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$f(\mathbf{x}) = \frac{1}{p}\sum_{i=1}^n |x_i|^p$, for $1 < p < \infty$. Find the conjugate of $f$, $f^*$.

My attempt: \begin{align} f^*(y) &= sup_{x \in \mathbb{R}^n} \{y^\intercal x - \frac{1}{p}\sum_{i=1}^n |x_i|^p\}\\ &= sup_{x \in \mathbb{R}^n} \{\|x\|_p\|y\|_q - \frac{1}{p}\|x\|_p^p\} &\text{Holder's Inequality: } y^\intercal x \leq \|x\|_p\|y\|_q \text{, where } \frac{1}{p} + \frac{1}{q} = 1\\ &= sup_{x \in \mathbb{R}^n} \{\|x\|_p(\|y\|_q - \frac{1}{p}\|x\|_p^{p-1})\}\\ &= \begin{cases} 0, & \text{if $\|y\|_q \leq \frac{1}{p}\|x\|_p^{p-1}$}.\\ \infty, & \text{otherwise}. \end{cases} \end{align}

I am not too sure if I am in the right direction. It looks incorrect because of the last line where $y$ is still depending on $x$?

Marwin
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1 Answers1

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Your last line is not correct. You have to maximize $a(b-\frac 1 p a^{p-1})$ over all $a \geq 0$ where $b=\|y\|_q$. The derivative w.r.t. $a$ is $(b-\frac 1 p a^{p-1})-a\frac 1 p (p-1)a^{p-2}=b-a^{p-1}$. The maximum value is attained when $a=b^{\frac 1 {p-1}}$. Luckily $x_i=sgn(y_i) |y_i|^{q/p}$ gives a vector where this property holds and you also have equality in Holder's inequality in your earlier step. So this choice of $x$ indeed gives you the value of $f^{*}(y)$.

[Here $sgn (y_i) =1$ if $y_i \geq 0$ and $-1$ if $y_i <0$].

Kavi Rama Murthy
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  • Hi, may I ask how do you arrive at $x_i = sgn(y_i)|y_i|^{q/p}$?

    Also, is it right for me to use Holding's inequality so early on because it is an inequality instead of equality as shown later.

     – Marwin Mar 12 '20 at 11:39
  • The choice of $x_i$ comes from the well known condition for equality in Holder's inequality. It so happens in this case, using Holder's inequality at an early stage does not give you an unreasonably large bound; the bound you get is actually attained, so everything is fine. @Marwin – Kavi Rama Murthy Mar 12 '20 at 11:42