Suppose that $f(z)$ is a continuous complex valued function on a disc such that
the integral $ \int_C f dz = 0 $ for every contour C that is the boundary
of a square in the disc. Prove that $f$ must be analytic.
I am trying to adapt other proofs of Morera's theorem to this case of square paths, but without much success. The general strategy I believe should be a construct an antidervative by using the fact that we can integrate along squares, but I cannot seem to get started.
Best Answer
We can patch 2 adjacent square paths to get a $1:2$ rectangle path. In the same manner, we can get $1:n$ and $m:n$ rectangle path on which $\int_C f(z)dz = 0$. Then, by approximation, we can get any rectangular path of real ratio from those of rational ratios, and by continuity of $f$, it holds $$\int_{\partial R}f(z)dz = 0$$ for any rectangle $R\subset \mathbb{D}=\{z\;|\;|z|<1\}$. Using this, define anti-derivative $F$ as $$ F(x+iy) = \int_{\gamma_1} f(z)dz + \int_{\gamma_2} f(z)dz $$ where $\gamma_1(t) = tx, 0\le t\le 1$ and$\gamma_2(t) = x+ity, 0\le t\le 1$. Then we can see that $F$ satisfies the Cauchy-Riemann equation, and thus $F$ is complex differentiable and $F'=f$.