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$$f(x,y) = \frac{ax+by}{1+cx+dy}; a,b,c,d>0$$

Also please suggest an easy way to determine the convexity of such functions? I would also appreciate if I can numerically verify it quickly (instead of analytical methods).

Sagnik
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    how would it be possible to numerically verify something that must hold at uncountably infinitely many points? – MT_ Mar 12 '16 at 04:04

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The function is undefined along the line $1+cx+dy=0$ with all its values negative on one side of the line and positive on the other. So if you pick points on the surface on either side of that line, say $(x_1,y_1)$ and $(x_2,y_2)$ then the line segment connecting $(x_1,y_1,f(x_1,y_1))$ and $(x_2,y_2,f(x_2,y_2))$ cannot lie entirely above the surface. The surface is asymptotic to the plane $1+cx+dy=0$ so any segment connecting points of the surface on opposite sides of that plane necessarily pass through the plane.

John Wayland Bales
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    Even with all variables and parameters positive, the function is simply not convex, not even necessarily convex in one variable with the other one fixed. For example, for $y>0$ fixed, the second derivative $\frac{\partial^2f}{\partial x^2} f(x,y)$ has the same sign $cby-a(1+dy)$, which could be either positive or negative. – Greg Martin Mar 12 '16 at 04:34
  • @GregMartin Please note that a,b,c,d are constants. For the double derivative to be positive $y>\frac{a}{cb-ad}$. So a can be made very small so that the inequality always satisfies. – Sagnik Mar 15 '16 at 04:52
  • @JohnWaylandBayles Please assume $0 \leq x,y \leq 1$. – Sagnik Mar 15 '16 at 04:59
  • Is $a$ a constant, or can $a$ be made very small? That's self-contradictory. The point is, there are (many) values of $a,b,c,d$ so that the function simply is not convex. – Greg Martin Mar 15 '16 at 05:39
  • Thanks, @GregMartin. Maybe I should rephrase my question later. – Sagnik Mar 16 '16 at 17:50
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The function is not convex. If it was, you would have that

$f((1-t)(x_1,y_1)+t(x_2,y_2))\leq (1-t)f(x_1,y_1)+tf(x_2,y_2)$ for any $(x_1,y_1),(x_2,y_2)$ and $t\in [0,1].$

Take the pairs $(\alpha,1)$ and $(1,\alpha)$ for $\alpha\neq 1.$ If you do the calculations you will arrive at a non-negative quadratic polynomial of second degree in terms of $t.$ This polynomial has as roots the numbers $0, 1.$ Since the coefficient of $t^2$ is positive (it is equal to $(1-\alpha)^2(ac+bc+ad+bd)$), you arrive at a contradiction.

Andreas K.
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