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I have a question understanding the proof of Theorem 5.4 in Stein Complex analysis. The statement of the theorem reads:

Let $F(z, s)$ be defined for $(z, s) \in \Omega \times [0, 1]$ where $\Omega$ is an open set in $\mathbb{C}$. Suppose $F$ satisfies the following properties:

  1. $F(z, s)$ is holomorphic in $z$ for each $s$.
  2. $F$ is continuous on $\Omega \times [0, 1]$ (jointly continuous).

Then the function $f$ defined on $\Omega$ by $$ f(z) = \int ^1 _0 F(z, s) ds $$ is holomorphic.

I will skip over the unnecessary details. The problem I have is on the following step:

Let $D$ to be any disc whose closure is contained in $\Omega$... A continuous function on a compact set is uniformly continuous, so if $ϵ > 0$ there exists $δ > 0$ such that $$ \sup _{z∈D} |F(z, s_1) − F(z, s_2)| < ϵ \quad \text{whenever} \quad |s_1 − s_2| < δ. $$

I do not understand why this inequality holds for all $z \in D$. In my mind, using the uniform continuity in $s$, the choice of $\delta$ will depend on $z$, so the inequality cannot be claimed holding for all $z$. It could be the case that the uniform continuity in the proof refers to that of $(z, s)$ instead of just $s$, but in that case it is unclear how the norm is even defined.

Any help is appreciated.

ムータンーオ
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  • See these postings P1, P2. – Mittens Jun 30 '23 at 17:08
  • Thank you, but from the posts you suggested I cannot really see an argument related to the $\epsilon-\delta$ of uniform continuity. Also, Stein suggested this proof to avoid the exchange of integrals, so I would really appreciate if the argument does not involve LDCT. – ムータンーオ Jun 30 '23 at 17:49
  • I think this question has arisen before. Here is another posting directly related to this Theorem – Mittens Jun 30 '23 at 17:56
  • I also saw that post, but the OP of that post asked a further step in the proof which I am comfortable with. The only part I am confused is the supremum in the $\epsilon - \delta$. – ムータンーオ Jun 30 '23 at 18:05
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    It is using the joint continuity. The set $D\times [0,1]$ is compact, so $F$ is uniformly continuous on the set. The norm on $D\times [0,1]$ doesn't matter since since they will be equivalent. The norm $|(z,s)| = |z| + |s|$ will work. – Trevor Norton Jun 30 '23 at 18:11
  • But even with uniform continuity of $F$, we can only conclude that $|F(z_1, s_1) - F(z_2, s_2)| < \epsilon$ when $|(z_1, s_2) - (z_2, s_2)| < \delta$. The best we can say is the $\epsilon - \delta$ holds in a ball centered at any $z \in D$, which may not cover the whole $D$. Does that mean we need to cover $D$ using such balls then use a finte subcover argument? – ムータンーオ Jun 30 '23 at 18:17
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    Just set $z_1 = z_2 = z$. Then we have $|F(z, s_1) - F(z,s_2)|< \epsilon$ if $|s_1 - s_2|<\delta$, which is exactly what we want. – Trevor Norton Jun 30 '23 at 18:20
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    Thank you. I think I get it now. I must have confused myself going from two variables to one. – ムータンーオ Jun 30 '23 at 18:32

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