Let's say we have a rational function f (i.e polynom divided by polynom) , and assume that f has no poles in the upper plane { z ; I m z ≥ 0 }we have to prove that: s u p { | f ( z ) | ; I m z ≥ 0 } = s u p { | f ( z ) | ; I m z = 0 }1. I assume that we are dealing in cases where the suprimum exists.2. every approach I tried, I can't really understand what is the significant of function being rational. I know, for example, that there is finite number of zero's, and finite number of poles, but I can't see how it is helpful.

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Let&#039;s say we have a rational function   f (i.e polynom divided by polynom) , and assume that   f has no poles in the upper plane   {  z  ;  I  m  z  ≥  0  }we have to prove that:  s  u  p  {      |    f  (  z  )      |    ;  I  m  z  ≥  0  }  =  s  u  p  {      |    f  (  z  )      |    ;  I  m  z  =  0  }1. I assume that we are dealing in cases where the suprimum exists.2. every approach I tried, I can&#039;t really understand what is the significant of function being rational. I know, for example, that there is finite number of zero&#039;s, and finite number of poles, but I can&#039;t see how it is helpful.

antantil0 · Accepted Answer

We know that the function is holomorphic on the upper half plane and thus does not have any local maxima on the interior of the upper half plane (maximum principle). Since   f is rational, it&#039;s behavior at infinity is easy to deal with. Let&#039;s consider the cases:First case:   f  (  z  )  →  ∞ as   z  →  ∞. Then the supremeum on the real line and on the upper half plane   H are both   ∞.Next case:   f  (  z  )  →  c  ∈      C   as   z  →  ∞. If       |    c      |    =      sup          R            |    f      |   then it is also       sup          H            |    f      |  . Otherwise,       |    c      |    &amp;lt;      sup          R            |    f      |   and there is a a big ball   B. where       |    f      |    &amp;lt;      sup          R            |    f      |   outside   B. But       B    ¯    ∩  H is compact and so   f achieves is max on the boundary. The max doesn&#039;t occur on the upper semicircle by construction of   B, so it must occur on       R    ∩      B    ¯  . This max is then the max on   H of       |    f      |   because       |    f      |   is smaller than       |    c      |   outside   B. Hence      max    H        |    f      |    =      max          H      ∩              B        ¯                  |    f      |    =      max                  R            ∩              B        ¯                  |    f      |    =      max          R            |    f      |

Let's say we have a rational function f (i.e polynom divided by polynom) , and assume that

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