What is the difference among pseudoconvex, quasiconvex, and convex functions?

Question

I am looking for a simple intuitive explanation of pseudoconvex, quasiconvex, and convex functions (or sets) rather than rigorous mathematical definition

Answer 1

The epigraph of a convex function is a convex set. The epigraph is the set of points above and including the graph of the function. So we can understand convex functions by looking at their graphs. Convexity is a geometric property so it's worth looking at.

A convex set is one such that if I take any two points in the set, the line segment connecting these two points is convex. So if I have two real vectors $x$ and $y$ the convex combination of $x$ and $y$ is $tx + (1-t)y$ for $0 \leq t \leq 1$, which is just that line segment in vector notation. This captures the idea that the boundary of a convex set is all sort of curving in the same way. For a convex function that direction will always be upward. Note that the function might be going down, like say $y=-x$, but that function is convex because it's "curving" upward as it has zero curvature. One of the most important and simple convex sets is a half plane. A circle is also convex set, but is not the epigraph of a single variable function.

A function is quasiconvex if all the sublevel sets are convex. The sublevels sets are domain of the graph when restricted to points below some value. It sounds a little complicated but the graph is once again the key. Take the graph and draw some horizontal line at some place on the graph. If we restrict our attention to the points on the graph below that line only some values of $x$ will be represented. If all those values form a convex set, the sublevel set associated with that line is convex. Now check every possible horizontal line and if they're all convex, the function is qusiconvex.

Intuitively this means the graph is sort of bounded on one side by a hyperplane but it's not too wiggly. Another way to think about it is that you can travel along lines in the domain of the function without leaving its range. Say I take the graph of $y=x^2$ cut it in half vertically, and swap the two halves. This function changes curvature at the pinch point so it's not convex, but the sublevel sets are all intervals around the origin so it is quasiconvex.

Now pseudoconvex functions are similar to quasiconvex function but we further impose that the must be smooth on those sublevel sets. Smoothness is related to differentiability of the function but geometrically it means more or less what it says, no sharp edges or breaks. So the example I gave with the two halves of $y=x^2$ swapped would not be pseudoconvex because it has that pinch point, but if I round it off there instead it would be. They're nice because you can again bound them by half-planes but this time we can use the power of calculus to help.

This gives us a chain of implications with convex implying pseudoconvex, and pseudoconvex implying quasiconvex, but not that other way around. You may have noticed we're bounding things by half-planes a lot like the lines for the sublevel sets or the curvature condition and that's not a coincidence. It's often how you find extrema, which is important for optimization problems. I hope that at least gives you some picture of what's happening.

Answer 2

I will follow an analytical approach. I consider the definition of convexity (by Jenssen's inequality) known. I also take all functions defined on $\mathbb{R}^{n}$, because I am not sure if and how pseudoconvexity and quasiconvexity is defined in infinite-dimensional spaces.

Now a differentiable function is defined as pseudoconvex if-f for all $x,y\in X$ we have that

$<\bigtriangledown{f(x)}, y-x>\,\geq\,0$ implies $f(y)\geq\,f(x)$.

If a function is convex, then it is pseudoconvex. The proof is simple. The subdifferential of a differentiable convex function is its gradient. But

$\partial{f(x)}=\left\{x^{\star}:f(y)-f(x)\geq\,<x^{\star},\,y-x> ,\forall y \right\} $ and hence

$f(y)-f(x)\,\geq\,0$ if $<\bigtriangledown{f(x)},\,y-x>\,>\,\geq\,0$.

The inverse is not true as we can see in the example $f(x)=x+x^{3}$.

Now we have that if f is pseudoconvex we have that

for any $x,y,\, f(y)<f(x)\,\Rightarrow\,\bigtriangledown{f(x)}(y-x)<0$. Quasi-convexity is defined (in an equivalent to the definition way) by :

see https://hal.archives-ouvertes.fr/hal-00678475/document

For any $x,y$ we have $f(y)<f(x)\,\Rightarrow\,\bigtriangledown{f(x)}(y-x)\,\leq\,0$.

So pseudoconvexity clearly implies quasiconvexity.

An example of a function which is quasiconvex but not pseudoconvex is

$f(x)=\dfrac{e^{x}}{x^{2}+1}+\dfrac{1}{e^{x}}$.