How can I prove Morera's theorem?
Morera's Theorem: If a function $f(z)$ is continuous in a simply connected region and its integration over a closed path gives $0$, then the function is analytic.
[Math] Morera’s theorem
complex-analysis
Related Solutions
It's enough to show that the integral of $f$ along every circle is $0$. If the circle does not cross or touch one of the axes, you've got it. If crosses an axis, approximate it by two simple closed curves: one of the follows an arc of the circle until it's very close to the coordinate axis, then moves along a line close to the axis until it reaches the arc, then moves along that arc. The other does the same on the other side of the axis. The integrals along those two lines approximately cancel each other, because you assumed continuity, and the approximation can be made as close as you want by making the lines close enough to the axis, again because you assumed continuity. So in the limit, they cancel each other and the whole thing approaches the integral along the circle. So $$\lim\limits_{\text{something}\to\text{something}} 0=\text{what}?$$
If it merely touches an axis without crossing, the problem is simpler. If it crosses both axes, it's more complicated in details, but conceptually pretty much the same.
If I understand correctly, "the complement of a function $x(y)$" means the complement of its graph, that is, the set $\{x+iy:x\ne x(y)\}$. You assume that $F$ is continuous in some domain $\Omega$ (possibly $\mathbb C$) and is holomorphic (=analytic) on $\Omega\setminus \Gamma$, where $\Gamma$ is the graph of some continuous function. The question is: does it follow that $F$ is holomorphic in $\Omega$?
(Stated more succinctly: is $\Gamma$ removable for continuous holomorphic functions?)
In general, the answer is no. Here is why.
Let $x(y)$ be a continuous function on a closed interval whose graph $\Gamma$ has Hausdorff dimension $D>1$. Such functions exist: for example, see reference 6 in the Wikipedia article on the Weierstrass function. Pick $1<d<D$. By Frostman's lemma, there exists a finite positive measure $\mu$ with support contained in $\Gamma$ and with the growth bound $\mu(B(z,r))\le r^{d}$ for all $z\in \mathbb C$ and $r>0$. Define $$F(z)=\int_{\mathbb C}\frac{1}{z-\zeta}d\mu(\zeta)$$ By construction, the function $F$ is holomorphic on $\mathbb C\setminus \Gamma$. It also tends to zero as $z\to\infty$. Since $F$ is not identically zero, it cannot be extended to a holomorphic function on $\mathbb C$; such an extension would contradict Liouville's theorem.
It remains to show that $F$ is continuous at every point $p\in \Gamma$. Pick a small $\delta>0$ and write $$F(z)=\int_{B(p,\delta)}\frac{1}{z-\zeta}d\mu(\zeta)+\int_{B(p,\delta)^c}\frac{1}{z-\zeta}d\mu(\zeta)=:F_1(z)+F_2(z) $$ The function $F_2$ is holomorphic in $B(p,\delta)$, and therefore continuous there. For $z\in B(p,\delta)$ the first term can be estimated from above by switching to polar coordinates and integrating by parts: $$ |F_1(z)|\le \int_{B(p,\delta)}\frac{1}{|z-\zeta|}d\mu(\zeta) = \int_0^\infty \mu(B(z,\rho)\cap B(p,\delta))\,\frac{d\rho}{\rho^2} \\ \le \int_0^\infty \min(\rho,\delta)^d \,\frac{d\rho}{\rho^2} = \int_0^\delta \rho^d \,\frac{d\rho}{\rho^2}+\int_\delta^\infty \delta^d \,\frac{d\rho}{\rho^2}=C\delta^{d-1} $$ Since $\delta^{d-1}\to 0$ as $\delta\to 0$, the function $F_1$ is continuous at $p$. In fact, the estimate shows that it is Hölder continuous with exponent $d-1$. This relation between Hölder exponent and Hausdorff dimension is not accidental: see this thread and references quoted there.
However, the answer is yes if $x(y)$ is a Lipschitz function. That is, Lipschitz graphs are removable for continuous holomorphic functions. More generally, curves of finite length are removable for continuous holomorphic functions: this a theorem of Painlevé. See Analytic capacity and measure by Garnett (Springer LNM series, 297).
Best Answer
Morera´s theorem states:Let the function $f: \mathbb{C} \longrightarrow \mathbb{C}$. be continuous in a simply connected open set S. If $$\oint_Cf(z)dz=0$$ for a simple closed path $C$ that lies on $S$ then $f$ is an analytic function in $S$
Let $z_0$ in $S$ and define for all $z$ in $S$ $$F(z)=\int_{z_0}^{z} f(z)dz$$ Look that $F(z)$ do not depend on the path of the curve between the integration intervals, so the function $F(z)$ is well defined. Now we need to prove that $F(z)$ is differentiable at any point $z$ in $S$, such that it derivative is $f$, by definition, $F$ and all its derivatives are analytic over $S$. now let $h$ be a complex number such that $z+h$ is in $S$, then $$\dfrac{F(z+h)-F(z)}{h}=\dfrac{1}{h}\int_{z}^{z+h} f(y)dy$$, note now that $f(z)$ is in terms of the variable $y$, so $$|\dfrac{F(z+h)-F(z)}{h}-f(z)|=|\dfrac{1}{h}\int_{z}^{z+h}(f(y)-f(z))dy|$$ The integration variable $y$ lies on the path defined by the last integral, so if $h$ tends to 0, $y$ tends to $z$ by basic complex analysis.Now , how $f$ is continuous at $z$: $f(y)$ tends to $f(z)$ as $y$ tends to $z$, so here you use epsilon, delta definition and show that
$$\lim_{h\to0}{\dfrac{F(z+h)-F(z)}{h}}=f(z)$$