# Delta-convex functions and inner products

A delta-convex (d.c.) function is one which can be written as the difference of two convex functions.

The space of d.c. functions includes all C2 functions, and is interesting because it allows many notions from differential geometry to be generalized to objects with "corners". For example, the "tangent cone" of directional derivatives exists at every point, and is actually a tangent plane except on a set of measure 0 (the corners). This means that the gradient exists except on a set of measure 0, and indeed so does the Hessian.

In general, there is no canonical way to decompose a d.c. function into a convex and concave part. However, there are some notable exceptions (from here on, "decompose" means "decompose in a canonical way"):

• f : ℝ → ℝ is d.c. iff it is the integral of a function of locally bounded variation. Since such functions can be decomposed into monotone nondecreasing and nonincreasing parts, f itself can be decomposed into the integrals of said parts.

• In the particular case where f : ℝ → ℝ is C2, we simply decompose f'' into non-negative and non-positive parts, and then integrate twice.

• A (homogeneous) quadratic function f : ℝn → ℝ is completely characterized by its Hessian; f is convex (resp. concave) iff its Hessian is positive (resp. negative) semi-definite. The inner product on ℝn decomposes any symmetric linear map into positive and negative semi-definite parts, so this gives a decomposition of f into convex and concave parts.

The last two examples suggest a common theme for C2 functions: if we can decompose the Hessian into positive and negative semi-definite parts, then we can decompose the function itself into convex and concave parts.

The decomposition of symmetric linear maps is associated with an inner product on the space of said maps, defined as

(A,B) = tr(AB).

Call the set of positive semi-definite maps P (so the negative semi-definite maps are -P). Then the inner product has the property that (A,B) ≥ 0 for any A, B ∈ P with equality iff all nonzero eigenspaces of A are orthogonal to those of B.

So, saying that any symmetric map C can be decomposed into positive and negative semi-definite parts is equivalent to saying that C can be written uniquely as A - B where A, B ∈ P are orthogonal. We can find A and B by noting that A + B ∈ (P + C) ∩ (P - C), a convex set, and it has minimal norm among elements of this set. (Technically we only need this norm, and not an inner product, but the properties of the inner product feel very desirable.)

If f = g - h where g, h are convex C2 functions whose Hessians are pointwise orthogonal, then g and h are certainly unique; and it would be nice to think of this decomposition as canonical. One way to make this rigorous might be to define an inner product on C2 functions, and find g + h by analogy with the above.

My initial instinct was to integrate the pointwise inner product of the Hessians. However, this generally doesn't give a finite value, even in one dimension. Since d.c. functions can be characterized as having "gradients of locally bounded variation (LBV)" - in the sense that the restriction of such a function to any line has LBV derivative - I thought of limiting the inner product to those functions whose gradients were globally BV. Such functions look a lot like tangent cones, from a great enough distance, and feel like they should be important in their own right. However, even for these functions the inner product defined above is finite only on finite-radius tubular neighbourhoods of curves whose derivatives are themselves BV.

Now I'm looking at an even more specific space of functions, those which are actually (everywhere) equal to their tangent cone at the origin. I'm convinced that these functions are key to understanding the functions of globally BV gradient, which in turn might eventually lead to insight into general d.c. functions. But so far, I've not got anywhere.

My question, then: does anyone else have any insight into the problem of canonically decomposing d.c. functions? Maybe even know some results? Am I wasting my time completely - should I move on to trying to generalize some basic results about manifolds and the calculus of variations, which was my original goal?

Any input will be greatly appreciated.

## 1 Answer

Note that the space of decompositions is convex; i.e., if function $f$ has two decompositions, say as $a_1-b_1$ and $a_1-b_2$ then $$[t\cdot a_1+(1-t)\cdot a_2]-[t\cdot b_1+(1-t)\cdot b_2]$$ is also a decomposition of $f$.

So, you can canonize the decomposition which minimize some linear functional; for example the sum of the integrals of second derivatives of $a$ and $b$ in all directions.

This should give unique answer up to an affine function (at least in the generic case). It also seems to agree with your examples.

• Thanks very much. However, I'm not quite clear how you're defining your example functional - I'm struggling to think of an interpretation which is both finite for functions such as quadratics, whose gradient is not globally BV, and also well-defined for arbitrary functions whose gradient is globally BV. – Robin Saunders Jun 2 '13 at 15:58
• Yes, I thought about functions defined on a bounded domain. The canonization seems to depend on the choice of the domain and you may get more canonizations by taking weighted integrals. For the functions on $\mathbb R^n$, one might prove that there can be no canonical choice which invariant under translations. (It should be essentially the same argument as for non-existence of generalized limit for all {0,1}-sequences.) – Anton Petrunin Jun 3 '13 at 7:39
• Nevertheless, there are some functions for which a canonical choice does exist. For example, if f is a sum of functions of each component of x, then f can be decomposed componentwise (this covers quadratics, for example). Is that as far as we can go, or could there be a bigger space of functions with canonical decompositions? – Robin Saunders Jun 4 '13 at 23:42
• It seems that for these functions the canonization does not depend on the choice of domain, am I right? – Anton Petrunin Jun 5 '13 at 7:21
• As far as I can tell, yes (up to an affine function). – Robin Saunders Jun 5 '13 at 17:11