Consider a function of two variables $f(x,y)$, where $x\in [0,\infty)$ and $y\in [0,1]$, which is a convex combination of two functions $f_a(x)$ and $f_b(x)$. Both, $f_a(x)$ and $f_b(x)$, are increasing and concave in variable $x$, saturate to $1$ as $x\to \infty$, and are such that $f_a(x)\leq f_b(x)$ for all $x$. Thus, the function $f$ has the form$$ f(x,y) = y f_a(x) + (1-y) f_b(x).$$
The conditions on $f_a$ and $f_b$ imply that $f$ is also positive, concave, increasing, and saturates to $1$ asymptotically in $x$.
We are interested in the relation between $x$ and $y$ when the partial derivative of $f$ with respect to $x$, denoted by $f'_x(x,y)$ is equal to a fixed positive constant $c$ i.e., what is $x(y)$ such that $f'_x(x,y) = c$? More specifically, we are interested in knowing the curvature of $x(y)$.
An example that illustrates this:
$$
f_a(x) = 1-Ae^{-x} \hspace{1em} f_b(x) = 1-Be^{-x}
$$
such that $A>B$. Then $f'_x(x,y) = c$ is given by —
$$
(yA+(1-y)B)e^{-x} =c
$$ From which we can solve for $x(y)$ to get —
$$
x(y) = \ln\bigg(\frac{yA+(1-y)B}{c}\bigg)
$$
This is concave in $y$. Similarly, one can think of examples where $x(y)$ is convex in $y$. However, are there functions $f_a$ and $f_b$ such that resulting $x(y)$ obtained is neither concave nor convex in $y$? How could we go about finding such functions?
Edit: After unsuccessfully trying of cook up (counter)examples, I am now pursuing the following approach:
Since we are interested in the relation between $x$ and $y$ when $f'_x(x,y) = c$, we can write
$$ y f_a'(x) + (1-y) f_b'(x) = c$$
which simplifies to give
$$ y(x) = \frac{c – f_b'(x)}{f_a'(x)-f_b'(x)}.$$
This is $y(x)$, and to learn about its curvature, we need to study its second derivative. The second derivative is a messy looking equation, but it can be seen that the sign of the second derivative will depend of the sign and magnitudes of terms like $c-f_b'(x)$, $f_a'(x)-f_b'(x)$, and the third derivatives of $f_a(x)$ and $f_b(x)$. So it seems likely that one could think of constraints such that $y''(x)$ will be positive for some values of $x$ and negative for some. But then, would the same hold $x(y)$? (we will have to worry about the existence of the inverse under those constraints, of course.)
Best Answer
After trying for some time, I came up with the following example. It is not a particularly nice one, but unfortunately it is the best I've been able to find.
Take $f_b(x)=1$ for all $x$ and $$f_a(x)=1-\frac{3\pi}{8\sqrt 7}+\frac{1}{56}\left(6\sqrt 7\arctan\left(\frac{2x-3}{\sqrt 7}\right)+14\ln(x)-7\ln\left(x^2-3x+4\right)\right).$$ Now, it is an easy computation that $$\lim_{x\to\infty}f_a(x)=1.$$ The derivative of $f_a$ in $(0,\infty)$ reads $$f_a'(x)=\frac{1}{x\left(x^2-3x+4\right)}$$ which is positive and strictly decreasing in $(0,\infty)$. Thus, $f_a$ is increasing and concave.
We choose now $c=\frac{1}{4}$ and get $$y(x)=\frac{1}{4f'_a(x)}=\frac{x^3-3x^2+4x}{4},$$ $$y'(x)=\frac{3x^2-6x+4}{4}>0$$ and $$y''(x)=\frac{3}{2}(x-1).$$ Thus, $y(x)$ is strictly increasing but neither convex nor concave in any open neighbourhood of $x=1$.
Finally, we can realize that the following Lemma holds:
Lemma: Let $g:(0,\infty)\to\mathbb{R}$ be twice differentiable and strictly increasing. Then:
Proof:
Applying this Lemma to $y(x)$ we can conclude that $x(y)$ is strictly increasing but neither concave nor convex in $(0,1)$.