0
$\begingroup$

Let $D=\{z\in\mathbb{C}: |z|<1\}$.

$C(\bar{D})=\{f:\bar D\longrightarrow \mathbb{C}: f \;\text{is continuous on}\; \bar{D}\}$

$A(D)=\{f\in C(\bar{D}): f \;\text{is analytic in} \;D\}$ Can you give me example of a function which is in $C(\bar{D})$ but not in $A(D)$? A function that is not analytic exactly at the boundary but continuous.

$\endgroup$
  • 1
    $\begingroup$ How about $z\mapsto\overline{z}$? $\endgroup$ – SmileyCraft Jan 10 at 16:03
0
$\begingroup$

Any function that is not differentiable will do. For instance $f(z)=|z|$.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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