Let $V\subseteq \mathbb C$ be an open connected set , $f:V \to \mathbb C$ be a holomorphic function . Then is the function $g:V \times V \to \mathbb C$ defined as $g(x,y):=\dfrac{f(x)-f(y)}{x-y} $ , for $x \ne y , x,y \in V$ and $g(x,x)=f'(x) , \forall x \in V$ continuous ? I can prove that for $(x,y) \in V \times V$ with $x \ne y$ , $(x,y)$ is a continuity point of $g$ , but I am having trouble showing the continuity of $g$ on diagonal points . Please help . Thanks in advance
$f:V \to \mathbb C$ holomorphic , then $g(x,y):=\dfrac{f(x)-f(y)}{x-y} $ , $x \ne y , x,y \in V$ and $g(x,x)=f'(x) , \forall x \in V$ is continuous?
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complex-analysis
continuity
several-complex-variables
holomorphic-functions
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0Use the mean value theorem on the real and imaginary part should work. – 2017-02-17
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0http://math.stackexchange.com/questions/71613/proving-a-function-is-continuous-for-a-fixed-variable, http://math.stackexchange.com/questions/18838/continuous-complex-function-2-variables, http://math.stackexchange.com/questions/1623135/showing-a-complex-valued-function-of-two-variable-is-continuous-conway-volume-i – 2017-02-17
1 Answers
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Your function is in fact holomorphic in the two variables. You may see this by writing: $$ g(x,y) = \int_0^1 f'(y+t(x-y)) \; dt $$ valid whenever the segment $[y;x]=\{tx + (1-t)y : 0\leq t \leq 1\} \subset V$. The integrand is (locally) holomorphic in both $x$ and $y$.