Do line integrals in a convex domain vanish for a function that is only continuous?

Question

Let $D$ be a convex domain, $\gamma \subset D$ a rectifiable closed path and $f(z): D \rightarrow \mathbb{C}$ continuous. Does $$\int_\gamma f(z)dz = 0$$ ?

I think there must be a counterexample (thinking of Cauchy), but my go-to non-holomorphic continuous function the modulus $||$ does in fact satisfy this condition for circles.

Is there a nice counterexample or some other way to argue this isn't true? Or is it in fact true?

Answer 1

Is there a nice counterexample ...

If $\gamma$ is a Jordan curve then $$ \int_\gamma \overline z dz = 2iA $$ where $A$ is the area enclosed by the curve, see for example this answer.

... or some other way to argue this isn't true?

If $f: D \to \Bbb C$ is continous, and $\int_\gamma f(z) dz = 0$ for every closed piecewise $C^1$ curve $\gamma$ in $D$ then $f$ is holomorphic in $D$ – that is Morera's theorem.

Therefore, conversely, if $f$ is not holomorphic then there must be closed path $\gamma$ in $D$ such that $\int_\gamma f(z) dz \ne 0$.

Answer 2

Take $f(z)=\bar{z}$, the complex conjugate, and take the unit circle as $\gamma$ Then

$$\int_\gamma \bar{z}\,dz=i\int_0^{2\pi}e^{-it}e^{it}\,dt=2\pi i$$

Answer 3

$$f(z)=\overline z\;\;\text{is continuous but}\;\;\oint_{|z|=1}\overline z\,dz\neq 0$$

as you can easily check.

Answer 4

No. Let $D= \mathbb C$, $\gamma(t)=e^{it}$ for $t \in [0,2 \pi]$ and $f(z)= \overline{z}.$

Then $\int_\gamma f(z)dz = 2 \pi i.$