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$local and global uniqueness theorems on fractional integro$

254 SUBHAS S. BHOOSNURMATH, MILIND N. KULKARNI AND VAISHALI PRABHUSuppose, if possibleT(r,f) < T(r,g),then, by (32), we get,T(r,g) ≤ n + 2 T(r,f) + S(r,f),n − 2T(r,g) ≤ T(r,f) + S(r,f),which is a contradiction, henceT(r,g) ≤ T(r,f).Now, using (30) we get,nT(r,f) ≤ 4 − [Θ(∞,f) + Θ(0,g) + Θ(0,f) + Θ(∞,g)] T(r,f) + S(r,f),which gives(n − 4 + ∆)T(r,f) ≤ S(r,f),where ∆ = Θ(∞,f) + Θ(0,g) + Θ(0,f) + Θ(∞,g). Since n ≥ 7, k ≥ 1. There<strong>for</strong>eT(r,f) ≥ S(r,f),which is a contradiction. HenceFG ≡ 1,is not possible.Hence, F ≡ G. We have,(F ∗ ) ′ = FP(z),and henceSimilarly,As F ≡ GF = (F ∗ ) ′P(z) .G = (G∗ ) ′P(z) .(F ∗ ) ′ = (G ∗ ) ′ ,
MEROMORPHIC FUNCTIONS CONCERNING POLYNOMIALS 255and henceF ∗ = G ∗ + c,where c is a constant. And henceT(r,f) = T(r,g) + S(r,f). (34)Suppose c ≠ 0, by Second Fundamental Theorem(T(r,G ∗ 1) (1)) = (n + 1)T(r,g) < N r,G ∗ + N r,G ∗ + N(r,G ∗ ) + S(r,g)(+ C≤ N r, 1 ) ( )1(+N r,g g n +N(r,g)+N r, 1 ) ( )1+N r,−af f n +S(r,f)− a≤ 5T(r,f) + S(r,f),which is a contradiction to (25), there<strong>for</strong>e c = 0, and hencef n+1 { 1f n −F ∗ = G ∗ ,f − afn+1n + 1 = g − agn+1n + 1 ,a }n + 1= g n+1 { 1g n −a }.n + 1Now, leth = f g .If h = 1, then f ≡ g.Suppose h ≠ 1, thenh n+1 { 1f n −h n+1 { 1g n h n −a }n + 1a }n + 1= 1g n −= 1g n −hg n − ahn+1n + 1 = 1g n −1(h − 1) =agn n + 1an + 1 ,an + 1 ,an + 1 ,{h n+1 − 1 } ,=⇒ g n = n + 1a{ } h − 1h n+1 − 1and f n = h nn + 1a{ } h − 1h n+1 ,− 1
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