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I have seen the following argument:

If $F: \mathbb{R}^n \to \mathbb{R}$ is a convex function, then there exist a Borel function $\lambda\colon\mathbb{R}^n \to \mathbb{R}^n$ bounded on compact subsets, and such that \begin{equation} F(w) \ge F(z) + \langle \lambda(z), w-z\rangle \end{equation} for every $z,w \in \mathbb{R}^n.$

Why is this true? Can you give one answer or a reference? if you know something similar, please tell me. Thank you.

dfeuer
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user29999
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    Maybe you can think of this as saying that convex functions are differentiable? Just a feeling. – awllower Jul 14 '13 at 00:09
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    Did you mean $F:\mathbb{R}^n\mapsto\mathbb{R}$? – S.B. Jul 14 '13 at 00:40
  • Ok, sorry. I will correct. – user29999 Jul 14 '13 at 00:43
  • @awllower Convex functions are not necessarily differentiable. The inequality in question doesn't say that the graph of $F$ in $\mathbb R^{n+1}$ has a tangent hyperplane. Rather, it says that, through any point $(z,F(z))$, there is a hyperplane that lies entirely below the graph of $F$ ("below" in the weak sense of $\leq$, not $<$). – Andreas Blass Jul 14 '13 at 06:00
  • @AndreasBlass Thanks for the reminder! I know that the statement is absolutely wrong, while, in some sense, it serves to tell us that the convex functions are very restrictive. In fact, I remember seeing some proof that convex functions are "almost" differentiable. But I forgot the definition of the italicized term, so I just gave a suggestion, as one possible direction. And no confusion is intended. Thanks again for your attention. – awllower Jul 14 '13 at 06:05
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    $\lambda$ is called a sub-gradient. For $n = 1$ it is Borel measurable simply because it is non-decreasing. In higher dimensions it may be a little more complicated. – Niels J. Diepeveen Jul 15 '13 at 00:20

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