Solve for $f$ in an ODE involving two points: $f'(x)-f'(y)\geq g(x,y)$ where $g$ is a known function. 
Prove that, for any $g$, there exists $f$ satisfying an ordinary differential inequity: $$f'(x)-f'(y)\geq g(x,y)$$ Where $g$ is a known function and $g(x,y)=-g(y,x)$.

$f$ is a real function. $x,y$ are real numbers. $g$ is a differentiable real function with two arguments. $g$ is also a monotonic function: increasing with the first argument and decreasing with the second argument. $g$ is defined on connected, convex, bounded, closed sets $C^2$. $f$ is defined on $C$.

Motivation: The ODE involving one point like $f^{(n)}(x)=g(x)$ is well studied. But the simplest ODE with two points: $f'(x)-f'(y)= g(x,y)$ seems never be studied?!
My Try: First try to solve the equity case, ODE: $f'(x)-f'(y)= g(x,y)$. But this case seems not too complicated. With some works you could see $g$ has to satisfy that there must exists $h$ such that:
$$g(x,y)=h(x)-h(y).$$
And $f$ is just the integration of $h$. So, for any $g$, $f$ can be non-existence in this ODE.
The true hard part is to solve the inequity.
Another approach is to treat it as a functional equation: $F(x)-F(y)\geq g(x,y)$.
The third approach would be transform it like: $\int_y^x f''(z)dz\geq g(x,y)$,
Then $\int_y^{y+\epsilon} f''(z)dz\geq g(y+\epsilon,y)$
Then $\lim_{\epsilon\to 0}\epsilon f''(y)\geq g(y+\epsilon,y)$
Since $g(y,y)=0$,
Then $$\lim_{\epsilon\to 0}\frac{g(y+\epsilon,y)}{\epsilon}=g_1(y,y)$$
So $f''(y)\geq g_1(y,y)$.
In fact, as $\int_y^xf''(z)dz\geq g(x,y)-g(y,y)=\int_y^xg_1(z,y)dz$.
So for all $y$ and all $z>y$, $f''(z)\geq g_1(z,y)$.
Let the upper bound $\sup\{g_1(z,y)||y\in C\}=U(z)$. The upper exists since $C$ is compact.
Then, let $f''(z)=U(z)$ and solve for $f$.
It seems solves the original ODIE?
 A: I'm afraid there are too many things mixed together, here. There are functional equations (i.e. equations where the unknown is a function), differential and integral equations would be examples important for applications, while things like $$f(x-y)\,f(x+y)=f(x)^2-f(y)^2$$ are rather exotic, reserved for contest questions and their likes.
In your case, $$f'(x)-f'(y)=g(x,y)\tag{F'}$$ would rather be a general functional equation, essentially, it is $$h(x)-h(y)=g(x,y)\tag{H}$$ with the additional requirement that $h$ be a derivative (meaning it satisfies the intermediate value theorem, so we're not just speaking of integration, here).
Unfortunately, your assumptions aren't very clearly stated, it starts with "any $g$" (that wouldn't work, obviously), soon followed by $$g(x,y)=−g(y,x)\tag{G}$$ and later by
"$g$ is a differentiable real function with two arguments. $g$ is also a monotonic function: increasing with the first argument and decreasing with the second argument" (why?).
Now you'll say "wait, I want a functional inequality:" $$h(x)-h(y)\le g(x,y)\tag{H'}.$$ but as a matter of fact, your assumption (G) reduces that to (H): exchanging $x$ and $y$ gives $$h(y)-h(x)\le g(y,x),$$ i.e.
$$h(x)-h(y)\ge -g(y,x)=g(x,y).$$
So we're left with (H). Under which assumptions is that solvable? Obviously, it would be necessary that $$g(x,y)+g(y,z)=h(x)-h(y)+h(y)-h(z)=h(x)-h(z)=g(x,z),$$ i.e. $$g(x,y)+g(y,z)=g(x,z)\tag{G'}.$$ That's also sufficient, because it implies $$g(x,y)=g(x,z)-g(y,z)=g(x,z_0)-g(y,z_0)=h(x)-h(y)$$ for any constant $z_0\in C$.
Speaking of $C$: you assume $C^2$ ($=C\times C$, I hope) to be "connected, convex, bounded, closed", but trust me, "convex" implies "connected", so $C$ must be convex, and since later in your post, we learn that $C$ is assumed to be compact, you could as well say $C=[a,b]$ from the very beginning.
In a word: for the solvability of your inequality/equation, you need (G'), together with some regularity assumption so that $h(x)=g(x,z_0)$ can be a derivative, right?
BTW, while (G') seems to be explicit in terms of $g$ (unlike "there exists a function $h$ so that..."), watching somebody checking this for, say $g(x,y)=\sin(x+y)\,\sin(x-y)$ would be... erm... entertaining, most likely. =D
