Let me define it first
A $\Phi:\mathbb{R}\to \mathbb{R^+}$ be a convex function that satisfies the following properties
(1) $\Phi(-x)=\Phi(x)$
(2) $\Phi(0)=0$
(3) $\lim_{x\to\infty}\Phi(x)=+\infty$
the for each such function we have another associated convex function $\Psi:\mathbb{R}\to\mathbb{R^+}$ which is defined as $\Psi(y)=\sup\{x|y|-\Phi(x):x\geq0\}$ for all $y\in \mathbb{R}$ and satisfies the same above properties (1), (2), and (3).
My question: Let $\Phi(x)=|x|^p,p\geq1$ so we can see that it satisfies the property (1), (2), and (3) now I want to find $\Psi(y)$ for this so according to the definition we have
$\Psi(y)=\sup\{x|y|-|x|^p,x\geq0\}$ for all $y\in \mathbb{R}$ , I am stuck here that how do I find the supremum of this function, what I know is to find the supremum of the function we differentiate it and equate it to zero then we find the second order derivative and check for values of $x$ but how do I differentiate here?
Ok, since $x \geq 0$ we have $|x| = x$ and hence $$ \sup\limits_{x\geq 0}\{x|y|-|x|^p\} = \sup\limits_{x\geq 0}\{x|y|-x^p\}. $$ If $p>1$, the expression under the supremum decreases unboundedly if $x\to\infty$, hence the supremum is either attained at the left boundary, or at an extremum point. At the left boundary the expression is $0$, whereas to find extremum point you need to find zeros of the derivative wrt $x$. There is just one at $(|y|/p)^{\frac1{1-p}}$. Can you work out the rest?