设 f(z)=z+∑α(?)z^n 为单位园|z|<1内解析且平均单叶,记其族为 M 又设{f(z)/z}~λ=1+sum from n=1 to ∞D_n(λ),λ>0,本文明了:定理一 若 f∈M,λ>0,则sum form k=1 to ∞{(?)D_k(λ)|-|D_(k-1)(λ)||/d_k(λ)}~2≤An,n=2,3...设 f(z)=z+∑α(?)z^n 为单位园|z|<1内解析且平均单叶,记其族为 M 又设{f(z)/z}~λ=1+sum from n=1 to ∞D_n(λ),λ>0,本文明了:定理一 若 f∈M,λ>0,则sum form k=1 to ∞{(?)D_k(λ)|-|D_(k-1)(λ)||/d_k(λ)}~2≤An,n=2,3,…其中 A 为绝对常数.d_k(h)=h(h+1)…(h+k-1)/k!当λ=1/2,f∈s 时为 I.V.Milm 所证明.定理二 若f∈M 并(?)(1-r)~2M(r,f)=a≠0,则(?)1/n sum from k-1 to n{[||D_(?)(λ)|-|D_(k-1)(λ)||]/[d_k(2λ-1)]}=B<∞λ>展开更多
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文摘设 f(z)=z+∑α(?)z^n 为单位园|z|<1内解析且平均单叶,记其族为 M 又设{f(z)/z}~λ=1+sum from n=1 to ∞D_n(λ),λ>0,本文明了:定理一 若 f∈M,λ>0,则sum form k=1 to ∞{(?)D_k(λ)|-|D_(k-1)(λ)||/d_k(λ)}~2≤An,n=2,3,…其中 A 为绝对常数.d_k(h)=h(h+1)…(h+k-1)/k!当λ=1/2,f∈s 时为 I.V.Milm 所证明.定理二 若f∈M 并(?)(1-r)~2M(r,f)=a≠0,则(?)1/n sum from k-1 to n{[||D_(?)(λ)|-|D_(k-1)(λ)||]/[d_k(2λ-1)]}=B<∞λ>
