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

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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.

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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.

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