It is known that:
(i) There exists a locally Lipschitz function on an Euclidean finite-dimensional space which is almost everywhere (almost everywhere means "except a set with Lebesgues measure zero") Frechet differentiable but is not almost everywhere strictly differentiable.
(ii) Any convex function on an Euclidean finite-dimensional space is strictly differentiable except a set of the first category.
(iii) There exists a set of the first category on $\mathbb R$, the complement of which has Lebesgues measure zero.
I have a question: Does there exist a convex function on an Euclidean finite-dimensional space which is not almost everywhere strictly differentiable?
Note that for a convex function, strictly differentiable at a point means that its subdifferential of convex analysis reduces to a singleton.
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