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Jamiya Weber

Jamiya Weber

Answered question

2022-06-13

Let h ( t ) : [ 0 , 1 ] C be a continuous function. Prove that f : C [ 0 , 1 ] C defined by f ( z ) = 0 1 h ( t ) z t is holomorphic.

My attempt:
lim z 0 f ( z + z 0 ) f ( z 0 ) z = lim z 0 0 1 h ( t ) z + z 0 t 0 1 h ( t ) z 0 t = lim z 0 0 1 ( ( z 0 t ) h ( t ) ( z + z 0 t ) ( z 0 t ) ( z + z 0 t ) h ( t ) ( z 0 t ) ( z + z 0 t ) ) d t z = lim z 0 0 1 z h ( t ) ( z + z 0 t ) ( z 0 t ) d t z = lim z 0 0 1 h ( t ) ( z + z 0 t ) ( z 0 t ) d t = 0 1 h ( t ) ( z 0 t ) 2 d t
Where I'm not sure about the last equality.
Is my solution correct? Any help would be appreciated.

Answer & Explanation

Donavan Scott

Donavan Scott

Beginner2022-06-14Added 22 answers

Let g be a function of two variables, say t in [ 0 , 1 ] and z in some disk about a point z 0 in the complex plane. We say g ( t , z ) g ( t , z 0 ) uniformly as z z 0 if
sup t [ 0 , 1 ] | g ( t , z ) g ( t , z 0 ) | 0
as z z 0 . If this condition holds, we can bring the limit in z inside the integral:
lim z z 0 0 1 g ( t , z ) d t = 0 1 g ( t , z 0 ) d t .
The triangle inequality for integrals implies
| 0 1 g ( t , z ) d t 0 1 g ( t , z 0 ) d t | = | 0 1 [ g ( t , z ) g ( t , z 0 ) ] d t | 0 1 | g ( t , z ) g ( t , z 0 ) | d t sup t [ 0 , 1 ] | g ( t , z ) g ( t , z 0 ) | ,
and the upper bound approaches 0 as z z 0 by hypothesis.

In your situation we can fix z 0 and take
g ( t , z ) = h ( t ) ( z t ) ( z 0 t ) .
If, say, the distance from z 0 to the interval [ 0 , 1 ] is r > 0, and if we assume | z z 0 | < r / 2, then g ( t , z ) g ( t , z 0 ) uniformly in the above sense as z z 0 . (This is not exactly your notation, but I hope expresses the ideas more clearly.)

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