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Suppose $f : \Omega \to \mathbb{C}$ is holomorphic and fix $z_0 \in \Omega$. We claim that for any $\epsilon > 0$, there exists some $R > 0$ such that $z,w \in B_{R}(z_0)$ and $z \neq w$ implies that $$\left\vert \frac{f(z) - f(w)}{z - w} - f'(z_0) \right\vert < \epsilon.$$ We expand the numerator as a power series to write \begin{equation*} \begin{aligned} \frac{f(z) - f(w)}{z - w} &= \frac{1}{z - w}\sum_{n=1}^{\infty}a_n[(z - z_0)^n - (w - w_0)^n] \\ &= \sum_{n=1}^{\infty}a_n\frac{(z - z_0)^n - (w - w_0)^n}{z - z_0 - (w - z_0)} \\ &= f'(z_0) + \sum_{n=2}^{\infty}a_n(Z^{n - 1}W + Z^{n - 2}W^2 + \cdots + ZW^{n - 1}), \end{aligned} \end{equation*} where $Z = z - z_0$ and $W = w - z_0$. Now $a_1 = f'(z_0)$, so to complete the proof we just have to show that the limit of the series in the last line above is $0$ as $z,w \to z_0$. For $z,w$ in a closed disk of radius $R$ centered at $z_0$ contained in $\Omega$, we can estimate the series as $$\left\vert \sum_{n=2}^{\infty}a_n(Z^{n - 1}W + Z^{n - 2}W^2 + \cdots + ZW^{n - 1}) \right\vert \leq \sum_{n=2}^{\infty}\vert a_n\vert n R^n.$$ The series on the right converges absolutely and uniformly on the disk since $$f'(z_0) = \sum_{n=1}^{\infty}a_nnz^{n - 1}$$ converges absolutely on any closed disk centered at $z_0$ contained in $\Omega$. Thus we may substitute $R = 0$ to find that the limit is $0$. This completes the proof.

Clearly this proof relies only on the fact that $f$ is analytic, not necessarily complex valued. I would be interested in a counter example for the real case, where I suspect this no longer holds.

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  • $\begingroup$ What does $\lim \limits_{\color{red}{z,w \to z_0}}\frac{f(z) - f(w)}{z - w} = f'(z_0)$ mean? $\endgroup$ – Git Gud Nov 29 '16 at 22:38
  • $\begingroup$ For every $\epsilon > 0$ there exists $\delta > 0$ such that if $\vert z - z_0 \vert < \delta$ and $\vert w - z_0 \vert < \delta$, then $$\left\vert \frac{f(z) - f(w)}{z - w} - f'(z_0) \right\vert < \epsilon.$$ $\endgroup$ – Ethan Alwaise Nov 29 '16 at 22:40
  • $\begingroup$ There must be more than this. Otherwise you could take $|w-z_0|<\delta$ and $z=w$, yet the difference quotient is undefined ($0/0$). $\endgroup$ – MPW Nov 29 '16 at 22:46
  • $\begingroup$ Then can we simply specify that $z,w$ should be distinct? $\endgroup$ – Ethan Alwaise Nov 29 '16 at 22:57
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    $\begingroup$ For the real case, I believe your condition is true provided $f$ is continuously differentiable. $\endgroup$ – GEdgar Nov 29 '16 at 22:58
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Consider

$$f(x) = \begin{cases}\quad 0 &, x = 0 \\ x^2\cos \frac{\pi}{x} &, x \neq 0.\end{cases}$$

Then $f$ is differentiable on $\mathbb{R}$, with $f'(0) = 0$, but for $y = \frac{1}{n}$ and $z = \frac{1}{n+1}$, we have

$$\frac{f(y) - f(z)}{y-z} = (-1)^n\frac{\frac{1}{n^2} + \frac{1}{(n+1)^2}}{\frac{1}{n} - \frac{1}{n+1}} = (-1)^n \biggl(\frac{n+1}{n} + \frac{n}{n+1}\biggr),$$

which doesn't converge.

If however we require $y$ and $z$ to lie on different sides of $x$, then we have

$$\lim_{y,z \to x} \frac{f(y) - f(z)}{y-z} = f'(x)$$

whenever $f$ is differentiable at $x$, since then the difference quotient $\frac{f(y) - f(z)}{y-z}$ is a convex combination of the one-sided difference quotients $\frac{f(y) - f(x)}{y-x}$ and $\frac{f(x) - f(z)}{x-z}$.

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