Show that $f\in \mathcal{L}^p$ and $||f||_p \leqslant M.$

Question

Let $(X, \mathcal{B}, \mu)$ be a measured space, $1<p<\infty$, and $q$ the conjugate exponent of $p$ : $\left(\dfrac{1}{p}+\dfrac{1}{q}=1\right)$.

1. Show that $f \in \mathcal{L}^p \implies$ $\displaystyle{||f||_p=\sup\left\{ \left|\int_X f(x)g(x)d\mu(x) \right| ; ||g||_q \leqslant1\right\}}$.

2. We assume here that the measure $\mu$ is $\sigma$-finite. Let $f : X\to \mathbb{C}$ a mesurable function. we assume that there exists $M\geqslant0$ such that $\displaystyle{\left|\int_X f(x)g(x)d\mu(x) \right| \leqslant M}$ for all $g \in \mathcal{L}^p$ with $||g||_q\leqslant 1$. Show that $f\in \mathcal{L}^p$ and $||f||_p \leqslant M.$

My attempt:

1. on the one hand \begin{align*} \left|\int_X f(x)g(x)d\mu(x) \right| & \leqslant \int_X|f(x)g(x)|d\mu(x) \\ & \leqslant \left(\int_X|f(x)|^pd\mu(x) \right)^{\frac{1}{p}} \left(\int_X|g(x)|^q d\mu(x) \right)^{\frac{1}{q}}\\ & =||f||_p||g||_q \\& \leqslant ||f||_p \end{align*} on the other hand

Let $g_0(x)= \dfrac{|f(x)|^p}{|f(x)|}$ if $f(x)\not=0$, $g_0(x)=0$ if $f(x)=0$. so $g_0(x)f(x)=|f(x)|^p$, for all $x\in X$, then $|g_0|^q= |f|^{(p-1)q}=|f|^p$, so $g_0 \in \mathcal{L}^q$, since $f\in \mathcal{L}^p$.

let's pose now $g=\dfrac{g_0}{||g_0||_q}$, we notice that $||g||^q=1$, and $$\int_Xf(x)g(x)d\mu(x) =\int_X f(x)\dfrac{g_0(x)}{||g_0||_q^q}d\mu(x)=\int_X\dfrac{|f(x)|^p}{||g_0||_q^q}= \dfrac{||f||_p^p}{||f||_p^{\frac{p}{q}}} = ||f||_p.$$ 2. I got stuck here !

Any help is highly appreciated.

Answer 1

For part 2., one can do without $\sigma$-finiteness and in place assume that $\{|f|\neq0\}$ is $\sigma$-finite.

Assume $1<p<\infty$. For any $E\in\mathscr{F}$ with $E\subset\{|f|\neq0\}$ and $\mu(E)<\infty$, we will show that $\|f\mathbb{1}_E\|_p\leq M_f$. This would imply that $f\in\mathcal{L}_p$ and that $\|f\|_p\leq M_f$, because of the assumption that $\{|f|\neq0\}$ is $\sigma$-finite. The reverse inequality follows from H"older's inequality.

Let $f_n$ be a sequence of simple functions such that $|f_n|\leq|f|$ and $f_n\rightarrow f$. Then $h_n=f_n\mathbb{1}_E$ belongs to $\mathcal{L}_s$ for all $s>0$, $|h_n|\leq |f|\mathbb{1}_E$ and $h_n\rightarrow f\mathbb{1}_E$. If $$ \phi_n=\frac{\overline{f}}{|f|}\frac{|h_n|^{p-1}}{\|h_n\|^{p-1}_p} \mathbb{1}_E, $$ then $\|\phi_n\|_q=1$ and $$ \|f\mathbb{1}_E\|_p\leq\liminf_n\|h_n\|_p=\liminf_n\int |\phi_n h_n| \,d\mu\leq \liminf_n\int \phi_n f \,d\mu\leq M_f. $$