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Define $ \Phi: G \times G \to \mathbb{C} $ by $ \Phi(z,w) = [f(z)-f(w)]/[z-w] $ if $ z \neq w$ and $ \Phi(z,z) = f'(z) $. Show that $\Phi(z,w)$ is continuous.

I want to try to show the function is continuous for $ z \neq w $.

I saw this question ask somewhere on here, but the answer usually skips this part since they state that it is very easy, but I don't see why since I don't have much experience with function of several variable.

Suppose $ z \neq w $. For $\epsilon > 0$, I need to show there exist $ \delta >0$ such that $ || \vec{z}-\vec{z_o}|| = \sqrt{ |z-z_o|^2+|w-w_o|^2} < \delta$ implies $| \Phi(\vec{z})-\Phi(\vec{z_o}) | < \epsilon ~$ where $\vec{z}=(z,w), \vec{z_o}=(z_o,w_o)$.

Now, $$| \Phi(\vec{z})-\Phi(\vec{z_o}) | = \bigg| \frac{f(z)-f(w)}{z-w}-\frac{f(z_o)-f(w_o)}{z_o-w_o} \bigg|$$

By controlling the size of $|z-z_o|,|w-w_o|$, I can control $|f(z)-f(z_o)|,|f(w)-f(w_o)|$. But in the norm for $\Phi$, I need to control the size of $|f(z)-f(w)|,|f(w)-f(w_o)|$ though. I try combine the two fraction together but I end up with messier equation. Did I do something wrong or did I get the def of continuity wrong?? thank you.

Khoa ta
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2 Answers2

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In fact continuity at points $(z,w)$ with $z\ne w$ is trivial; it's true assuming just that $f$ is continuous. Continuity at points $(z,z)$ is much trickier. From your post it sounds as though you believe you've handled the points $(z,z)$; I conjecture that you've actually just proved continuity in each variable separately at such points. Separate continuity at $(z,z)$ follows by definition just from the existence of $f'(z)$, while joint continuity at $(z,z)$ requires coninuity of $f'$. (So in particular if you solution for continuity at $(z,z)$ does not use the fact that $f'$ is continuous it's wrong.)

A little trick to establish continuity at $(p,p)$: Choose $r>0$ so that $D(p,r)\subset G$. Show that if $|z-p|<r$ and $|w-p|<r$ then $$\Phi(z,w)=\int_0^1f'(tz+(1-t)w)\,dt.$$(That holds for $z=w$ and also for $z\ne w$.)

It's fairly easy to use that representation of $\Phi$ and the conituity of $f'$ to show that $\Phi$ is continuous in $D(p,r)\times D(p,r)$. For example, if $z_n\to z$ and $w_n\to w$ you can show that $f'(tz_n+(1-t)w_n)\to f(tz+(1-t)w)$ uniformly for $t\in[0,1]$...

David C. Ullrich
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Since you consider $z \neq w$, we may assume $|z-w|$ and $|z_0-w_0|$ are bounded away from zero. One way to show continuity is: \begin{align*} \left|\frac{f(z)-f(w)}{z-w} - \frac{f(z_0)-f(w_0)}{z_0-w_0}\right| &\leq \left|\frac{f(z)-f(w)}{z-w} - \frac{f(z_0)-f(w_0)}{z-w}\right| + \left|\frac{f(z_0)-f(w_0)}{z-w} - \frac{f(z_0)-f(w_0)}{z_0-w_0}\right| \\ &= \frac{1}{|z-w|}\left|(f(z)-f(z_0)) - (f(w)-f(w_0))\right| + |f(z_0)-f(w_0)|\left|\frac{1}{z-w} - \frac{1}{z_0-w_0}\right| \\ &\leq \frac{1}{|z-w|}\left(|f(z)-f(z_0)| + |f(w)-f(w_0)|\right) + \frac{|f(z_0)-f(w_0)|}{|z-w||z_0-w_0|}\left(|z-z_0| + |w-w_0|\right). \end{align*} For $|z-z_0|$ and $|w-w_0|$ getting arbitrarily small, since $f$ is continuous already this shows that the right hand side is vanishing.

Dead-End
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  • Thank you for the answer. This is my first time I saw an epsilon delta proof for a function of two variables, I know it's obvious to you so I appreciate your help. – Khoa ta Jan 23 '16 at 06:51