Let $f$ be continuous on the complex plane and analytic on the complement of the coordinate axes. Show that $f$ is analytic everywhere. Hint: Morera's theorem. I think that I need to show that the integral is zero almost everywhere so by Morera's theorem it's analytic, but I'm not sure how to do that. Any help please?
[Math] Prove analyticity by Morera’s theorem
complex-analysis
Related Solutions
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).
Morera's theorem, when properly(1) stated, is indeed the exact converse of Goursat's theorem.
Theorem (Morera): Let $\Omega\subset\mathbb{C}$ open, and $f\colon\Omega\to\mathbb{C}$ a continuous function. If for all triangles $T\subset\Omega$ whose interior is also contained in $\Omega$ $$\int_T f(z)\,dz = 0,$$ then $f$ is holomorphic in $\Omega$.
Instead of triangles, one could of course also use rectangles, or other polygons. And actually, we could drop the condition that the interior of the triangle be contained in $\Omega$ and be left with a still true, but arguably less useful result, since we would then have a sufficient but not necessary condition (consider $1/z$ on $\mathbb{C}\setminus \{0\}$ to have function satisfying the condition as stated, but not the stronger condition one obtains by dropping "whose interior is also contained in $\Omega$").
The condition entails the existence of local primitives of $f$, i.e. every $z\in \Omega$ has a neighbourhood $U$ such that $f = F'$ for a holomorphic function $F$ on $U$. Thus $f$ is holomorphic on $U$ (the derivative of a holomorphic function is again holomorphic), and since holomorphicity is a local property, $f$ is holomorphic on $\Omega$.
To establish the existence of local primitives, one considers (for example) for $z_0 \in \Omega$ a disk $U = D_r(z_0) \subset \Omega$, and on $U$ the function $F(z) = \int_{z_0}^z f(\zeta)\,d\zeta$. The vanishing of the integral of $f$ over triangles whose interior is contained in $\Omega$ then yields $F(z) - F(w) = \int_w^z f(\zeta)\,d\zeta$, from which $F' = f$ follows with the continuity of $f$.
(1) The term "properly" means "properly for this purpose", or "adequately to show it is the converse of Goursat's theorem" here. Stating it for simply connected domains or disks is not wrong.
There is one downside to stating it explicitly for simply connected domains, however. Often, people aren't aware of the local character of the theorem, and consider the simple connectedness as essential for the validity of the theorem. The essential point is the locality, that one considers not the entire domain $\Omega$, but a small convex neighbourhood $U\subset \Omega$ of a point $z\in \Omega$ to construct the local primitive.
Best Answer
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.