Continuous functions with convex level sets

Assume that $f:\mathbb{R}^{2}\to \mathbb{R}$ is a continuous function such that each level set $f^{-1}(c)$ is a convex set. To what extent such functions are studied?

In particular:

1. Is there a partial or total ordering on $\mathbb{R}^{2}$ such that all functions with this property, must be monotone?

2.Is it true to say that $f$ is differentiable, almost every where?

These two questions are motivated by the fact that this property for continuous functions $f:\mathbb{R} \to \mathbb{R}$ is equivalent to monoton property and every monoton function from $\mathbb{R}\to \mathbb{R}$ is almost every where differentiable.

Note: As usual, a convex set in the plane is a subset $C$ such that $\forall a,b\in C$ and $\forall t\in [0,1]$ we have $ta+(1-t)b\in C$

• What do you mean by convex? Jan 17, 2015 at 19:18
• I think a positive answer to (2) would follow from the statement: "If $A\subseteq\mathbb{R}^2$ is such that every set of the form $\{x: (x, y)\in A\}$ or $\{y: (x, y)\in A\}$ has measure 0, then $A$ has measure 0." However, I'm also fairly certain that statement is false. Jan 17, 2015 at 20:35
• @DenisSerre, only for continuous $\phi$; and such $\phi$ are differentiable a.e. Or am I missing something? Jan 17, 2015 at 21:19
• @AlexDegtyarev the usual convexity as in the plane geometry. Jan 18, 2015 at 4:53
• @DenisSerre I think this is a classic theorem that an increasing function is almost every where differentiable and we have $\int_{a}^{b}f'\leq f(b)-f(a)$. I learned it from Royden book Jan 18, 2015 at 7:48

Let's call the functions defined by Ali Taghavi to be sliced functions: a continuous function $\ f:\mathbb R^2\rightarrow\mathbb R\$ is called sliced $\ \Leftarrow:\Rightarrow\ \ \forall_{c\in\mathbb R}\ f^{-1}(c)\$ is convex.

NOTATION: $$\ [x;y]\ :=\ \{(1\!-\!t)\cdot x\ +\ t\cdot y\ :\ 0\le t\le 1\}\$$ for arbitrary $\ x\ y\in \mathbb R^n\$ and $\ n=0\ 1\ 2\ \ldots.\$ Thus $\ [x;y]=[y;x]\$ in every dimension including $\ n=1$.

THEOREM 0   Let $\ f:\mathbb R^2\rightarrow \mathbb R\$ be sliced. Then $\ f^{-1}(C)\$ is convex for every convex $\ C\subseteq\mathbb R$.

PROOF   Let $\ C\subseteq \mathbb R\$ be convex. Let $\ a\ b\in C.\$ We want to show that

$$\ [a;b]\ \subseteq\ C$$

If $\ f(a)=f(b)\$ then the above holds due to the convex part of the definition of a sliced function.

Now, assume $\ f(a)\ne f(b).\$ Then $\ f^{-1}([f(a);f(b)])\ \cap\ [a;b]\$ is a closed subset of $\ [a;b].\$ Next, consider arbitrary $\ x\ y \in [a;b]\cap f^{-1}([f(a);f(b)],\$ and $\ x\ne y.\$ If $\ f(x)=f(y)\$ then again

$$[x;y]\ \subseteq f^{-1}(x)\ \subseteq\ f^{-1}([f(a);f(b)])$$

And if $\ f(x)\ne f(y)\$ then, due to continuity of $\ f\$ (and of the nature of $\ \mathbb R$) we have

$$f([x;y])\ \supseteq\ [f(x);f(y)]$$

Thus there exists $\ w\in[x;y]\$ such that

$$f(w)\ =\ \frac {f(x)+f(y)}2$$

We see that $\ w\in(x;y)\$ belongs to the interior of the interval, and $\ w\in f^{-1}[f(a);f(b)].\$ This shows that $[a;b]\cap f^{-1}([f(a);f(b)])\$ is dense in $\ [a;b],\$ hence $\ [a;b]\subset f^{-1}(C).\$ END of PROOF

After this exercise we get:

THEOREM 1   Let $\ f:\mathbb R^2\rightarrow \mathbb R\$ be sliced. Then there exists $\ (a\ b)\in\mathbb R^2\setminus\{(0\ 0)\}\$ such that:

$$\forall_{(x\ y)\,\ (x'\ y')\,\in\,\mathbb R^2 }\ \ \left(\ a\cdot x+b\cdot y = a\cdot x'+b\cdot y'\ \ \Rightarrow\ \ f(x\ y)=f(x'\ y')\ \right)$$

PROOF   The case of a constant function is trivial. Otherwise there exists $\ h\in\mathbb R\$ such that both sets $\ f^{-1}((-\infty\;h))\$ and $\ f^{-1}((h;\infty))\$ are non-empty. Then these two sets are disjoint open half-planes, i.e. they are non-empty, open and convex, and the complement of each of them is non-empty, closed and convex. Thus there exist $\ s\ t\in\mathbb R\$ and $\ (a\ b)\in\mathbb R^2\setminus\{(0\ 0\}\$ such that $\ s<t\$ and

$$f(x\ y) = h\quad\Leftrightarrow\quad s\ \le\ a\cdot x+b\cdot y\ \le\ t$$

Then this $\ (a\ b)\$ is the one required by the theorem. END of PROOF

This is about all about the general structure of the sliced functions.

A sliced function is constant in the direction perpendicular to the vector $\ (a\ b)\in\mathbb R\setminus\{(0\ 0)\}\$ (see above), and it is monotone along that direction: let $\ \phi:\mathbb R\rightarrow\mathbb R\$ be given by:

$$\forall_{u\in\mathbb R}\ \ \phi(u)\ :=\ f\left(u\cdot(a\ b)\right)$$

Then $\ \phi\$ is monotonne. Thus each slide function is differentiable almost everywhere.

• thanks for the answer. could you please more explain on the last part of your proof: what is $\phi$ and how is your final conclusion? Jan 18, 2015 at 4:39
• Are you using some type of geometric Han Banach theorem? could you please more explain? Jan 18, 2015 at 6:19
• I'll add some explanations. This is all elementary. In this finite-dimensional problem there is no need for Hahn-Banach, it's be an overkill. However, this results actually work for any locally convex topologically linear space--your assumption is that strong. Jan 18, 2015 at 8:04
• I think you mean $\phi(u)=f(u.(a,b))$ Jan 18, 2015 at 11:05

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?

• In fact, I'm pretty sure that no partial order with all antichains countable ("c.c.c") can have the desired property; I thought I saw the proof a minute ago, but now I can't quite put it together. Jan 17, 2015 at 19:40
• I deleted a silly response to question (2) - I missed the "almost everywhere" bit. Jan 17, 2015 at 20:33
• thanks for your elegance argument for non existence of total order. Jan 18, 2015 at 4:52
• I thank you again for your very interesting idea. I am sorry that I can not accept two answer meta.mathoverflow.net/questions/1491/… Jan 18, 2015 at 12:45
• I can not resist to express my admiration again to your interesting idea. Jan 19, 2015 at 7:06