sol108c4 - (1 As a uniform limit of holomorphic functions...

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(1) As a uniform limit of holomorphic functions is holomorphic, we see that f is holomorphic on compact subsets of U . As holomorphicity is a local property and any point in U is contained in some compact subset of U , we find that f is holomorphic on U . If all f j have a removable singularity at p , we know there exist holomor- phic extensions ˆ f j : D ( p, r ) C . Now we find (where we can interchange limit and integral as convergence is uniform on ∂D ( p, r/ 2)) lim j →∞ ˆ f j ( z ) = lim j →∞ 1 2 πi Z ∂D ( p,r/ 2) f j ( ζ ) ζ - z = 1 2 πi Z ∂D ( p,r/ 2) lim j →∞ f j ( ζ ) ζ - z = 1 2 πi Z ∂D ( p,r/ 2) f ( ζ ) ζ - z for z D ( p, r ). In particular we see that we can define ˆ f , holomorphic on D ( p, r/ 2) by this integral, which is an extension of f in p . Thus f has a removable singularity at p . If all f j have a pole at p we can have all kinds of singularities for f , to wit (choose p = 0 and r = 1 for simplicity): f j = 1 /jz 0 uniformly on compacts of U , while f ( z ) = 0 has a removable singularity; f j = 1 /z 1 /z uniformly on compacts of U , while f ( z ) = 1 /z has a pole at 0; f j = j n =0 1 j ! z - j e 1 /z uniformly on compacts of U , while e 1 /z has an essential singularity at 0; Likewise if all f j have an essential singularity, all kinds of singularities are possible for f : f j = e 1 /z /j 0 uniformly on compacts of U ; f j = 1 /z + e 1 /z /j 1 /z uniformly on compacts of U ; f j = e 1 /z e 1 /z uniformly on compacts of U ; And finally; yes it is possible to have infinitely many f j with a pole, and infinitely many f j with an essential singularity. For example f j = 1 /jz for odd j and f j = e 1 /z /j for even j converges uniformly on compacts of U to f ( z ) = 0. (2) (a) Note that the poles of f ( z ) z/ sinh( z ) 2 are located at z = kπi for k Z .
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