Kind of equality on the sphere in Holder inequality Lets suppose p > 1 and q as $\frac{1}{p} + \frac{1}{q} = 1$
$g$ is function of $L^q(\Omega, A, \mu) $
We suppose that exists $M$ positive number as : $M = sup\left \{ \int fg d\mu  ;  f \in \epsilon(A) ;  \left \| f \right \|_p = 1 \right \}$
$\epsilon(A)$ is the set of staircase functions
The goal is to show that : $\left \| g \right \|_q = M $
With Holder inequality, we show easily that $\left \| g \right \|_q \geq  M $
In other sens, I'm searching for range of function $(g_n)$ in $\epsilon_+(A)$ to show that : $ \int \left | g \right |^q d\mu \leq M^q $ (then we can conclude easily)
Any idea ?
 A: Let $p > 1$ and $q$ as $\frac{1}{p} + \frac{1}{q} = 1$
Let $g$ be a function of $L^q(\Omega, A, \mu) $
Suppose that exists $M$ positive number as :
$$M = \sup\left \{ \int fg d\mu  ;  f \in \epsilon(A) ;  \left \| f \right \|_p = 1 \right \}$$
$\epsilon(A)$ is the set of simple functions. It is clear that $M \geqslant 0$.
Lemma:
$$M = \sup\left \{ \int fg d\mu  ;  f \in \epsilon(A) ;  \left \| f \right \|_p \leqslant 1 \right \}$$
Proof:
Let $M' = \sup\left \{ \int fg d\mu  ;  f \in \epsilon(A) ;  \left \| f \right \|_p \leqslant 1 \right \}$.  It follows immediately that $M' \geqslant M$.
If $M' > M$ then there is $f_1 \in \epsilon(A) ;  \left \| f_1 \right \|_p \leqslant 1$ such that $\int f_1g\: d\mu > M \geqslant 0$. So $0< \|f_1\|_p \leqslant  1$. Let $f_2=\frac{f_1}{\|f_1\|_p}$. It is immediate that $\|f_2\|_p=1$ and
$$\int f_2g\: d\mu = \frac{1}{\|f_1\|_p}\int f_1g\: d\mu \geqslant \int f_1g\: d\mu > M $$
Contradiction, since $M = \sup\left \{ \int fg d\mu  ;  f \in \epsilon(A) ;  \left \| f \right \|_p = 1 \right \}$. So $M'= M$. $\square$
Now we want to prove that : $ \| g \|_q = M $
From Holder inequality, we show easily that $\| g \|_q \geqslant  M $.
So we need to prove that $\left \| g \right \|_q \leqslant  M$.
Now, if $\|g\|_q=0$, then $M=0$ and it is trivial that $\|g\|_q=M$. Let us suppose  $\|g\|_q>0$
Now, we have two cases: $q>1$ and $q=1$.
Case $q>1$.
Let
$$h = sgn(g) \frac{|g|^{q-1}}{\|g\|_q^{q/p} }$$
where $sgn(g) = \chi_{\{x: g(x)>0\}} - \chi_{\{x: g(x)<0\}} $.
Since $q=(q-1)p$, it follows that $h\in L^p(\Omega, A, \mu) $ and $\|h\|_p=1$.
Moreover, since $q-(q/p)=1$, we have
$$\int h g \:d\mu = \frac{1}{\|g\|_q^{q/p}} \int |g|^q d\mu =\|g\|_q^{q-(q/p)} = \|g\|_q$$
Now, let $\{f_n\}_n$ be a sequence of simple function converging such that $\{f_n^+\}_n$ converges monotonically from below to $h^+$  and $\{f_n^-\}_n$ converges monotonically from below to $h^-$.
It follows that $\|f_n\|_p \leqslant \|h\|_p =1$  and
$$\lim_n \int f_n g d\mu = \int h g \:d\mu = \|g\|_q$$
But, for all $n$, $\lim_n \int f_n g d\mu \leqslant M$. So $\|g\|_q \leqslant M$. So $\|g\|_q = M$.
Case $q=1$.
Let
$$h = sgn(g) $$
where $sgn(g) = \chi_{\{x: g(x)>0\}} - \chi_{\{x: g(x)<0\}} $.
It follows that $h\in L^\infty(\Omega, A, \mu) $ and $\|h\|_\infty=1$.
Moreover, we have
$$\int h g \:d\mu = \int |g| d\mu =\|g\|_1$$
Now, let $\{f_n\}_n$ be a sequence of simple function converging such that $\{f_n^+\}_n$ converges monotonically from below to $h^+$  and $\{f_n^-\}_n$ converges monotonically from below to $h^-$.
It follows that $\|f_n\|_1 \leqslant \|h\|_1 =1$  and
$$\lim_n \int f_n g d\mu = \int h g \:d\mu = \|g\|_1$$
But, for all $n$, $\lim_n \int f_n g d\mu \leqslant M$. So $\|g\|_1 \leqslant M$. So $\|g\|_1 = M$.
