Assume the function $f: \mathbb R^2 \rightarrow \mathbb R$ satisfies the property $f(x+s,y+t) \geq f(x,y) -s^2-t^2$ for every $(x,y) \in \mathbb R^2$ and every $(s,t) \in \mathbb R^2$. Prove that $f$ must be a constant function.
I am trying to learn partitions but I am not sure how to use partitions with more than one variable. Could someone please give me a detailed proof for this question? Thank you for your help in advance!
So $f(x,y)-f(x+s,y+t)\leq s^2+t^2$. Let $x':=x-s$ and $y':=y-t$ to get:
$f(x'-s,y'-s)-f(x',y')\leq s^2+t^2$, and making a substitutions $s,t\rightarrow -s,-t$ implies
$$|f(x,y)-f(x+s,y+t)|\leq s^2+t^2.$$
Now let $t=0$ and divide both sides by $s$ and take the limit as $s\rightarrow 0$ to get that $f_x(x,y)=0$ for all $x,y$. Repeat this procedure with $s=0$ to get $f_y(x,y)=0$. The conclusion follows.