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Q. Consider $f := \mathbb{R}^n → R$ defined by $$f(x) = 1/p |x_i|^p$$ for $1 < p < ∞$. Find $f^∗$

I came across other questions such as that of finding the conjugate of $f(x) = \frac{1}{2}||x||^2$. However, I am not really sure of whether it is similar for this question. I tried to find the conjugate $f^*(y) = sup_x(x^Ty - \frac{1}{p}\sum_{i=1}^n |x_i|^p) $ and then I get $f^*(y) = \sum_{i=1}^n y_i^2 - \frac{1}{p}\sum_{i=0}^n |y_i|^p $. But I am not sure whether this is correct and where exactly it is defined since $1<p<\infty$.

Any help is appreciated.

eun ji
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1 Answers1

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It is just a matter of grinding through the computation.

Let $\phi^*(y) = \sup_{x \in \mathbb{R}} xy -{1 \over p} |x|^p$ and note that $f^*(y) = \sum_k \phi^* (y_k)$, so we can focus on $\phi^*$.

Note that $\phi^*(-y) = \phi^*(y)$, so we can assume $y \ge 0$.

Let $\eta(x,y) = xy -{1 \over p} |x|^p$, and note that if $x \ge 0$ then $\eta(x,y) \ge \eta(-x,y)$, hence we can assume that $x \ge 0$ as we look for a solution. Furthermore, $\lim_{x \to \infty} \eta(x,y) = -\infty$, so we know the $\sup$ is attained for some finite $x$. Hence we look for solutions of $y-x^{p-1} = 0$ to get $x = \sqrt[p-1]{y}$ and so $\phi^*(y) = {1 \over q}y^q$, where ${1 \over p } +{1 \over q} = 1$.

Consequently, $\phi^*(y) = {1 \over q} |y|^q$ and so $f^*(y) = {1 \over q} \sum_k |y_k|^q$.

copper.hat
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