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From Wikipedia, I have learned that the Fenchel conjugate of $f(x)=\frac{|x|^p}{p}$, where $0<p<\infty$, is

\begin{equation} f^\ast(x^\ast)=\frac{1}{q}|x^\ast|^{q}, 1<q<\infty, \text{where}\; \frac{1}{p}+\frac{1}{q}=1. \end{equation}

From the definition of the Fenchel conjugate, I know

\begin{equation} f^\ast(x^\ast)=\underset{x \in X}{\sup} \{\langle x^\ast, x \rangle - f(x)\}= \underset{x \in X}{\sup} \left\{\langle x^\ast, x \rangle - \frac{|x|^p}{p} \right\} \end{equation}

After that I am stuck and couldn't figure out how $f^\ast(x^\ast) = \frac{1}{q}|x^\ast|^{q}$.

1 Answers1

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Well, assume $x^*, x$ are positive, differentiate the RHS, to get $(p-1)x^{p-1}/p = x^*$. Now plug the value of $x$ into the sup, and voila

Igor Rivin
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