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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

Taylorien
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  • It 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). – copper.hat Mar 26 '14 at 16:10

2 Answers2

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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 $D$.

copper.hat
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    thank you, i start to think that this book is actually THE reference for convex analysis. – Taylorien Mar 26 '14 at 16:50
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    Yes, 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. – copper.hat Mar 26 '14 at 16:56
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It's ofcouse weaker than Rademacher theorem. But here is a direct proof with some measure theory. (Convex Functions, Monotone Operators and Differentiability first chapter , R. Phelps, Exercise 1.17) The function $x \to d^{+}f(x)(e_k)$ is pointwise limit of continous functions, so its Borel measurable. The set where a convex function $f$ in finite dimensions is not differentiable is exactly the set of points for which the gradient doesn't exists(see Exercise 1.15 (b)). Denoting the coresponding set in each direction $B_k=\{x \in D : \frac{\partial f}{\partial x_k}(x) \text{doesn't exists}\}$ is Borel measurable. But for the correspoding theorem in $\mathbb R^1$ we know that $\operatorname{meas}(B_k)=0.$ Thus from Fubini theorem $$ \operatorname{meas}(B_k)=\int_{\mathbb{R}^n} \chi_{B_k}(x)dx=\int_{\mathbb{R}^{n-1}} \left(\int_{\{\lambda e_k\}} \chi_{B_k}(x)dx_k \right)dx_1\ldots dx_{k-1} dx_{k+1}\ldots dx_n=0. $$

Petar
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