enter image description here

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?


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.