Re: Alex' comment on the OP, I assume "convex" w/r/t subsets of $\mathbb{R}^2$ is meant in the usual sense.

Call a function as in the OP tame. Here's an answer to (1):

First, I claim there is no total order $\prec$ on $\mathbb{R}^2$ such that any tame function is monotone with respect to $\prec$. To see this, consider the four points $a=(0, 1)$, $b=(1, 0)$, $c=(0, -1)$, and $d=(-1, 0)$. Then we must have:

• Either every point on $l_1$ is $\prec$ every point on $l_2$ ("$l_1\prec l_2$"), or vice versa, and

• Either every point on $l_3$ is $\prec$ every point on $l_4$ ("$l_3\prec l_4$"), or vice versa.

(To see this, consider the tame functions $f(x, y)=x+y$ and $g(x, y)=x-y$.) But now we reach a contradiction: suppose WLOG $l_1\prec l_2$ and $l_3\prec l_4$ - then $c\prec a$ since $l_1\prec l_2$, but $a\prec c$ since $l_3\prec l_4$.

As for partial orders, every function whatsoever is monotone w/r/t the discrete order; so maybe some constraint on the considered partial orders should be imposed?

As for question (2), if I read it correctly the answer is trivially "no": consider your favorite monotonic, continuous, nowhere-differentiable function $g: \mathbb{R}\rightarrow\mathbb{R}$, and let $f(x, y)=g(x)$.

Re: Alex' comment on the OP, I assume "convex" w/r/t subsets of $\mathbb{R}^2$ is meant in the usual sense.

Call a function as in the OP tame. Here's an answer to (1):

First, I claim there is no total order $\prec$ on $\mathbb{R}^2$ such that any tame function is monotone with respect to $\prec$. To see this, consider the four points $a=(0, 1)$, $b=(1, 0)$, $c=(0, -1)$, and $d=(-1, 0)$. Then we must have:

• Either every point on $l_1$ is $\prec$ every point on $l_2$ ("$l_1\prec l_2$"), or vice versa, and

• Either every point on $l_3$ is $\prec$ every point on $l_4$ ("$l_3\prec l_4$"), or vice versa.

(To see this, consider the tame functions $f(x, y)=x+y$ and $g(x, y)=x-y$.) But now we reach a contradiction: suppose WLOG $l_1\prec l_2$ and $l_3\prec l_4$ - then $c\prec a$ since $l_1\prec l_2$, but $a\prec c$ since $l_3\prec l_4$.

As for partial orders, every function whatsoever is monotone w/r/t the discrete order; so maybe some constraint on the considered partial orders should be imposed?

Re: Alex' comment on the OP, I assume "convex" w/r/t subsets of $\mathbb{R}^2$ is meant in the usual sense.

Call a function as in the OP tame. Here's an answer to (1):

First, I claim there is no total order $\prec$ on $\mathbb{R}^2$ such that any tame function is monotone with respect to $\prec$. To see this, consider the four points $a=(0, 1)$, $b=(1, 0)$, $c=(0, -1)$, and $d=(-1, 0)$. Then we must have:

• Either every point on $l_1$ is $\prec$ every point on $l_2$ ("$l_1\prec l_2$"), or vice versa, and

• Either every point on $l_3$ is $\prec$ every point on $l_4$ ("$l_3\prec l_4$"), or vice versa.

(To see this, consider the tame functions $f(x, y)=x+y$ and $g(x, y)=x-y$.) But now we reach a contradiction: suppose WLOG $l_1\prec l_2$ and $l_3\prec l_4$ - then $c\prec a$ since $l_1\prec l_2$, but $a\prec c$ since $l_3\prec l_4$.

As for partial orders, every function whatsoever is monotone w/r/t the discrete order; so maybe some other constraint should be imposed?

As for question (2), if I read it correctly the answer is trivially "no": consider your favorite monotonic, continuous, nowhere-differentiable function $g: \mathbb{R}\rightarrow\mathbb{R}$, and let $f(x, y)=g(x)$.

Re: Alex' comment on the OP, I assume "convex" w/r/t subsets of $\mathbb{R}^2$ is meant in the usual sense.

Call a function as in the OP tame. Here's an answer to (1):

First, I claim there is no total order $\prec$ on $\mathbb{R}^2$ such that any tame function is monotone with respect to $\prec$. To see this, consider the four points $a=(0, 1)$, $b=(1, 0)$, $c=(0, -1)$, and $d=(-1, 0)$. Then we must have:

• Either every point on $l_1$ is $\prec$ every point on $l_2$ ("$l_1\prec l_2$"), or vice versa, and

• Either every point on $l_3$ is $\prec$ every point on $l_4$ ("$l_3\prec l_4$"), or vice versa.

(To see this, consider the tame functions $f(x, y)=x+y$ and $g(x, y)=x-y$.) But now we reach a contradiction: suppose WLOG $l_1\prec l_2$ and $l_3\prec l_4$ - then $c\prec a$ since $l_1\prec l_2$, but $a\prec c$ since $l_3\prec l_4$.

As for partial orders, every function whatsoever is monotone w/r/t the discrete order; so maybe some constraint on the considered partial orders should be imposed?

As for question (2), if I read it correctly the answer is trivially "no": consider your favorite monotonic, continuous, nowhere-differentiable function $g: \mathbb{R}\rightarrow\mathbb{R}$, and let $f(x, y)=g(x)$.