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I've seen the following question and solution

Question When is the epigraph of a function a convex cone?

Solution If the function is convex and positively homogeneous $(f(\alpha x) = \alpha f(x)$ for $\alpha \geq 0$).

The solution I came up with was that the function must be the pointwise maximum of a finite number of linear functions, i.e.

$$ f(x) = max(f_1(x), ..., f_n(x)) $$

where $f_1$, ..., $f_n$ are linear.

This lead me to wonder, is the set of functions defined by the pointwise maximum of a finite number of linear functions equivalent to the set of convex positively homogeneous functions? If not, what is wrong with my solution above? Does this equivalence hold for a countably infinite set of linear functions?

Edit

Since this question is derived from epigraphs, I am interested in functions $f: \mathbb{R}^M \rightarrow \mathbb{R}$ and similarly for $f_1, ..., f_n$.

mgilbert
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2 Answers2

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your claim is not true. Take $f(x,y)= \sqrt{x^2 + y^2}.$

$f$ is convex and homogeneous function, but not convex piece-wise function. But any convex homogeneous function can be expressed as the pointwise supremum of linear function. See Variational analysis by Rockafellar, Theorem 8.13

It is not difficult to adopt the proof for homogenous convex function. Note that for convex homogenous function you may assume those affine functions supporting $f$ pass through origin, this makes affine functions indeed linear! Secondly you can pick just at most countable many of them since $f$ is continuous and $Q^n$ is dense in $R^n.$ Therefore any convex homogenous function, $f: R^n \rightarrow R$ can be equivalently written in the following form:

$$ f(x) = \sup_{n \in N} \{f_n(x)\} $$ where $f_n$ is linear for all $n\in N.$

Red shoes
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  • This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review – lioness99a May 30 '17 at 08:13
  • What do you mean? This is a counter-example, a convex homogeneous function which is not piecewise . So OP's claim is wrong. You recommend correct answer to be deleted but wrong answers stay here like below !! – Red shoes May 30 '17 at 08:23
  • First off, what is the 'this' you refer to? Then this should be posted as a comment underneath it. – lioness99a May 30 '17 at 08:43
  • @lioness99a OP is asking for a proof or a counter example. and I gave him/her a counter example! Do you know anything about this topic? If not , I can't convince you! Just for your information: this is the best possible answer to this question in this planet. – Red shoes May 30 '17 at 09:07
  • In the version of Variational Analysis that I am looking at 8.3 is a definition of subgradients. Is the statement of the proof you are referring to the following?

    A proper, lsc, convex function $f : \mathbb{R}^n \rightarrow \overline{\mathbb{R}}$ is the pointwise supremum of its affine supports

     – mgilbert May 30 '17 at 13:08
  • I am a bit unclear how $f(x,y) = \sqrt{(x^2 + y^2)}$ is a counterexample. This defines a cone, which is the limit of $n$-pyramid's who's bases are defined by regular polygon's with $n$ sides. These pyramids can be defined by the pointwise maximum of countably infinite planes through the origin? – mgilbert May 30 '17 at 14:24
  • @mgilbert if you look at your own question, you never talked about limit, You said is the set of functions defined by the pointwise maximum of a finite number of linear functions equivalent to the set of convex positively homogeneous functions? YOU said finite number. So this is a counterexample. but if you omit the word finite and consequently consider $\sup$ instead of $\max$ then it is true. look at the theorem. The claim hold for infinitely linear function but you must change $\max$ to $\sup$ since there is no guarantee the Maximum exists – Red shoes May 30 '17 at 18:42
  • Yes, the second part of the question, albeit not bold, asks about countably infinite sets of functions. I just wanted some clarification since this did not address that part. In relation to the theorem, could you please clarify on my first comment? – mgilbert May 30 '17 at 22:52
  • @mgilbert yes exactly that theorem you pointed out. (Theorem 8.13) I added some sentences to my answer, it will answer your first comment. – Red shoes May 30 '17 at 23:42
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You see in a moment that, in $\mathbb{R}^n$ with $n\geq 2$, your result cannot be correct.

Consider, for example, the convex and positively homogeneous function $f(x) = |x|$. This function is of class $C^1(\mathbb{R}^n\setminus\{0\})$, whereas the pointwise maximum of a finite family of linear functions cannot be differentiable on that set. (Roughly speaking, look at the graphs of the $f_i$: at "lines of junction" the function is not differentiable.)

Rigel
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  • I'm slightly confused by your counterexample here, if I take $max(x, -x)$, which is the pointwise maximum of two linear functions, this equals $|x|$? – mgilbert May 24 '17 at 16:41