In this question: Legendre transform and Young's Inequality, given $ f \colon \mathbb{R}^n \to \mathbb{R} $, $ f(x) = \frac{1}{p}|x|^{p} $ for $ 1 < p < \infty $, the Legendre transform of $ f $ which I'll denote $ f^* $ (and is defined by $ f^*(x) = \sup_{y \in \mathbb{R}^n} \{x \cdot y - f(y) \}$ for $f$ convex) is shown to be equal to $ f^*(x) = \frac{1}{q}|x|^q $, where $ \frac{1}{p} + \frac{1}{q} = 1$.
Is it possible to prove $ f^*(x) = \frac{1}{q}|x|^q $ without using Young's Inequality so that Young's Inequality may be proved using the fact that $ f^*(x) = \frac{1}{q}|x|^q $? I am stuck on showing $f^*(x) \leq \frac{1}{q}|x|^q $ without using Young's Inequality. Thank you very much for your help.
Yes, it is possible. To find the sup, take the $Y$-derivative (gradient) $$ \nabla_Y(X\cdot Y-1/p\|Y\|^p)=X-\|Y\|^{p-2}Y $$ The sup is achieved (given the convexity) at just one point, namely where $X-\|Y\|^{p-2}Y=0$.
At that point, $X\cdot Y=\|Y\|^p=\|X\|^{p/(p-1)}$ and $$ X\cdot Y-1/p\|Y\|^p=(1-1/p)\|Y\|^p=1/q\|X\|^{1/q}. $$