For $p>1$, $f(x)=\frac{\|x\|^p}{p}$ is a convex function. In fact, $$f(\alpha x + (1-\alpha) y) = \frac{\|\alpha x + (1-\alpha) y\|^p}{p} \leq \alpha \frac{ \|x \|^p}{p} + (1-\alpha) \frac{\|y\|^p}{p}= \alpha f(x)+(1-\alpha) f(y),$$ because for $\alpha \in [0,1]$ $\alpha^p\leq \alpha$ for all $p>1$.
Now I would like to compute the conjugate function $f^*(y)$ of the function $f$ for $p>1$; where: $f^{*}(y)=\sup_{x} (\left< y,x\right> - f(x))$.
For the case $p=2$, we have $$\begin{align} f^{*}(y) &=\sup_{x} (\left< y,x\right> - f(x)) \\ &= \sup_x ( \left< y,x\right>-\|x\|^2/2) \\ &= \sup_x (\left< y,x\right>-\|x\|^{2} /2 - \|y\|^2/2 + \|y\|^2 /2) \\ &= \sup_x (\|y\|^2 /2 - (x-y)^2 /2)\\& = \|y\|^2 /2. \end{align}.$$ Thank you in advance
This result is a scholium of a proof of Hölder's inequality via Young's inequality. Start with strict concavity of the logarithm: $$\ln(tx+(1-t)y)\geq t\ln x + (1-t)\ln y$$ for $x>0$, $y>0$, with equality iff $x=y$. Setting $x=a^p$, $y=b^q$, $t=1/p$, and $1-t=1/q$ and exponentiating, we have shown Young's inequality $$ab\leq\frac{a^p}{p}+\frac{b^q}{q}$$ where equality holds iff $a^p=b^q$. Let $a=\lvert f\rvert$, $b=\lvert g\rvert$ for measurable functions $f,g$ on some measure space $X$. We have $$fg\leq \lvert fg\rvert\leq\frac{\lvert f \rvert^p}{p}+\frac{\lvert g\rvert^q}{q}\text{,}$$ the former inequality being an equality exactly when $f$ and $g$ have the same sign, the latter whenever $\lvert f\rvert^p=\lvert g\rvert^q$. Using integration, $L^p$ norms, etc., with respect to some measure $\mu$, we obtain the Fenchel inequality $$\langle f,g\rangle \leq \tfrac{1}{p}\lVert f\rVert_p^p+\tfrac{1}{q}\lVert g\rVert_q^q$$ with equality achieved iff $f=\lvert g\rvert^{q/p}\mathrm{sgn}\, g$ almost everywhere.
Altogether, we have shown that, for $f\colon L^p(X,\mu)$, $g\colon L^q(X,\mu)$, $\tfrac{1}{p}+\tfrac{1}{q}=1$, $p>1$, $$\tfrac{1}{q}\lVert g\rVert_q^q\geq\langle f,g\rangle-\tfrac{1}{p}\lVert f\rVert_p^p$$ and that equality is actually achieved for some $f$. Thus, $$\tfrac{1}{q}\lVert g\rVert_q^q=\sup_{f:L^p(X,\mu)}\left(\langle f,g\rangle-\tfrac{1}{p}\lVert f\rVert_p^p\right)\text{.}\quad\blacksquare$$