Let $X$ a Banach space, $f:U\subseteq \mathbb{C}\to X$ an holomorphic function non constant and $N:=\left\{z\in\mathbb{C}:f(z)=0\right\}$. Show that $N$ is closed and discrete.

My idea:

The function $f$ is holomorphic, particulary $f$ is continuos. Then $f^{-1}(\{0\})$ is colsed. Now, suppose that $N$ is not discrete, that is, $N$ has a limit point $\omega\in N$. The function $g=0$ satisfy that $$f(x)=g(x),\hspace{0.2cm}\text{if }x\in N.$$ Finally, $N$ has at least one limit point and cause of the identity principle $$f(x)=g(x)=0\hspace{0.1cm},\text{if }x\in U.$$


  1. Is it right?
  2. Can I use the hypothesis $f$ non zero instead of non constant?


  • 1
    $\begingroup$ 1.Yes, that looks right and 2. yes, if $f$ is not the zero function and constant $N=\emptyset$ which is closed and discrete. $\endgroup$ – Peter Melech Sep 7 '18 at 9:54
  • $\begingroup$ Hint $:$ Since $f$ is non constant the zeros of $f$ has no limit point by identity theorem. $\endgroup$ – Dbchatto67 Sep 7 '18 at 9:55
  • $\begingroup$ 1. Right , 2. Right too. $\endgroup$ – Dbchatto67 Sep 7 '18 at 9:57

Identity Principle is for complex valued functions. Here you have to take $x^{*} \in X^{*}$ (where $X^{*}$ is the dual of $X$) and apply Identity Principle for $x^{*} \circ f$ to show that $x^{*} \circ f (z)=0$ for all $z$ for $x^{*}$ (if $N$ has a limit point). This implies $f(z)=0$ for all $z$.

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  • $\begingroup$ there is an identity principle for Banach valued functions , see, e.g. ssmr.ro/bulletin/pdf/52-3/Dragomir.pdf theorem 12 $\endgroup$ – Peter Melech Sep 7 '18 at 10:15
  • $\begingroup$ Fine. What I did was to prove Identity Theorem for Banach valued holomorphic functions using the result for complex valued functions. You can ignore my answer, but there may be others who don't know the result for Banach valued case. $\endgroup$ – Kavi Rama Murthy Sep 7 '18 at 10:17
  • $\begingroup$ Could You add to Your answer the definition of $x^{\ast}$ and $B^{\ast}$? $\endgroup$ – Peter Melech Sep 7 '18 at 10:20
  • $\begingroup$ I suppose You mean $X$ and $X^{\ast}$ instead of $B$...?I upvoted. $\endgroup$ – Peter Melech Sep 7 '18 at 10:28

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