I have three convex functions $f_1(x,y), f_2(x,y),$ and $f_3(x,y)$. I know that $f_1(x,y)$ is non decreasing on both $x$ and $y$, and $f_3(x,y)$ is non increasing on both $x$ and $y$. $f_2(x,y)$ is just convex. To put it more formally:
$$\forall \delta_x,\delta_y\geq 0, f_1(x+\delta_x, y+\delta_y)\geq f_1(x,y)$$
$$\forall \delta_x,\delta_y\geq 0, f_3(x+\delta_x, y+\delta_y)\leq f_3(x,y)$$
Now I want to prove that all three functions pairwise intersect in $[0,1]\times [0,1]$
I am able to show that (these are assumptions) $$f_1(0,0)\leq f_2(0,0) \leq f_3(0,0)$$
And
$$f_1(1,1)\geq f_2(1,1) \geq f_3(1,1)$$
However, I am not sure whether this is sufficient to conclude that they all intersect or not (Intuitively because $f_1$, $f_3$ are increasing and decreasing respectively, and because they switch order then it makes sense that they intersected with an analogous to IVT).
I also did some python simulations and they always intersect, just not sure how to show that.
You do not need any further assumptions than your inequalities and continuity of $f_1, f_2, f_3$. Let $1 \leq i < j \leq 3$ and consider $f_i$ and $f_j$. Then $f_j(0,0) - f_i(0,0) \geq 0$ by the first set of inequalities, and $f_j(1,1) - f_i(1,1) \leq 0$ by the second set. Because the function $h(t) = f_j(t,t) - f_i(t,t)$ is continuous, by the (one-dimensional) intermediate value theorem, it must be zero somewhere, so $f_i$ and $f_j$ intersect.