Here is the question:
Give an example of a measure space $(X, \mathfrak{M}, \mu)$ for which the Riesz Representation Theorem does extend to the case $p=\infty.$
Here is a trial for a solution:
Taking X to be a single point, call it $x_{0}$ . So $L^\infty(X,\mu)$ consists of the space of all bounded measurable functions $f : X \to \mathbb{R}$. The example is the space consisting of one point $X = \{x\}$ with the counting measure $\mu$, then we want to prove that the dual of $L^\infty equals L^1$. Now we want to satisfy the requirements of Riesz RT statement in Royden "Real analysis 4th edition":
$p=\infty$ and $q = 1$ the conjugate of $p$. every function $f$ in $L^1(\{x\}, \mu )$ will be a constant function say $c$ . Also, since by definition $L^\infty(X,\mu)$ consists of the space of all essentially bounded measurable functions $f : X \to \mathbb{R}$ then every $g \in L^\infty(X,\mu)$ will be a constant function, say $ g=c' .$ Now by Riesz RT we know that for $f \in L^1(\{x\}$ the linear functionals on $L^\infty(X,\mu)$ are those of the form $T_f(g)= \int_X f(x) g(x) d\mu(x)$, $ \forall g \in L^{\infty}(X,\mu)$ and $T_{f}: L^{\infty}(X,\mu) \rightarrow \mathbb{R}.$\
Now,we need to construct a function$ f \in L^1(X,\mu)$ such that $T_f(g)= \int_X f(x) g(x) d\mu(x)$ for all $g \in L^\infty(X,\mu)$ is onto. since the integral in our case will give us 2 constants times the $\mu (\{x\})$ which is 1. then our function can be any constant function say $f = c/d$ where $d$ is the constant representing the function $g.$
But we received these comments:
you need to start with a linear dual say $S$ (as the book named it) on $L^{\infty}$, then construct $f\in L^1$ and finally show that $S=T_f.$
My Questions:
1-In the proof of RRT, we used Radon Nikodym derivative, what will we do in our case instead of this?
2-Could anyone help me adjust the above trial according to include the proof of onto?
I don't think there is a need to try and mimic a complicated proof since your measure space is incredibly simple.
So we have $X = \{x\}$ and $\mu$ is simply the counting measure. First, we have to show that if $T : L^\infty(X) \to \mathbb{C}$ is a continuous linear functional. Then observe that $$ T(f)=\int_X fg\,\mathrm{d}\mu $$ where $g(x) = T(\mathbf{1})$ which is clearly in $L^1(X)$. To see that the above inequality is true, we first note that every function in $L^\infty(X)$ is simply a constant multiple of $\mathbf{1}$. Therefore, for any $f\in L^\infty(X)$ we have $f \equiv \lambda\mathbf{1}$ where $\lambda = f(x)$ is a constant. Therefore $$ T(f) = T(\lambda\mathbf{1}) = \lambda T(\mathbf{1}) = \lambda g(x) =f(x)g(x) = \int_X fg \,\mathrm{d}\mu. $$ Note that the second equality comes from linearity of $T$.
Conversely, let $g$ be a function in $L^1(X)$. Then it is clear that $$ T(f) \overset{def}{=} \int_X fg\,\mathrm{d}\mu $$ is a continuous linear functional. Linearity follows from the linearity of the integral. Continuity holds since $$ |T(f)| = \left\lvert\int_X fg\,\mathrm{d}\mu\right\rvert =\left\lvert f(x) g(x) \right\rvert = |g(x)|\cdot |f(x)| = |g(x)|\cdot ||f||_{L^\infty(X)}. $$ So, for the constant $C = |g(x)| < \infty$, we have $T(f)\leq C||f||_{L^\infty(X)}$ which shows that $T$ is continuous.
Note: You could also note that $L^p(X)$ is the same space with for any $1\leq p \leq \infty$. Since the Riesz representation theorem holds for $p=2$, it must hold for all $1\leq p \leq \infty$.