I am looking for an example of a function ( with adequate restrictions on its domain if necessary ) such that it is a pseudo -concave function but not a concave function.
I was reading about the pseudo concave function and concave functions but couldn't understand properly what the additional conditions on the hessian matrices would mean for the shape of the function. Hence I am searching for an example to understand better.
Although one-dimensional functions would be nice , but I am especially looking for a 3-D functions
Not every quasi-concave function is concave. Here is an example for that:
Define a function $G$ with domain $\mathbb R^+$(the positive reals) such that $G(x) = x^2$. We have that $G$ is an increasing function on $\mathbb R^+$. Therefore, $G(y) \geq G(x) \Rightarrow y\geq x$ and hence for any $t\in [0,1], ty+(1-t)x \geq x$. Since $G$ is increasing, it follows that $G(ty+(1-t)x) \geq G(x).$ Thus $G$ is quasi-concave. Also, note that $G(2)=4$ and $G(0)=0$, but that $$G(\frac{1}{2}2 +\frac{1}{2}0)=G(1)=1 \leq \frac{1}{2}G(2) + \frac{1}{2}G(0) = 2$$ This cannot be possible if $G$ is concave. So we have our required function.
EDIT: Consider, for example, the function $f(x,y) = xy$ defined on the set of pairs of nonnegative real numbers. This function is quasiconcave (its upper level sets are the sets of points above rectangular hyperbolae), but is not concave (for example, $f(0, 0) = 0, f(1, 1) = 1$, and $f(2, 2) = 4$, so that $f(\frac{1}{2}(0, 0) + \frac{1}{2}(2, 2)) = f(1, 1) = 1 < 2 = \frac{1}{2}f(0, 0) + \frac{1}{2}f(2, 2))$.