I have seen this statement: The dual space $C^{*}\left[a,b\right]$ of $C\left[a,b\right]$ is isometrically isomorphic to $BV_{0}\left[a,b\right]$. where $$BV_{0}\left[a,b\right]=\left\{ \alpha\in BV\left[a,b\right]|\alpha\left(a\right)=0,\alpha\text{ right-continuous on}\left(a,b\right)\right\} $$
Now I want to show:
For $T\in C^{*}\left[a,b\right]$, if there exists a $\alpha\in BV_{0}\left[a,b\right]$ such that $T=T_{\alpha}$. That is, $$T\left(f\right)=T_{\alpha}\left(f\right)=\int_{a}^{b}fd\alpha$$
for all $f\in C\left[a,b\right]$. then $\left\Vert T\right\Vert =T_{a}^{b}\left(\alpha\right)$. And if there exists $\alpha,\beta\in BV_{0}\left[a,b\right]$ such that $T=T_{\alpha}=T_{\beta}$. Then $\alpha=\beta$ on $\left[a,b\right]$.
I have prove that $\left\Vert T_{\alpha}\right\Vert \le T_{a}^{b}\left(\alpha\right)$, but I have trouble proving $\left\Vert T_{\alpha}\right\Vert \geq T_{a}^{b}\left(\alpha\right)$ and the uniqueness, could anyone give some help?
If you start with $T \in C[a,b]^*$, then you can extend $T$ to $S \in L^{\infty}[a,b]* $ in such a way that $\|S\| = \|T\|$. Then $S$ is defined on step functions such as $\chi_{[a,b]}$ and $\chi_{[a,b)}$, etc.. For any positive integer $N$, let $$ x_0 = a, \; x_1 = a+\frac{b-a}{N}, x_2=a+2\frac{b-a}{N},\cdots, x_N=b. $$ For any $f\in C[a,b]$ define $f_N$ to be $$ f_N = f(x_0)\chi_{[x_0,x_1)}+f(x_1)\chi_{[x_1,x_2)}+\cdots+f(x_{N-1})\chi_{[x_{N-1},x_N]}. $$ Then $f_N\rightarrow f$ in $L^{\infty}[a,b]$ as $N\rightarrow\infty$ because $f\in C[a,b]$. Define $\alpha(t)=S(\chi_{[a,t)})$ for $a \le t < b$ and let $\alpha(b)=S(1)$. Then $\alpha$ is of bounded variation and there are unimodular constants $c_n$ such that $$ \sum_{n=1}^{N}|\alpha(t_n)-\alpha(t_{n-1})|=\sum_{n=1}^{N-1}c_nS(\chi_{[t_{n-1},t_n)})+c_NS(\chi_{[t_{N-1},t_{N}]}) \\ = S\left(\sum_{n=1}^{N-1}c_n\chi_{[t_{n-1},t_{n})}+C_N\chi_{[t_{N-1},t_N]}\right) $$ Therefore, $$ \sum_{n=1}^{N}|\alpha(t_n)-\alpha(t_{n-1})| \le \|S\|=\|T\|. $$ So $\alpha$ is of bounded variation and $V_{a}^{b}(\alpha) \le \|T\|$. Furthermore, $$ T(f)=\lim_{N} S(f_N)=\lim_N \sum_{n=1}^{N}f(x_n)\Delta\alpha_n = \int_{a}^{b}f(t)d\alpha(t) \\ |T(f)| \le \|f\|_{C[a,b]}V_{a}^{b}(\alpha) \\ \|T\| \le V_{a}^{b}(\alpha). $$ So $\|T\|=V_a^b(\alpha)$. Modiffying $\alpha$ to be left- or right-continuous will not increase its variation, or the representation of $T(f)=\int_{a}^{b}fd\alpha$. So it still follows that $\|T\|=V_a^b(\tilde{\alpha})$, where $\tilde{\alpha}$ is left- or right- continuous.