A variation of the Holder's inequality Can someone please give me an idea how to go about solving this problem?
If $f\in L^p(\Omega)$ show that 
$$\| f\|_p = \sup\left| \int_\Omega fg dx\right| = \sup \int_\Omega |fg|dx $$
where the supremum is taken over all $g \in L^q(\Omega)$ such that $\|g \|_q\leq 1$ where $ 1\leq p,q\leq\infty$ and $\frac{1}{q}+\frac{1}{p}=1$.
 A: Let us first check the second equality. Indeed, we always have 
$$
\left|\int_\Omega fg\right|\leq\int_\Omega |fg|,
$$
so 
$$
\sup\left|\int_\Omega fg\right|\leq\sup\int_\Omega |fg|.
$$
Now, given $g\in L^q$, define a function $\alpha=\frac{fg}{|fg|}$ (make it zero where $fg=0$). It is easy to see that $\alpha$ is measurable and that $|\alpha|=1$ whenever $\alpha\ne0$. Let $\Omega'=\{fg\ne0\}$. Then 
$$
\int_\Omega |fg|=\int_{\Omega'} f\,\left(\frac g\alpha\right)=\left|\int_\Omega f\,\left(\frac g\alpha\,1_{\Omega'}\right)\,\right|,
$$
the second equality since the middle number is positive by the first equality. As $|\alpha|=1$ where $fg\ne0$, we have that $\frac g\alpha\,1_{\Omega'}\in L^q$. So 
$$
\sup\left|\int_\Omega fg\right|\geq\sup\int_\Omega |fg|.
$$
To prove the first equality, Hölder gives you that $\int_{\Omega}|fg|\leq\|f\|_p$ for all $g$ with $\|g\|_q=1$. Now note that 
$$\tag1
\|f\|_p^p=\int_\Omega |f|^p=\int_\Omega |f|\,|f|^{p-1}=\int_\Omega |f h|,
$$
where $h=|f|^{p-1}$. And 
$$\tag2
\|h\|_q^q=\int_\Omega (|f|^{p-1})^q=\int_\Omega |f|^{q(p-1)}=\int_\Omega|f|^p=\|f\|_p^p.
$$
Thus $h\in L^q$ and $\|h\|_q=\|f\|_p^{p/q}$. Define $g=h/\|f\|_p^{p/q}$. Then $\|g\|_q=1$ and 
$$\tag3
\int_\Omega |fg|=\frac{\|f\|_p^p}{\|f\|_p^{p/q}}=\|f\|_p^{p-p/q}=\|f\|_p.
$$
So the supremum is achieved, and $\|f\|_p=\sup_g \int_\Omega |fg|$.
