I encountered a strange function $F \colon ]-1/2,1/2 \;[ \; \to \mathbb{R}$, defined on the open three-dimensional unit cube centered on the origin and taking real values. The function has some symmetry properties. It is not difficult to see that it is zero if any two of its inputs agree. I want to know if this is the only situation when it can be zero.
The function is a sum of three commutators.
$$ F(m_1,m_2,m_3) = \\ \log (1/2 + m_2) \log (1/2 - m_1) - \log (1/2 + m_1) \log (1/2 - m_2) \\ + \log (1/2 + m_1) \log (1/2 - m_3) - \log (1/2 + m_3) \log (1/2 - m_1) \\ + \log (1/2 + m_3) \log(1/2 - m_2) - \log (1/2 + m_2) \log (1/2 - m_3) $$
Some remarks:
- $F(-m) = -F(m)$. This implies that $F(0) = 0.$
For any odd permutation $\sigma$ we have $F(\sigma(a,b,c)) = -F(a,b,c)$. This implies that if $m_j=m_k$ for $j \neq k$, then $F(m) = 0$.
Another way of ordering the inputs of the function gives $$ F(m_1,m_2,m_3) = \\ \log (1/2 + m_2) \log (1/2 - m_1) + \log (1/2 + m_1) \log (1/2 - m_3) + \log (1/2 + m_3) \log(1/2 - m_2) \\ - [\log (1/2 + m_1) \log (1/2 - m_2) + \log (1/2 + m_2) \log (1/2 - m_3) + \log (1/2 + m_3) \log (1/2 - m_1)], $$ which emphasizes that there are in some sense two cyclical structures, one positive and negative, going in different directions. The positive one goes $m_2 \to m_1 \to m_3 \to m_2$ and the negative goes $m_1 \to m_2 \to m_3 \to m_1$. I do not know if this has any deeper meaning or consequences.
Numerics tell that the function is zero only when at least two of its arguments agree. Since the function is continuous, it has definite sign in each open set where all the three arguments are different.
Question: How to prove this analytically?