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In Appendix B of Evans' Partial Differential Equations the fact is stated that if $f:\mathbb{R}^n\longrightarrow \mathbb{R}$ is convex then for each $x\in\mathbb{R}^n$ there exist $r\in\mathbb{R}^n$ such that the inequality $f(y)\geq f(x)+r\cdot (y-x)$ holds for all $y\in\mathbb{R}^n$.

How can this be proved?

Thank you.

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This is actually true in much greater generality. See for example Theorem 39 and replace Hahn-Banach with the separating hyperplane theorem (which is the finite dimensional version of Hahn-Banach) for your proof – Chris Janjigian May 1 '12 at 4:43
up vote 2 down vote accepted

Here is an outline:

Consider the epigraph of $f$ at the point $(x,f(x))$. This point lies on the boundary of the epigraph. Since $f$ is convex, and finite-valued on all of $\mathbb{R}^n$, $f$ is continuous, so the epigraph is closed and convex. Use the Hahn Banach separation theorem to find a functional $\phi$ on $\mathbb{R}^n \times \mathbb{R}$ that 'separates' the point $(x,f(x))$ from the rest of the epigraph (ie, $(y,\alpha)$ such that $f(y) \leq \alpha$). I write separates in inverted commas because the separation is in a limiting sense, ie, $$\phi((y,\alpha)) \geq \phi((x,f(x))), \; \; \forall y, \forall \alpha \geq f(y).$$ Using the representation theorem, we can write $\phi((z, \beta)) = <r', z>+c\beta$, so the separation result can be rewritten as $$<r',y> + c \alpha \; \geq \; <r',x> + c f(x), \; \; \forall y, \forall \alpha \geq f(y)$$ It is straightforward to show that $c > 0$, so we can divide across by $c$, choose $\alpha=f(y)$, and let $r= \frac{1}{c} r'$ to get $$f(y) \geq f(x)+<r, y-x>.$$ The set of $r$ satisfying this inequality is known as the subdifferential of $f$ at $x$, and is usually denoted by $\partial f(x)$.

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