# When Biconvex function is Pseudoconvex function?

Is a Biconvex function f(x,y):=yg(x), (where g is a convex function, y>=0), Pseudoconvex function?

I am a bit confused by the wording. The title says when biconvex is pseudoconvex, but in the description asks whether a biconvex function is pseudoconvex. I am answering assuming you are asking the latter, i.e, whether biconvex functions are pseudoconvex.

I guess, the answer is not always. Consider the function $$f(x,y) = y x^2$$. Firstly, recall that every pseudoconvex function is a quasiconvex function (definition: all sub-level sets are convex sets). So if we show that $$f(x,y)$$ is not quasiconvex, then $$f(x,y)$$ is not pseudoconvex. Consider the points:

1. $$x_1$$ = -0.5, $$y$$ = -1, so $$f(x_1, y)$$ = -0.25.
2. $$x_2$$ = 0.5, $$y$$ = -1, so $$f(x_2, y)$$ = -0.25.
3. $$x_3$$ = 0, $$y$$ = -1, so $$f(x_3, y)$$ = 0.

Consider the sub-level set $$C = \lbrace{(x,y) | f(x, y) <= -0.2 \rbrace}$$. Clearly $$(x_1, y), (x_2, y)$$ lie in $$C$$, whereas their convex combination $$(x_3, y) = \dfrac{1}{2}(x_1, y) + \dfrac{1}{2}(x_2, y)$$ does not belong to $$C$$, so $$C$$ is not a convex set.

The example in the other answer (by batwing) is not a counterexample to the question as it currently stands because it does not satisfy the conditions that $$g$$ is non-decreasing and $$y\geq 0$$. (These conditions were added after the answer was posted.) To satisfy these two conditions, we need to restrict the domain to $$x\geq 0$$ and $$y\geq 0$$, which rules out the three points considered in the answer.

An easy counterexamples is as follows. Take $$g(x)=x$$, then $$f(x,y)=xy$$. There is a stationary point at $$(0,0)$$, but it is not a global minimizer as $$f(x,y)<0$$ for $$x<0$$ and $$y>0$$. This tells us that $$f$$ is not pseudoconvex, because for pseudoconvex functions stationary points are global minimizers.