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I can prove that the function is continuous on $(z,w)$ where $z\neq w$ by showing for any sequence $(z_n,w_n)\rightarrow (z,w)$,

$lim $ $\phi((z_n,w_n))\rightarrow\phi((z,w))$

I can do this since $\exists N\forall n\geq n (Z_n\neq W_n)$

Hence, $lim $ $\phi((z_n,w_n))=lim \frac{f(z_n)-f(w_n)}{z_n-w_n}= \frac{f(z)-f(w)}{z-w}$

However, Im stuck at showing continuity when z=w since the function is not always of the same form. I can't seem to provide epsilon proof as well.

Any solutions?

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If $(z_n,w_n) \to (z,z)$ then $z_n \to z$ and $w_n \to z$. Assume first that $z_n\neq w_n$ for all $n$. Now $\frac {f(z_n)-f(w_n)} {z_n-w_n}=\frac 1 {z_n-w_n} \int_L f'(\zeta) \, d\zeta$ where $L$ is the line segment from $z_n$ to $w_n$. Continuity of $f'$ makes it easy to see that $\frac {f(z_n)-f(w_n)} {z_n-w_n} \to f'(z)$: $|\frac 1 {z_n-w_n} \int_L f'(\zeta) \, d\zeta - f'(z)|=|\frac 1 {z_n-w_n} \int_L \{f'(\zeta)-f'(z)\} \, d\zeta $. Given $\epsilon >0$ choose $\delta >0$ such that $|f'(\zeta)-f'(z)|<\epsilon$ if $|\zeta -z| <\delta$. Note that if $n$ is sufficiently large then $|\zeta -z| <\delta$ for all points $\zeta$ on the line $L$. Hence $|\frac 1 {z_n-w_n} \int_L f'(\zeta) \, d\zeta - f'(z)| <\epsilon$ for $n$ large enough. Now if $z_n=w_n$ for all $n$ then $\phi (z_n,w_n)=f'(z_n) \to f'(z)=\phi (z,z)$. Finally any sequence $(z_n,w_n)$ converging to $(z,z)$ can be split into (at most) two subsequences of the two types we have considered.

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  • $\begingroup$ what if $z_n=w_n$? Im sorry could you expand your proof slightly? I just can't seem to get it. $\endgroup$ – Jhon Doe Sep 5 '18 at 9:08
  • $\begingroup$ Good point! I have edited my answer now. $\endgroup$ – Kavi Rama Murthy Sep 5 '18 at 9:24
  • $\begingroup$ Just checking is my proof for z not equals w correct? $\endgroup$ – Jhon Doe Sep 6 '18 at 7:21
  • $\begingroup$ @JhonDoe Sure. For $z\neq w$ your argument is fine. $\endgroup$ – Kavi Rama Murthy Sep 6 '18 at 7:23

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