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How does one show that $\phi(x)$ convex and twice differentiable implies that $x\phi(\frac{y}{x})$ is convex on the plane $x>0$? Thanks.

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Did you try differentiating twice? –  Jonas Meyer Aug 13 '11 at 21:12
    
Writing phi as a supremum of affine functions yields a simple proof which explains why the restriction to x>0 is needed (and why the hypothesis that phi is twice differentiable is not). –  Did Aug 13 '11 at 22:30
    
Thanks guys, I have managed to solve it now :) –  Pretzel Aug 14 '11 at 2:42
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Could you add your proof? –  draks ... Apr 5 '12 at 18:22

1 Answer 1

The constant $y$ is simply scaling, so it can be replaced by $1$ without losing generality. If $\phi$ is twice differentiable, one can compute $$ \begin{split} \frac{d^2}{dx^2} x\phi(x^{-1})& =\frac{d}{dx}(\phi(x^{-1})-x^{-1}\phi'(x^{-1})) \\ &=-x^{-2}\phi'(x^{-1})+x^{-2}\phi'(x^{-1} )+x^{-3}\phi''(x^{-1}) \\ & =x^{-3}\phi''(x^{-1}) \end{split} \tag1$$ ... however this is not very enlightening, since the cancellation comes as a miracle. The approach suggested by Did is better: since $\phi$ is the envelope of some family of affine functions $x\mapsto ax+b$, the transform $x\mapsto x\phi(x^{-1})$ is the envelope of some family of functions of the form $$x\mapsto x(ax^{-1}+b)=a+bx$$

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