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I have been trying to solve this question and managed to prove subquestion (a). But I have no idea about proving the other two subquestions. Can anyone help?

Note:

The question is from Stephen Boyd's Convex Optimization book - Additional Exercise 2.31.

2.31 Suppose the function $h:\mathbb{R}\to\mathbb{R}$ is convex, nondecreasing, with $\mathbf{dom}\,h=\mathbb{R},$ and $h(t)=h(0)$ for $t\le 0.$

(a) Show that the function $f(x)=h(\|x\|_2)$ is convex on $\mathbb{R}^n.$

(b) Show that the conjugate of $f$ is $f^*(y)=h^*(\|y\|_2).$

(c) As an example, derive the conjugate of $f(x)=(1/p)\|x\|_2^p$ for $p>1,$ by applying the result of part (b) with the function $$h(t)=\frac1p \, \max\{0, t\}^p=\begin{cases}\frac1p\,t^p\quad& t\ge 0\\ 0 &t<0.\end{cases} $$

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Johnnylin
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  • Can you clarify what the problem means by "conjugate"? It evidently does not mean complex conjugate, so it must mean something else. – Adrian Keister Sep 21 '18 at 23:36
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    @AdrianKeister Convex conjugate – angryavian Sep 21 '18 at 23:37
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    If $f:\mathbb R^n \to \mathbb R \cup { \infty}$ is a convex function, then the convex conjugate of $f$ is the function $f^(z) = \sup_x , \langle z, x \rangle - f(x)$. Significance:* The function $x \mapsto \langle z, x \rangle - f^*(z)$ is the best affine minorant of $f$ with slope $z$. – littleO Sep 21 '18 at 23:50

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$$f^*(y) := \sup_{x \in \mathbb{R}^n} \{y^\top x - h(\|x\|)\} = \sup_{c \ge 0} \sup_{z : \|z\|=1} \{y^\top (cz) - h(c)\}.$$

The maximizing $z$ in the inner supremum is $z = y / \|y\|$ (why?) which yields $$\sup_{c \ge 0} \{c\|y\| - h(c)\} =: h^*(\|y\|).$$

angryavian
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