I am confused by how $V(g) = \Vert f \Vert ≤ V(\hat g),$ with $f$ being a bounded linear functional, come about via (2.48) and (2.49) on page 49 of Schechter's Principle of Functional Analysis.
First the total variation V(g) is defined as
$$V(g) = \sup \sum_{i=1}^n \vert g(t_{i})-g(t_{i-1}) \vert$$
And of course (2.48) and (2.49)
(2.48) $$\vert \int_{a}^b x(t)dg(t) \vert ≤ \Vert x \Vert V(g) $$ (2.49) $$V(g) = \Vert f \Vert$$
Based on my understanding, $\hat g$ comes about in order to prove uniqueness of g (see page 48).
First suppose on the contrary $\exists g_1, g_2$ w. bounded variation s.t. $$f(x) = \int_{a}^b x(t)dg_{i}(t), x \in C[a,b]$$
With $g=g_1-g_2,$ (2.50) comes about $$0 = \int_{a}^b x(t)dg_{i}(t), x \in C[a,b]$$
And after about 1.5 pages of tip-towing around (the M-beast),$\hat g$ is introduced as a normalized version of g.
And to make my head spinning even faster, at last the conclusion is going to be $V(g) = \Vert f \Vert = V(\hat g)$ by arguing that $V(\hat g) < V(g)$???
I am self-studying (the book is online) and found this section really hard to follow/swallow. Any tips/comments are appreciated. Thanks!