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Suppose $F(x,y;k)=f(x,y)+kg(x,y)=0$ uniquely defines the solution $y(x;k)$ for $x\in \mathbb{D}$, a compact domain, and $0\leq k \leq 1$ is a parameter. We know that for $k=0,1$, $y(x;0)$ and $y(x,1)$ are strictly increasing and strictly concave functions.

Then, can we say something about the property of $y(x;k)$ for $0<k<1$? For example, is $y(x;k)$ strictly increasing and concave?

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At least each $y(x;k)$ is strictly increasing if $\partial_yF(x,y;k)\neq0$ for $k=0,1$ (so that the implicit function theorem can be applied to both $y(x;0)$ and $y(x;1)$). This can be seen in the implicit derivative $$ y'(x;k) = -\frac{\partial_x F}{\partial_y F} = -\frac{\partial_x f+k\partial_x g}{\partial_y f+k\partial_y g}. $$ The denominator is nonvanishing by the assumption that $\partial_yF(x,y;k)\neq0$ for $k=1,2$. Since both $y(x;0)$ and $y(x;1)$ are strictly increasing, the numerator and denominator both have constant sign, and each $y(x;k)$ is strictly increasing.

Strict concavity is trickier. If some of the intermediate functions failed to be strictly concave, there would be $x$ and $k$ so that $\partial_ky''(x;k)=0$ (since $y''(x;k)<0$ for $k=0,1$). The resulting equation seems messy, so I won't push it any further.

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    $\begingroup$ Thanks for your solution. I am still a bit confused on the increasing property. The implicit derivative holds on the curve $(x,y(x;k))$, depending on $k$. By knowing the sign on the curves $(x,y(x;0))$ and $(x,y(x;1))$, how do we induce the sign on the curve $(x,y(x;k))$ [as e.g., $\partial_x f(x,y(x;k))$ is now evaluated at $(x,y(x;k))$ instead]? $\endgroup$
    – Allen
    Commented Aug 22, 2014 at 21:55
  • $\begingroup$ @Allen, I'm sorry for being slow. (Your first comment did reach me.) I'm a bit confused myself, frankly. Assuming the inverse function theorem is applicable throughout, the denominator can't vanish and thus has constant sign. I claimed that $\partial_xf+k\partial_xg$ having constant sign for $k\in\{0,1\}$ implies constant sign for $k\in[0,1]$. This seems to require more assumptions on $\partial_xf$ and $\partial_xg$, since they are indeed evaluated on different curves for different $k$. But if you work with the whole curve family $F=c$, $c\in\mathbb R$, at once, my argument works better. $\endgroup$
    – Joonas Ilmavirta
    Commented Aug 23, 2014 at 18:47

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