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Consider the following function: $f: \mathbb{R}^n\rightarrow\mathbb{R}$ $$f(x)=\max_{p\in\mathcal{P}} x^Tp$$ where $\mathcal{P}\in\mathbb{R}^n$ is a nonempty and compact set.

It is easy to see that $f(x)$ is a piecewise convex function but is it true that $f(x)$ has gradients almost everywhere? Any references and counterexamples are very welcome.

Mathexx
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1 Answers1

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It is not actually the case that $f$ is always a piecewise convex function. It depends completely on the set $\mathcal{P}$.

Onto your question, I assume by "almost everywhere" you mean that the set of non-differentiable points has measure $0$.

So, if $\mathcal{P}$ has measure $0$, then $f$ is differentiable "almost everywhere".

If $\mathcal{P}$ has non-zero measure, then it may be the case that $g$ is non-differentiable on a set with non-zero measure (some subset of $\mathcal{P}$ that has non-zero measure, perhaps $\mathcal{P}$ itself).

NicNic8
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  • But what does it mean for $\mathcal{P}$ to have zero measure? It's not a subset of $\mathbb{R}$, it's just a general set. – Sambo Aug 02 '19 at 00:50
  • I have modified the question and I think convexity should always hold this time. – Mathexx Aug 02 '19 at 06:25
  • @Sambo measure is not restricted to subsets of $\mathbb{R}$. Measure is the "size" of the set. It's a way of measuring how large it is. Some examples of measure are length, area, volume. Note that a set of isolated points has measure 0. – NicNic8 Aug 03 '19 at 00:36
  • @NicNic8 I know that, but prior to the edit $\mathcal{P}$ was an arbitrary set, for which it doesn't make sense to talk about measure. What's the measure of, say, $\aleph_1 \backslash \aleph_0$? – Sambo Aug 03 '19 at 13:32
  • @Sambo Even if it's arbitrary, if I can show that there are some sets where f is non-differentiable with non-zero measure, then it isn't differentiable "almost everywhere".

    If you have another definition for "almost everywhere" please specify it.

     – NicNic8 Aug 03 '19 at 23:05