If $f_x(x,y)>0$, $f_{xx}(x,y)<0$, $f_y(x,y)>0$, $f_{yy}(x,y)<0$ can $f_{xy} $ change sign? That is, suppose we have a continuous and (at least) twice differentiable function $f(x,y)$ which is

*

*increasing but concave in each of its individual arguments$^*$,

*

*Note that these derivatives are all assumed to be non-zero



*the domain of $f$ is $[0,C]$ with $C>0$ (if changing the domain matters please feel free to let me know. I'm interested)

Is it possible for the cross partial derivative to change signs (? (i.e. $\frac{\partial f}{\partial x \partial y} >0$ at some point but $<0$ at another point)

*

*If it is possible, what if we look along a line (i.e. fix an $x$ or $y$ value and look along the line from that point) (just answering the bold question is an acceptable answer)

If it is possible, can someone provide an example function? If not, I am looking for a proof (or hints for a proof, or some intuition)
*By increasing but concave in each argument I mean  $\frac{\partial f}{\partial x} >0$ and $\frac{\partial^2 f}{\partial x^2}<0$, and similar for $y$
Edit: I made question more slightly general. I apologize if someone was already typing an answer. (It is more general because if this edited question is true, then so is the original)
 A: For convenience, I will work on the domain $x,y > 0$. Of course you can shift this solution to get one in your desired domain.
There are no obvious relationships between the second order partials of a function in general that I'm aware of, so I'm going to guess there is an $f$ with the properties you desire.
Guess a solution of the form $f(x,y) = F(xy)$ where $F:\mathbb{R}_+ \to \mathbb{R}$. Then we compute
$$f_x = F'(xy)y,\qquad f_y=F'(xy)x,$$
$$
f_{xx} = F''(xy)y^2, \qquad f_{yy} = F''(xy)x^2
$$
$$
f_{xy} = F''(xy)xy + F'(xy).
$$
In order that $f_x, f_y > 0$ and $f_{xx}, f_{yy} < 0$, we restrict to $F$ such that $F' > 0$ and $F'' < 0$.
Note that $f_{xy}$ is a function of $xy$, so we really just need that $F''(t) t + F'(t)$ changes sign. Near $t=0$ (but positive), we have $F''(t) t + F'(t) \approx F'(t)> 0$, but when $t$ is large, we hope that $t F''(t)$ dominates, making the sum negative, which will be the case if $F'$ and $F''$ are approximately the same magnitude. At this point, a particularly easy choice stands out, $F(t) := -e^{-t}$ which has $F'(t) = e^{-t} > 0$ and $F''(t) = -e^{-t} < 0$, and $$F''(t) t + F'(t) = -e^{-t} t + e^{-t} = e^{-t}(1-t),$$
which changes sign at $t=1$.
Therefore, $f(x,y) = -e^{-xy}$ satisfies all your conditions and has $f_{xy} < 0$ if $xy > 1$, and $f_{xy} > 0$ if $xy < 1$.
By chance, this example also answers your second question. Even if you restrict to a line $x = x_0$ you can still choose $y$ with $x_0 y < 1$ or with $x_0 y > 1$ (and similar for fixed $y=y_0$).
