# Why is convexity more important than quasi-convexity in optimization?

In the mathematical optimization literature it is common to distinguish problems according to whether or not they are convex. The reason seems to be that convex problems are guaranteed to have globally optimal solutions, so one can use methods such as gradient (steepest) descent to find such solutions, but I am not convinced.

For example, the function $$|x|^{0.001}$$ is not convex (see the yellow-shaded area in the picture below) but it has a single global minimizer.

So why is convexity so important in optimization? Why is e.g. quasiconvexity often not enough?

• The function above is amenable to convex optimization after the transformation $x \mapsto 1/x$.
– Emre
May 17, 2012 at 23:34
• In a few words, convexity implies that a local minimum is a global minimum. May 18, 2012 at 0:18
• Tyrell Rockafellar: "...the great watershed in optimization isn't between linearity and nonlinearity, but convexity and nonconvexity".
– Emre
May 18, 2012 at 3:51
• Your example function is quasiconvex. May 18, 2012 at 15:06
• Fromthe Wikipedia article on quasiconvex functions one can infer the advantages and disadvantages of this class of functions relative to convex ones. One I find particularly striking is that the sum of two quasiconvex functions need not be quasiconvex.
– user856
May 18, 2012 at 15:57

There are many reasons why convexity is more important than quasi-convexity in optimization theory. I'd like to mention one that the other answers so far haven't covered in detail. It is related to Rahul Narain's comment that the class of quasi-convex functions is not closed under addition.

Duality theory makes heavy use of optimizing functions of the form $f+L$ over all linear functions $L$. If a function $f$ is convex, then for any linear $L$ the function $f+L$ is convex, and hence quasi-convex. I recommend proving the converse as an exercise: $f+L$ is quasi-convex for all linear functions $L$ if and only if $f$ is convex.

Thus, for every quasi-convex but non-convex function $f$ there is a linear function $L$ such that $f+L$ is not quasi-convex. I encourage you to construct an example of a quasi-convex function $f$ and a linear function $L$ such that $f+L$ has local minima which are not global minima.

Thus, in some sense convex functions are the class of functions for which the techniques used in duality theory apply.

Sophisticated optimization methods are developed to deal with problems where it is not obvious what the solution is. In general, we may not know much about the behavior of the cost function, so we cannot decide on an a-priori base whether there is a unique minimum or not. We need optimization methods to find the or a minimum.

For convex problems, the global behavior is determined by purely local properties. If a function is everywhere locally convex, it is globally convex too. So for such problems, local solutions are sufficient, which make them a lot easier to deal with.

• Thanks Michael, but when you said: For convex problems, the global behavior is determined by purely local properties. If a function is everywhere locally convex, it is globally convex too. So for such problems, local solutions are sufficient, which make them a lot easier to deal with. Wouldn't all this hold as well if I replace the word convex by quasiconvex? If so, why is convexity more important than quasiconvexity? May 19, 2012 at 18:44
• Almost. Every monotone function from the reals to the reals is quasiconvex. But a function can be locally monotone without beig globally monotone. I think it works with strictly quasi-convex functions though. May 19, 2012 at 19:44

Gradient methods are usually descent methods, so the cost function decreases at points that are non-optimal points. Many such methods also have the property that any accumulation point of the sequence generated cannot have a non-zero gradient (call these 'guaranteed' descent methods). (This awkward wording allows the cost function to be non-differentiable at a minimum.)

Here is a sufficient condition that applies to the problem above: Let the initial point be $x_0$, and let $L_{x_0} = \{x | f(x) \leq f(x_0) \}$. Suppose $L_{x_0}$ is compact and the gradient is non-zero at all points other than the minimizer. Then any 'guaranteed' descent methods will converge to the minimizer. (I'm assuming that the cost function $f$ is $C^1$ at non-optimal points.) The problem you have above satifies these conditions.

The second question is more of a discussion topic. Some points to consider:

Duality is another big plus for convex problems. Dual problems are often more amenable to solutions.

Problems can often be transformed into convex problems, posynomials being a good example.

Often convexity of the level sets is the important criterion. Various generalizations (eg, pseudo-, and quasi-convexity) can extend the usefulness of convexity to broader settings.

Like someone else said, convex problems often have a dual, which lets an algorithm solve the problem from two sides and declare the solution when the duality gap is sufficiently close to 0 (when the lower and upper bound on the solution converge). There are also fairly efficient methods and algorithms that can solve canonical convex problems which do not extend easily to general quasiconvex problems.

For strict quasiconvex problems, a gradient descent method should be sufficient, but it's a question of algorithmic efficiency. The algorithms for canonical convex problems are just much more efficient.

At least, that's my impression of why there's so much focus on convexity. However, one could certainly argue that quasiconvexity is often overlooked when it would be useful...

I'm not an expert, but just from eyeballing your function you can see that gradient descent would not converge to the global minimizer. Gradient descent looks for places where the gradient (which is just the derivative for 1-dimensional functions like the one you provide) is zero. In a convex function, as you approach the minimum, the gradient approaches zero (and vice versa). But you can see that, as you approach the minimum in your function, the derivative is getting further and further from zero.

If you try to do gradient descent on that function, you'll actually move further and further away from the minimum, since the gradient is minimized as you approach (positive or negative) infinity.

So at least one reason convexity is so important in optimization is that the global minimum is also the unique critical point (place where the gradient is zero), which allows you to search for one by searching for the other. In some sense, you pick a point, look around a little bit, and move in the direction that makes your gradient smaller. Repeat this enough and you'll get to a really small gradient, which corresponds to being close to the minimum.

• Generally gradient methods use some form of step size to ensure a descent. May 18, 2012 at 4:09
• -1: This is not how gradient descent works. In short, you perform successive line searches along the ray $x_n - \alpha \nabla f(x_n)$ with $\alpha$ positive, so you never go up the gradient, and use sufficient decrease criteria to ensure that $f(x_n-\alpha\nabla f(x_n))$ is (sufficiently) less than $f(x_n)$. So you do approach the minimum even on the nonconvex function in the question.
– user856
May 18, 2012 at 15:45
• I wrote that poorly. It should be "use sufficient decrease criteria to ensure that the value at the next iterate $f(x_n−\alpha\nabla f(x_n))$ is (sufficiently) less than the current value $f(x_n)$."
– user856
May 18, 2012 at 16:09
• MartianInvader, did you mix up gradient descent with the Newton method (en.wikipedia.org/wiki/Newton%27s_method_in_optimization)? The Newton method would have exactly the problem you are describing... (Indeed, there is variant of gradient descent which uses the Newton method to calculate the "ideal" step length. In fact, this would not work here... Of course, this only makes sense for a multi-dimensional function.) May 21, 2012 at 18:25