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So the question asked to prove the following: "Prove that if f is a continuous complex valued function in the open subset G of the complex plane, and if the integral of $f(z)dz = 0$ over every rectangle R, with edges parallel to the coordinate axes, contained with its interior in G, then f is holomorphic" (Sarason Complex Function Theory 2nd edition).

Following the logic the book used to show this was true for triangular regions, I came up with this seemingly roundabout way of proving it, which I know works, but seems very roundabout. Basically the book proves the other direction for triangles, so I can just modify that proof, and then the book uses the triangle theorem to prove Cauchy's Theorem for a convex region. It is easy to substitute the rectangle theorem, for the triangle theorem. While I'm glad this works, and its almost directly lifted from the textbook, it is over 2 pages long, so I was wondering if a more direct approach could be found.

Proof Summary: "Our approach will be to first that if f is holomorphic the integral around every rectangle of $f(z)dz$ equals 0. We will then apply this fact to prove Cauchy's Theorem for a convex region. Once this is complete we can note that this new proof of Cauchy's Theorem allows us to say that if f has integral zero around every rectangle contained in G, f has a primitive in G. If f has a primitive in G, it is holomorphic."

Edit: Also I was thinking that since we've gone through the proof in the textbook and it works for triangles, if we can somehow divide the rectangle into 2 triangles and show that the integral around each of these triangles is equal to zero, then the rest follows from the book's derivations based on triangles, though honestly this seems very difficult to do.

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The article Morera's theorem is relevant (in case you are not familiar with the contents). – Amitesh Datta Jul 24 '11 at 6:18
What are the rules of the game? E.g., are you allowed to use that the derivative of a holomorphic function is again holomorphic? – Christian Blatter Jul 24 '11 at 11:29
@Christian Blatter: We are allowed to make that assumption. We've covered up to Cauchy's Theorem for a convex region. Basically a list of topics that might be helpful that we've covered are: Power Series (not Laurent Series though). Basics of Complex Integration Cauchy's Theorem for a triangle/convex region/Circle To be covered before the assignment is due: Mean Value Property Goursat's Lemma Liouvillve's Theorem Fundamental Theorem of Algebra Identity Theorem Weistrass Convergence Maximum Modulus Principle Schwarz's Lemma – I Love Cake Jul 25 '11 at 0:12
up vote 6 down vote accepted

This is a local theorem, i.e., we don't have to bother about holes in $G$ and may assume $0\in G$. Define two functions $F$ and $G$ in the neighborhood of $0$ as follows: For $z=x+iy$ put $$F(z):=\int_0^x f(z')\ dz' +\int_x^{x+iy} f(z')\ dz'=\int_0^x f(t)\ dt +i \int_0^y f(x+it)\ dt\ ,$$ $$G(z):=\int_0^{iy}f(z')\ dz'+\int_{iy}^{x+iy} f(z')\ dz'=i \int_0^y f(it)\ dt +\int_0^x f(t+iy)\ dt\ .$$ This amounts to integrating $f(z')\ dz'\ $ along two different L-shaped paths from $0$ to $z$. As $\int_{\partial R}f(z)\ dz=0$ for all rectangles $R$ it follows that $F(z)=G(z)$, and this is true for all $z$ in the neighborhood of $0$.

By definition of $F$ one has $F_y(z)=i f(z)$, and from the definition of $G$ we conclude that $F_x(z)=G_x(z)= f(z)$. This means that (a) the function $F$ has continuous partial derivatives and (b) that these derivatives satisfy the Cauchy-Riemann equations. Therefore $F$ is a holomorphic function of $z$ in the neighborhood of $0$. Since we have $F'=F_x=f\ $ the function $f$ is holomorphic there also.

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Hi @Christian. I was wondering why it is that these two functions amounts to integration around two L-shaped paths from 0 to z. I drew out the paths taken, and it appears that the path for F(z) is the L-shaped path defined by the points (0,y), (0,0), and (x,0), while the second is the one define by the same points, thus we are integrating over half of the rectangle. Is it something in the way that you parameterized the curves? – I Love Cake Jul 25 '11 at 20:58
@I Love Cake: See my edit. – Christian Blatter Jul 26 '11 at 8:33

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