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let $f:\mathbb R^n \rightarrow \mathbb R$ be a convex function.

for arbitrary $p\in \mathbb R^n$, how can I find an affine function $\ell_p(x)$ such that

$$\ell_p(x)\leq f(x)\ \text{ and }\ \ell_p(p)=f(p)?$$

I could find it for $n=1$, but I don't know what to do when $n\geq2$.

456 123
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    Tangent line/plane, i.e. $f(p)+\text{Span}\left(\frac{\partial f}{\partial x_1}(p),\ldots,\frac{\partial f}{\partial x_n}(p)\right)$ if $p$ is a point of differentiability. – Jack D'Aurizio Mar 24 '18 at 17:54
  • I don't know how to prove it when f is not differentiable. – 456 123 Mar 24 '18 at 18:04
  • A convex function is differentiable almost everywhere! - https://math.stackexchange.com/questions/727789/convex-function-almost-surely-differentiable – Jack D'Aurizio Mar 24 '18 at 18:23
  • @JackD'Aurizio It is not enough. What happens at the non-differentiability points? See my answer. – szw1710 Mar 24 '18 at 20:56

1 Answers1

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Taking the axial sections we arrive at convex functions of one variable. A convex function of one variable admits one-sided derivatives at every interior point of a domain. That is why all one-sided parial derivatives do exist at $p$. It is enough to take a plane described by @Jack D'Aurizio with, say, left-hand-side partial derivatives at $p$.

EDIT

As @JonnMa said, it is not enough. See my comment below.

szw1710
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  • It is not clear taking $n$-axial sections is good enough. –  Mar 24 '18 at 21:02
  • Indeed there seems to be convex functions on $\mathbb R^2$ so that it is identically zero on the positive $x$ and $y$ axis but is negative somewhere. –  Mar 24 '18 at 21:04
  • @JohnMa Of course you are right. It is not enough because there in a support line on any axis only. It is far from the hyperplane. Nevertheless, there is a theorem stating that a subdifferential of a convex function is non-empty. This is what we want here. For a proof, everybody can see e.g Rockafellar "Convex analysis" book. – szw1710 Mar 24 '18 at 21:32