Riesz representation theorem, part of the proof (2).

Question

Let $(X, A,\mu)$ is a measure space and $G$ is a bounded linear functional on $L_p(X,A, \mu)$, if exist $g$ in $M(X,A)$ such that $$ G(f)=\int fgd\mu $$ for all $f$ in $L_p(X,A, \mu)$. Proof that $g\in L_q(X,A, \mu)$, whit $\frac 1 p+\frac 1 q=1$, and $\|G\|=\|g\|_p$.

Answer 1

I have solved my problem, we have two cases

• $(X,A, \mu)$ is $\sigma$-finite. For $|g|$, exists an increasing sequence $(\phi_n)$ of simple functions, such that converges to $|g|$. Since $\sigma$-finite, exists an increasing sequence of sets $(E_n)$ whit finite mesure such that $\bigcup_{n=1}^\infty E_n=X$. Define $g_n=\phi_n\cdot\chi_{E_n}$, we have that $(g_n)$ is an increasing sequence of simple function such that converges to $|g|$, even more \begin{align*} \int |\text{sig}(g)(g_n)^{q-1}|^p\;d\mu & = \int |(g_n)^{q-1}|^p\;d\mu \\ &= \int \chi_{E_n}|(\phi_n)^{q-1}|^p\;d\mu \\ &= \int_{E_n}|(\phi_n)^{q-1}|^p\;d\mu \\ & \leq \mu(E_n)\max(|(g_n)^{q-1}|^p)<\infty, \end{align*} hence $(g_n)^{q-1}\text{sig}(g)\in L_p$ for all $n$, even more \begin{align*} \int (g_n)^q\;d\mu &= \int (g_n)^{q-1}g_n\;d\mu\\ &\leq \int (g_n)^{q-1}|g|\;d\mu\\ &\leq \int (g_n)^{q-1}\text{sig}(g)g\;d\mu\\ &=G(\text{sig}(g)(g_n)^{q-1})\\ &\leq\|G\|\cdot\|\text{sig}(g)(g_n)^{q-1}\|_p\\ &= \|G\|\cdot\|(g_n)^{q-1}\|_p\\ &= \|G\|\left(\int |(g_n)^{q-1}|^p\;d\mu\right)^{1/p}, \end{align*} but $(q-1)p=q$, hence $$ \int (g_n)^q\;d\mu\leq \|G\|\left(\int (g_n)^q\;d\mu\right)^{1/p}, $$ where, $$ \left(\int (g_n)^q\;d\mu\right)^{1-1/p}=\|g_n\|_q\leq \|G\|. $$ by monotone convergence theorem, we have $$ \left(\int |g|^q\;d\mu\right)^{1/q}\leq \|G\|, $$ hence $g\in L_q$ and $\|g\|_q\leq \|G\|$.

• $(X,A, \mu)$ is an arbitrary space, we have that (see Elements of Integration - Bartle p. 92), exists a set $E\in A$ $\sigma$-finite such that: for all functions $f\in L_p$ such that $E\cap \{x\in X:f(x)\neq 0\}=\varnothing$, we have $G(f)=0$. For $f\in L_p$, we have that $E\cap \{x\in X:f(x)\chi_{X-E}(x)\neq 0\}=\varnothing$, hence: $$ \int_{X-E} fg\;d\mu=\int f\chi_{X-E}g\;d\mu=G(f\chi_{X-E})=0, $$ hence $g=0$ almost everywhere in $X-E$. Using the first part for the set $E$ we have $g\in L_q$ and $\|g\|_q\leq \|G\|$.

Even more, using Hölder inequality, we have that, for $f\in L_p$, $$ |G(f)|=\left|\int fg\;d\mu\right|\leq\int|fg|\;d\mu\leq \|f\|_p\|g\|_q, $$ i.e., $\|G\|\leq\|g\|_q$, hence $\|G\|=\|g\|_q$.