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The following is one version of Morera's theorem from complex analysis, as presented by Theodore W. Gamelin.

Theorem (Morera’s Theorem).
Let $f(z)$ be a continuous function on a domain $D$ (defined as simple connected open set). If
\begin{equation*} \int_{\partial R} f(z)\,dz = 0 \end{equation*} for every closed rectangle $R \subset D$ with sides parallel to the coordinate axes, then $f(z)$ is analytic on $D$.

Since continuity can be extended to the boundary of $D$ and we may still define the function, \begin{equation*} F(z) = \int_{z_0}^{z} f(\zeta) \, d\zeta, \quad z \in \bar{D}, \end{equation*} where $z_{0}$ is a fixed point in $D$, and the path of integration runs along a horizontal line and then a vertical line, whenever such a path is contained in $\bar{D}$. Now we can then attempt to differentiate $F(z)$. Indeed,
\begin{equation*} F(z + \Delta z) - F(z) = \int_{z}^{\,z+\Delta z} f(\zeta) \, d\zeta, \end{equation*} If $z$ lies on the boundary of $D$, then $\Delta z$ must be restricted to directions for which the segment $[z, z+\Delta z]$ is contained in $\bar{D}$.

Here arises my puzzle: the difference quotient
\begin{equation*} \frac{F(z+\Delta z) - F(z)}{\Delta z} \end{equation*} appears to exist even when $z$ is on the boundary of $D$. Since $f$ is continuous at $z$, this suggests that $F$ is analytic on $\bar{D}$, and therefore so is $f$. However, we know that analytic functions are defined only on open sets. Which step in this reasoning is mistaken?

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    $\begingroup$ analyticity on the boundary means something different (much stronger) than your statement, namely that there is an analytic function $F_1$ defined on an open set $\Omega$ containing $\bar{D}$ and equal to $F$ on $D$ (hence on $\bar D$); what you prove is just that $F$ extends to a (real or one dimensional) differentiable function on the boundary (if it's nice enough as your construction doesn't quite work always when the boundary of $D$ is spiky say) but that is far from complex differentiable/analytic; $\endgroup$ Commented Oct 17 at 2:39
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    $\begingroup$ see $\sum z^{2^n}/(2^n)^2$ which extends differentiably in this sense to the closed unit circle but is not complex analytic at any unit circle point $\endgroup$ Commented Oct 17 at 2:41
  • $\begingroup$ @Conrad what I learned is that analytic function is defined as continuously differentiable, isn’t it correct? $\endgroup$ Commented Oct 17 at 3:10
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    $\begingroup$ analytic functions are defined as continuously differentiable in an open plane set where the open condition is essential; here that's precisely the point, on the boundary you can show the differentiablity from the inside and from the boundary itself but not from the outside (again assuming a nice not too spiky boundary) so the "open" part fails in general and continuously differentiable doesn't imply analytic precisely like in the real one variable case where one can have $C^{\infty}$ functions that are not analytic $\endgroup$ Commented Oct 17 at 3:14
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    $\begingroup$ also, note that continuity in general doesn’t extend to the boundary (e.g $f(z)=\frac{1}{z}$ on $\Bbb{C}\setminus\{0\}$; or even worse, $\frac{1}{z}\sin\left(\frac{1}{z}\right)$, which doesn’t even have a limit, not even in the extended sense of being infinite). If you assume uniform continuity, then there is a continuous extension to the boundary. $\endgroup$ Commented Oct 17 at 4:07

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The function $$f(z)= \sum_{n=0}^\infty \frac{z^{2^n}}{2^n}$$ is analytic on the unit disc $\mathbb{D}=\{z\colon |z|< 1\}$ and it is continuous on $\overline{\mathbb{D}}=\{z\colon |z|\leq 1\}$ (the series converges absolutely on $\overline{\mathbb{D}}$). However $f$ is not differentiable (even in the real sense!) at any point of the unit circle $|z|=1$. To see that, note that if $f$ is differentiable at some $z=e^{ix_0}$, with $x_0\in \mathbb{R}$ then by the chain rule the function

$$x\in \mathbb{R} \mapsto f(e^{ix})$$ would be differentiable at $x=x_0$, but

$$f(e^{ix})= \sum_{n=0}^\infty \frac{\cos(2^n x)}{2^n} + i \sum_{n=0}^\infty \frac{\sin (2^n x)}{2^n},$$

and the real part of this function is an example of the famous Weierstrass functions, that are continuous but nowhere differentiable.

However this function $f$ satisfies all your assumptions: it is analytic on $\mathbb{D}$ and continuous on $\overline{\mathbb{D}}$. But the limit of the difference quotient does not exist along $\mathbb{S}^1$.

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