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 $D$.

Share:
3,170

Related videos on Youtube

Taylorien
Author by

Taylorien

Updated on August 13, 2022

Comments

  • Taylorien
    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

    • copper.hat
      copper.hat over 9 years
      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).
  • Taylorien
    Taylorien over 9 years
    thank you, i start to think that this book is actually THE reference for convex analysis.
  • copper.hat
    copper.hat over 9 years
    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.