I am learning convex analysis by myself and I need help.
How to show that if $X=U=\mathbb{R}$ and $f\left(x\right)=\frac{|x|^{p}}{p}$ then the convex conjugate $f^{*}\left(u\right)=\frac{|u|^{q}}{q}$ when $\frac{1}{p}+\frac{1}{q}=1$? There exists a particular technique that I have to apply in order to compute the convex conjugate?
For practical computation I would use the fact that $\nabla f^*$ is the inverse of $\nabla f$ (see here). By the chain rule, $$\nabla f(x) =|x|^{p-1} \nabla |x| = |x|^{p-1} \frac{x}{|x|}$$ which means the direction of $x$ stays the same but its length is raised to power $p-1$. The inverse of this map is $$u\mapsto |u|^{1/(p-1)} \frac{u}{|u|} = |u|^{q-1} \frac{u}{|u|}$$ Observing the similarity of two formulas, we arrive at $f^*(u)=|u|^q/q$.
Another nice way to show it is by using Young's inequality: For $x, \xi \in \mathbb R^n$, we have $$\langle\xi, x \rangle \le \frac{|\xi|^p}{p} + \frac{|x|^q}{q},$$ with equality if $|\xi|^p = |x|^q$. Therefore, $$f^*(\xi) = \sup_{x \in \mathbb R^n}\left(\langle\xi, x \rangle - \frac{|x|^q}{q}\right) = \frac{|\xi|^p}{p}.$$
