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Given:

  • $f$ and $g$ are lower-semicontinuous proper convex functions,
  • $x \in \text{ri dom}(f) \cap \text{ri dom}(g)$,
  • $h = f+g$,
  • $p \in \partial h(x)$,

Prove that there exist some $s \in \partial f(x)$ and $t \in \partial g(x)$ such that $p = s+t$, without using duality theory.

I've seen this using duals, but I'm looking for a proof relying solely on the primal definitions of subdifferential $$ s \in \partial f \;\; \Leftrightarrow \;\; f(y) \ge f(x) + s^t(y-x) \;\; \forall \;\; y$$ and convexity $$ f(\beta a + (1-\beta) b ) \le \beta f(a) + (1-\beta) f(b) \;\; \forall \;\beta \in [0,1] \;\; \text{and } a,b \in \text{dom} f$$

Kevin Holt
  • 2,574
  • What do you mean by "duality theory"? The sum-rule can be understood as a separation theorem of the epigraphs of $f$ and $g$. Does such a approach satisfy "without duality theory"? – gerw Dec 06 '14 at 08:33
  • Sorry, that was an ambiguous way of wording it. I just meant is there a direct proof using only the above two inequalities without invoking additional theorems? It seems like there should be, but I've never seen one. I was tinkering with it over the weekend and may have found a way. I don't know if anyone else would care about it though... – Kevin Holt Dec 08 '14 at 19:47
  • Then you could post your proof as an answer. I feel, however, that (in infinite dimension) any proof has to use Hahn-Banach (or the AC) at some point. Hence, it is not possible by only using the inequalities. – gerw Dec 09 '14 at 13:41
  • OK, thanks for the advice / feedback. It looks like I have something that works when either f or g is continuously differentiable. I had hoped for something less limiting, but if you're right then maybe it's a waste of time to keep looking more. Will post it soon. Thanks again. – Kevin Holt Dec 13 '14 at 04:14
  • @gerw could you please elaborate on your comment about the sum rule being understood as a separation theorem? (Feel free to use any duality ideas.) – littleO Jun 03 '15 at 08:14
  • @littleO: Yes, but this would be too long for a comment, I guess. Could you post a separate question? – gerw Jun 03 '15 at 10:36
  • @gerw Thank you, I just posted a separate question here: http://math.stackexchange.com/questions/1310481/understanding-the-subdifferential-sum-rule – littleO Jun 03 '15 at 10:45

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