$f:V \to \mathbb C$ holomorphic , then $g(x,y):=\frac{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?

Question

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

Use the mean value theorem on the real and imaginary part should work. — Hermès, Feb 17 '17 at 15:54
http://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 — Martin R, Feb 17 '17 at 19:08

score 1 · Answer 1 · answered Feb 17 '17 at 16:00

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 $y$.

$f:V \to \mathbb C$ holomorphic , then $g(x,y):=\frac{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?

1 Answers1