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 tWhere I'm not sure about the last equality.Is my solution correct? Any help would be appreciated.

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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  tWhere I&#039;m not sure about the last equality.Is my solution correct? Any help would be appreciated.

Donavan Scott · Accepted Answer

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        )      |    →  0as   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  &amp;gt;  0, and if we assume       |    z  −      z          0            |    &amp;lt;  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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