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.