Prove that $f^{*}(y) =\frac{\|y\|^q}{q}$; $\frac{1}{p}+\frac{1}{q}=1$? 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
 A: 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,
$$\tfrac{1}{q}\lVert g\rVert_q^q=\sup_{f:L^p(X,\mu)}\left(\langle f,g\rangle-\tfrac{1}{p}\lVert f\rVert_p^p\right)\text{.}\quad\blacksquare$$
