Sparsity is a very useful property of some Machine Learning algorithms. Such an algorithm yields a sparse result when, among all the coefficients that describe the model, only a small number are non-zero. This is typically associated with interesting properties such as fast evaluation of the model (see the reduced set methods for obtaining sparse kernel expansions), fast optimization (e.g. in SVM, many algorithmic approaches exploit this fact), statistical robustness (sparsity is usually associated to good statistical performance), or other computational advantages (e.g. ability to compute full regularization paths, for example in LASSO-style regression).

However I have not seen a clear explanation of this phenomenon. My feeling (I have no proof but it seems intuitively reasonable) is that sparsity is related to the regularity of the criterion to be optimized.

More precisely, the less regular the optimization criterion, the more sparse the solution may end up being.

The idea is that, for sparsity to occur, the value 0 has to play a special role, hence something unusual has to happen at the value 0. This something can be a discontinuity of the criterion or of one of its derivatives.

If the criterion is discontinuous at 0 for some variables, the solutions might get "stuck" in this value (provided it is a local miminum of course). If instead, the criterion is continuous but has a derivative which is discontinuous at 0, it means that the criterion is V-shaped at 0, so that solutions might be "trapped" at this point. If we continue the reasoning, we see that the "attraction" of the point 0 is less and less effective as the regularity increases. When the function is twice differentiable everywhere, there is not any reason for the solution to be "trapped" at 0 rather than ending up somewhere else.

This reasoning partly explains the sparsity of SVMs. Indeed, the standard L1-SVM (hinge loss) have a discontinuous criterion, while for L2-SVM (squared hinge loss), the criterion has a discontinuous derivative and finally, for the LS-SVM (squared loss), the criterion is twice differentiable. It turns out that the most sparse is the L1 version and then the L2 version, while for LS-SVM there is no sparsity at all.

The same reasoning applies when one compares penalized least squares regression: when the penalization is the L2-norm of the weights, there is no sparsity, while with the L1-norm, the sparsity occurs, and for the L0-norm there is even more sparsity.

I am wondering whether there is any mathematical treatment of these issues anywhere in the Machine Learning litterature. If anyone has a pointer, please let me know.

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