My book states that:
$f$ is a quasiconcave function on $U$ if for all $x,y \in U $ and $t \in [0,1]$:
$f(x) \geq f(y) \implies f(tx + (1 - t)y) \geq f(y)$
$f$ is a quasiconvex function on $U$ if for all $x,y \in U $ and $t \in [0,1]$:
$f(y) \leq f(x) \implies f(tx + (1 - t)y) \leq f(x)$
Now take a look to the graph below $f(x)=xy$. I know it should be quasi-concave, but given the definitions above I seems to be both quasi-convave and quasi-convex. What am I doing wrong?
Thanks!
I see incorrect answers on this topic everywhere and nobody is correcting them. In the case of z= f(x,y)= xy, the graph that you draw ( rectangular hyperbola) is a level curve and not the function of the graph. The graph is originally in 3-D. To check its properties ( mainly quasi-concavity and quasi- convexity), level curves are used. I see people getting confused among the two.
Property of monotonicity is a property of functions and not of level curves.
F(x,y) = xy is therefore not quasiconvex and only quasiconcave ( because the upper contour set is a convex set).
Nothing is wrong. Every monotone function is both quasiconvex and quasiconcave.
Indeed, the definition of quasiconvexity amounts to saying that on every closed interval, the function attains its maximum at an endpoint. Same for quasiconcavity, except replace maximum with minimum. Every monotone function has both of these properties.
The function $f(x,y) = xy$ is neither quasiconcave nor quasiconvex and you can show that by solving the bordered Hessian.
For a formal definition, let $f(x,y)$ be a function with continuous partial derivatives and continuous cross partial derivatives in a convex set $S$. Let $D_r$ be the determinant of its $rth$ order bordered Hessian. If $f$ is quasiconcave then $D_1 \leq 0 $ , $D_2 \geq 0$ and $D_n \leq 0$ if $n$ is odd and $D_n \geq 0$ if $n$ is even for all $x$ in $S$.
