, and q are positive integers, it follows from (31) that
(32)pqT(r,f)≤(∑i=1pkq+1mi+1+∑j=1q1nj+1+1l+1(1+k∑j=1q1nj+1)) ×T(r,f) +(kq+1)∑i=1p[N¯mi(r,a[i],f)+V(r,a[i],f)] +∑j=1q[N¯nj(r,b[j],f(k))+V(r,b[j],f(k))] +(1+∑j=1qknj+1)ll+1N¯l(r,f)+S(r,f).
Hence, (13) follows from (32).