Assume $f:[0,1] \times [0,1]$ is symmetric, i.e. $f(x,y) = f(y,x) \;\;\forall x,y \in [0,1]$. Assume further that $f$ is smooth, and that for every $x \in [0,1]$ the map $\phi_{x}(y):=f(x,y)$ attains its max at exactly one $y \in (0,1)$, say $y=y_{m}(x)$. (By symmetry we can then conclude that $\phi_{y}(x)$ attains its max at some unique $x_{m}(y) \in (0,1)$).
I'd like to conclude that the global max of $f$ on $[0,1] \times [0,1]$ occurs on the diagonal, i.e. at some $(x_{0},x_{0})$. But I'm not sure that this is true. The global max must occur at some ordered pair which is a max in both directions, i.e. some pair $(x, \;y_{m}(x)) =(x_{m}(y),\;y)$, but I'm having a hard showing that it lies on the diagonal for this reason. I want it to lie on the diagonal because that will make my calculations easier, and intuitively (and geometrically) it seems like it ought to.
I don't know whether to expect a proof or counterexample at this point, but I'd be very glad to hear either.
NOTE: The above posing of the problem is edited; I originally stated that the maxes of $\phi_{x}$ and $\phi_{y}$ were attained on $[0,1]$, not $(0,1)$. ellya posed a counterexample in which $(1,0)$ and $(0,1)$ were the locations of global maxes. But in my actual application, the maxes of $x \rightarrow f(x,y)$ and $y \rightarrow f(x,y)$ are constrained to lie in $(0,1)$, so I'm wondering if my conjecture holds in this case.
I believe this assumption is not true in general consider a symmetric function that attains it's maximums at the two points $(0,1),(1,0)$, such a function satisfies your assumptions, but violates the hypothesis.