Is the potential of a periodic conservative field periodic?

Question

Let $Y = [0,1]^3$ and consider a conservative vector field $F$. Denote its scalar potential by $\varphi$, i.e. $$ \nabla \varphi = F. $$ If $\varphi$ is $Y$-periodic it is clear that $F$ is periodic, too. Furthermore, we find $$ \int_Y F(x) \, dx = \int_Y \nabla \varphi \, dx = \int_{\partial Y} \varphi n \, dS(x) = 0. $$ (Here $n$ denotes the outer normal and we used the periodicity of $\varphi$ in the last equality.)

My question is, is the converse also true? To clarify my question, let me rephrase:

If $F$ is $Y$-periodic and $\int_Y F \, dx = 0$ can we conclude, that $\varphi$ is $Y$-periodic?

Answer 1

I found the answer myself: The converse holds true as well.

I will only sketch the proof in the following. A complete proof of a little more general result can be found in this thesis, see Lemma 2.10

Sketch of the proof:

1. Show that "$\varphi$ $Y$-periodic $\Leftrightarrow \int_0^1 F_i(y + t e_i) \, dt = 0, \; \forall y$ and $i$". Here $e_i$ is the $i$-th element of the canonical basis of $\mathbb{R}^3$ and $F_i = F \cdot e_i$ is the $i$-th component of $F$.
2. Expand $F_i$ in a Fourier series: $F_i = \sum_{k \in \mathbb{Z}^3} c_i^k e^{2 \pi i k \cdot y}$
3. To conclude note that $\int_0^1 F_i(y + t e_i) \, dt = c_i^0 = \int_Y F_i \, dy$.