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Let $G\subset \mathbb{C}$ be a non-empty open set and $f$ be a function holomorphic on $G$. Let $g: G\times G\to \mathbb{C}$ be a function defined as $$g(z,w)= \begin{cases} \frac{f(z)-f(w)}{z-w}, & z\ne w \\ f'(z), & z=w \end{cases}$$

I need to prove that $g$ is continuous on $G\times G$.

My approach:

Let $\varepsilon > 0$. Then $\exists \delta_1, \delta_2>0$ such that $\| (z,w)-(z_0,w_0) \|^2\le|z-z_0|+|w-w_0|<\delta_1+\delta_2$ implies that $|f(z)-f(z_0)|<\epsilon$, $|f(w)-f(w_0)|<\epsilon$, since $f$ is continuous on $G$.

Now, $\| g(z,w)-g(z_0,w_0) \|=\left| \frac{f(z)-f(w)}{z-w} - \frac{f(z_0)-f(w_0)}{z_0-w_0} \right|\le \left| \frac{f(z)-f(w)}{z-w}\right| + \left|\frac{f(z_0)-f(w_0)}{z_0-w_0} \right|$

Since $f$ is holomorphic on $G$, it is continuous on $G$, thus $\exists \delta'>0$ such that $| {f(z)-f(w)}|<\varepsilon'|z-w|<\varepsilon'\delta'$ whenever $|z-w|<\delta'$.

And here's where I'm stuck, because it's not clear how to deal with $\left|\frac{f(z_0)-f(w_0)}{z_0-w_0} \right|$, because $z_0, w_0$ are fixed points, so we can't just make them approach each other. Of course, my first thought was to somehow rearrange the inequality in such a way as to obtain $\left| \frac{f(z)-f(z_0)}{z-z_0}\right| + \left|\frac{f(w)-f(w_0)}{w-w_0} \right|$, but I don't see how to do so.

Your help would be appreciated.

sequence
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    http://math.stackexchange.com/q/18838/ – Jonas Meyer Mar 16 '17 at 03:04
  • @JonasMeyer I'm not quite sure about the approach mentioned there. How did the OP prove that $g$ is continuous when $z_0\ne w_0$? For him this may be trivial, but I do not see how this is proved. For the approach when $z_0=w_0$, it appears to be clear more or less. I think that because $z\ne w$ and $f$ is holomorphic, $\frac{f(z)-f(w)}{z-w}$ is also holomorphic, and hence continuous. When $z=w$, can we take the limit as $w \to z$ to deduce that, since $f$ is holomorphic, the limit is the derivative of $f$ at $z$? – sequence Mar 16 '17 at 04:54
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    I don't know for sure what "when $z=w$, can we take the limit as $w\to z$" means. The reason it is continuous at a point $(z_0,w_0)$ with $z_0\neq w_0$, is because differences of continuous functions and quotients of continuous functions are continuous. For continuity at points where $z_0=w_0$ there are two proofs (or sketches at least) in the other thread. It is clear from the definition of the derivative that $\lim\limits_{w\to z}\dfrac{f(z)-f(w)}{z-w}=f'(z)$, but that is not a two variable limit, so all it really tells us is what the limit must be if it exists. – Jonas Meyer Mar 16 '17 at 05:02
  • Your approach here will not work because you are bounding with a sum of absolute values of difference quotients that will not go to zero. It would be better to use theorems about arithmetic operations being continuous, or leave everything inside the absolute values until you find a way to break it into parts that go to 0. – Jonas Meyer Mar 16 '17 at 05:06

2 Answers2

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On the open set $\Omega = G \setminus \Delta_G$, both $z - w$ and $f(z) - f(w)$ are continuous and $z - w$ has no zeros, so $g(z,w)$ is continuous. So we only need to show that $g(z,w)$ is continuous at points in $\Delta_G$, i.e. $$\Vert (z,w) - (z_0,z_0) \Vert < \delta \implies \vert g(z,w) - g(z_0,z_0) \vert < \epsilon.$$

If $z = w$ this is easy, as then $$g(z,w) - g(z_0,z_0) = f'(z) - f'(z_0)$$ and $f'$ is continuous. If $z \neq w$, we have $$g(z,w) - g(z_0,z_0) = \frac{f(z) - f(w)}{z - w} - f'(z_0).$$ For this case write $$f(z) = \sum_{n=0}^{\infty}a(n)(z - z_0)^n.$$ Then \begin{equation*} \begin{aligned} \frac{f(z) - f(w)}{z - w} &= \frac{1}{z - w}\sum_{n=0}^{\infty}a(n)[(z - z_0)^n - (w - z_0)^n] \\ &= \frac{1}{(z - z_0) - (w - z_0)}\sum_{n=1}^{\infty}a(n)[(z - z_0)^n + (w - z_0)^n] \\ &= \sum_{n=1}^{\infty}a(n)[Z^{n - 1} + Z^{n-2}W + \cdots + ZW^{n-2} + W^{n-1}] \\ \end{aligned} \end{equation*} with $Z = z - z_0, W = w - z_0$. We have $$\frac{f(z) - f(w)}{z - w} - f'(z_0) = \sum_{n=2}^{\infty}a(n)[Z^{n - 1} + Z^{n-2}W + \cdots + ZW^{n-2} + W^{n-1}].$$

Thus if $\vert Z \vert, \vert W \vert < \delta$, then $$\left\vert \frac{f(z) - f(w)}{z - w} - f'(z_0) \right\vert < \sum_{n=2}^{\infty}n \vert a(n) \vert\delta^{n - 1} < \epsilon.$$

Ethan Alwaise
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  • Can you please explain how you went from $\frac{1}{(z - z_0) - (w - z_0)}\sum_{n=1}^{\infty}a(n)[(z - z_0)^n + (w - z_0)^n]$ to $ \sum_{n=1}^{\infty}a(n)[Z^{n - 1} + Z^{n-2}W + \cdots + ZW^{n-2} + W^{n-1}]$? – sequence Mar 16 '17 at 22:28
  • Why do you have it as $\frac{f(z) - f(w)}{z - w} = \frac{1}{z - w}\sum_{n=0}^{\infty}a(n)[(z - z_0)^n + (w - z_0)^n]$ and not as $ \frac{1}{z - w}\sum_{n=0}^{\infty}a(n)[(z - z_0)^n - (w - z_0)^n]$ (with the minus sign)? Also, how did you shift the index from $0$ to $1$ without any changes to the series? Was this a typo? – sequence Mar 16 '17 at 22:31
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    @sequence: apparently a typo, should be minus, and then the identity $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1})$ is used. (Similar to what you asked about in the other thread, except this answer gives more details.) – Jonas Meyer Mar 17 '17 at 01:24
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If $z_0\ne w_0$ there is an $\epsilon>0$ such that $U_\epsilon(z_0)$ and $U_\epsilon(w_0)$ are disjoint. One then has $$g(z,w)={f(w)-f(z)\over w-z}\qquad\forall\>(z,w)\in U_\epsilon(z_0)\times U_\epsilon(w_0)\ ,$$ hence $g$ is continuous at $(z_0,w_0)$.

If $z_0=w_0$ choose a convex neighborhood $U\subset \Omega$ of $z_0$. Fix two points $z$, $w\in U$, and consider the auxiliary function $$\phi(t):=f\bigl((1-t)z+tw\bigr)\qquad(0\leq t\leq 1)$$with derivative $$\phi'(t)=f'\bigl((1-t)z+tw\bigr)\>(w-z)\ .$$ One has $$f(w)-f(z)=\phi(1)-\phi(0)=\int_0^1\phi'(t)\>dt=(w-z)\int_0^1f'\bigl((1-t)z+tw\bigr)\>dt\ .$$ If $z\ne w$ we therefore have $$g(z,w)=\int_0^1f'\bigl((1-t)z+tw\bigr)\>dt\ ,\tag{1}$$ and if $z=w$ formula $(1)$ is trivially true. Since the right hand side of $(1)$ is continuous on $U\times U$ we are done.

Christian Blatter
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