Functional equation on unit square

Question

Suppose $F$ is a continuous function defined on the unit square $[0,1]\times[0,1]$ satisfying the following properties :

i) $ F(a,a)=0,$ for all $a\in[0,1],$

ii) $F(a,b)=-F(b,a)$ for all $(a,b)\in[0,1]\times[0,1],$

iii) $F(a,b)\geq0$ if $a\geq b$.

Then does $F(a,b)>0$ imply that $a>b$?

Edit 1. Suppose $F$ is differentiable on the interior of the unit square. Would this change the answer to the problem?

Answer 1

The claim is true, and you don't need continuity or differentiability. Prove by contraposition:

Claim: If $a\le b$ then $F(a,b)\le0$.

Proof: If $a\le b$ then $b\ge a$ so $F(b,a)\ge0$ by (iii). But $F(b,a)=-F(a,b)$ by (ii). Substituting, we get $-F(a,b)\ge0$, i.e., $F(a,b)\le0$.