Convex function almost surely differentiable.
3,170
For a proof that doesn't rely directly on the Rademacher theorem, try Rockafellar's "Convex analysis", Theorem 25.5.
From the book: Let $f$ be a proper convex function on $\mathbb{R}^n$, and let $D$ be the set of points where $f$ is differentiable. Then $D$ is a dense subset of $(\operatorname{dom} f)^\circ$, and its complement is a set of measure zero. Furthermore, the gradient mapping $x \mapsto \nabla f(x)$ is continuous on $D$.
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Taylorien
Updated on August 13, 2022Comments
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Taylorien about 1 year
If f: $\mathbb{R}^n \rightarrow \mathbb{R}$ is a convex function, i heard that f is almost everywhere differentiable. Is it true? I can't find a proof (n-dimentional).
Thank you for any help
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copper.hat over 9 yearsIt is true. One was to show it is via the Rademacher theorem (since finite valued convex functions defined on a open set are locally Lipschitz continuous).
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Taylorien over 9 yearsthank you, i start to think that this book is actually THE reference for convex analysis.
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copper.hat over 9 yearsYes, Rockafellar is a truly excellent reference. I like Boyd & Vandenberghe as well; Stephen's curious intellect and concrete connection of concepts to nice examples adds a dimension you won't find in Rockafellar.