Space of functions of bounded variation

Remarks

This is a normed space. The variation is subadditive and $V_a^b(t f) = \abs{t} V_a^b(f)$, which gives the triangle inequality and homogeneity. If $\norm{f}_{BV} = 0$, then $V_a^b(f) = 0$, so $f$ is constant, and $f(a) = 0$ gives $f = 0$. The space is complete, so it is a Banach space.

Example: $f(x) = x$ on $[0, 1]$. It is increasing, so $V_0^1(f) = f(1) - f(0) = 1$ and $\norm{f}_{BV} = 0 + 1 = 1$.

Non-example: $f(x) = x \sin(1/x)$ for $0 < x \le 1$, with $f(0) = 0$. It is continuous on $[0, 1]$, but its total variation is infinite, so it is not in $BV([0, 1])$.

An equivalent norm is $\sup_{x} \abs{f(x)} + V_a^b(f)$.