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Let $I$ be an open interval and let $f:I\to\mathbb R\:$ be a convex function with an invertible derivative $(f')^{-1}:=\phi$. Then $$f^*(y):=\text{sup}\ \{xy-f(x):x\in I\}=\phi(y)y-f(\phi(y))$$ for all $y\in f'(\mathbb R)$.

How do you prove this or where can I find a proof?

Bernard
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Filippo
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1 Answers1

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Since the $x$-derivative of $g:=xy-f$ is $y-f^\prime$, a stationary point occurs at $x=\phi(y)$. Since $g$ is concave, it follows that we have found its global maximum.

J.G.
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  • Thank you. But how can I make use of the convexity to prove that $\phi(y)$ is the maximum of $g$? $g$ should be concave, but so is the logarithm, which does not have a maximum. – Filippo Sep 06 '20 at 20:13
  • @Filippo But a stationary point of a concave function is a maximum thereof. – J.G. Sep 06 '20 at 20:29
  • Thank you for improving my edit :) – Filippo Sep 06 '20 at 20:56