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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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Wang posted an answer that every continuous convex function on a finite dimensional Euclidean space is almost everywhere strictly differentiable and provided some references and I would like to thank him for the very useful answer. The arguments for the above result can be found in the paper Borwein, J. M.; Moors, W. B., Essentially smooth Lipschitz functions. J. Funct. Anal. 149 (1997), no. 2, 305–351:

  1. On separable Banach spaces, the continuous convex functions are strictly differentiable everywhere except on a Haar-null set (page 312)

  2. In finite dimensions, the Haar-nulls sets coincide with the Lebesgue null sets (page 322).

For other facts on subdifferential, see S. X. Wang, Fine and pathological properties of subdifferentials (thesis)

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  • $\begingroup$ On can find in Qi, L., The maximal normal operator space and integration of subdifferentials of nonconvex functions, Nonlinear Anal. 13 (1989), 1003-1012. $\endgroup$ – Xuan Duc Ha Truong Mar 30 '18 at 4:26

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