From different books I realized two things: 1) dual space of $L_\infty[0,1]$ is the absolutely continuous functions or AC measures on $[0,1]$, 2) dual space of $C[0,1]$ is $BV_0$ function spaces or equally the bounded variation measures. The questions is how in this example Hahn - Banach theorem works? How can I understand 1) how it is not contradiction: if we are going to extend linear functional that is integration against some bounded variation function to $L_\infty[0,1]$, how it is gonna change to AC function? 2) Is there any proof of fact 2) using this Hahn-Banach theorem?
2026-04-05 14:54:16.1775400856
$L_{\infty}[0,1]$ and extended linear functionals from $C[0,1]$.
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The dual space of $L^{\infty}[0,1]$ consists of integration with respect to finitely-additive set functions of finite variation.
If you want to use the Hahn-Banach Theorem to determine the dual $C[0,1]^*$ of $C[0,1]$, then you can direct application of more difficult results for the dual of $L^{\infty}[0,1]$ by starting with a bounded linear functional $F$ on $C[0,1]$, and extending it to a bounded linear functional $\tilde{F}$ on $L^{\infty}[0,1]$ with the same norm by applying the Hahn-Banach Theorem. This can be done because $C[0,1]$ is a closed subspace of $L^{\infty}[0,1]$, and an isometric embedding. Then define $$ \rho(t)=\tilde{F}(\chi_{[0,t]}). $$ Then $\rho$ is a function of bounded variation. To see this let $\mathscr{P}=\{t_0,t_1,t_2,\cdots,t_n \}$ be a partition of $[0,1]$. Let $\alpha_j$ be a unimodular constant such that $|\rho(t_j)-\rho(t_{j-1})|=\alpha_j\{\rho(t_j)-\rho(t_{j-1})\}$. Then $$ \sum_{j=1}^{n}|\rho(t_j)-\rho(t_{j-1})| = \tilde{F}\left(\sum_{j=1}^{n}\alpha_j\chi_{(t_{j-1},t_j]}\right) \le \|\tilde{F}\|\|\sum_{j=1}^{n}\alpha_j\chi_{(t_{j-1},t_{j}]}\|_{\infty}=\|F\| $$ So $V_{0}^{1}(\rho) \le \|F\|$. If $t_j^* \in [t_{j-1},t_j]$ and if $f\in C[0,1]$, then $$ \lim_{\|\mathscr{P}\|\rightarrow 0}\left\|\sum_{j=1}^{n}f(t_j^*)\chi_{[t_{j-1},t_{j}]}-f\right\|_{L^{\infty}[0,1]}=0. $$ Therefore, $$ \lim_{\|\mathscr{P}\|\rightarrow 0}\tilde{F}\left(\sum_{\mathscr{P}}f(t_j^*)\chi_{[t_{j-1},t_j]}\right)=\tilde{F}(f)=F(f) \\ \lim_{\|\mathscr{P}\|\rightarrow 0}\sum_{j=1}^{n}f(t_j^*)\{\rho(t_{j})-\rho(t_{j-1})\}= F(f). $$ This last limit is the definition of the Riemann-Stieltjes integral of $f$ with respect to $\rho$, which gives $$ \int_{a}^{b}f(t)d\rho(t) = F(f). $$ You can use various constructions to lift this to a Riemann-Stieltjes integral, and to define a signed Borel measure $\rho$, but nothing is gained in determining a representation of $F$.