I have a function $F(x,y)$ which is continuous and analytic on the complement of a certain function $x(y)$. Is it possible to use Morera's theorem to show that it is analytic everywhere? Clearly, this approach can be used if it fails to be analytic on some linear function $x = a y + b$, for example (as far as I can guess, we could complexify $z = x + iy$ and use Morera's theorem by approximating the integrals over circles passing through the line until we showed analyticity everywhere). If $x(y)$ is also continuous (at least on a certain range of interest), could we use a similar method? If I'm wrong, can anyone think of convincing counterexamples?!
[Math] Using Morera’s theorem to prove analyticity
complex-analysis
Related Solutions
Here is what the hint (probably) means:
We only know that $f$ is continuous, but for what follows we need a smooth function ($C^\infty$ - although less smooth would be enough). To get it, take a smooth function $\phi(x,y)$ such that $\phi(x,y)=0$ if $x^2+y^2>1$ and $\int\phi dxdy=1$. Let $\phi_\epsilon(x,y)=\phi(x/\epsilon,y/\epsilon)/\epsilon^2$ (so that $\phi_\epsilon(x,y)=0$ if $x^2+y^2>\epsilon^2$ and $\int\phi_\epsilon dxdy=1$). Now set $f_\epsilon=f*\phi_\epsilon$, i.e. $f_\epsilon(x,y)=\int f(x-x',y-y')\phi(x',y')dx'dy'$. We have $f_\epsilon\to f$ uniformly as $\epsilon\to 0$, and $f_\epsilon$ are smooth.
If $\gamma$ is a contour bounding a domain $D$ then by Stokes theorem $\int_\gamma f_\epsilon dz = \int_D df_\epsilon\wedge dz = 2i\int_D\bar\partial f_\epsilon dx dy$. As $\int_\gamma f_\epsilon dz=0$ for all translations and rescaling of some $\gamma_0$ bounding some $D_0$, we get $\bar\partial f_\epsilon=0$ (e.g. by $0=\lim_{\delta\to 0} \int_{z_0+\delta D_0} \partial f_\epsilon dx dy/\text{Area}(\delta D_0)=\partial f_\epsilon(z_0)$), i.e. $f_\epsilon$ is holomorphic.
As $f_\epsilon\to f$ uniformly and $f_\epsilon$ are holomorphic, $f$ is also holomorphic.
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.
Best Answer
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).