Inequality on function variations

Question

Assume $f\in BV([a,b])$. Is the following true:

$$(f(b)=0)\Rightarrow (V[a,b] \leqslant V[a,b) + ||f||_\infty),$$

where $VI$ denotes the total variation of $f$ on $I$.

Answer 1

First observe tbat $Var_{[a,b)}f=\lim_{x\rightarrow b^{-}}Var_{[a,x]}f$. Let $x_0=a<x_1<...<x_{n-1}<x_n=b$ be a partition of $[a,b]$. Note that $$\sum_{i=1}^n|f(x_i)-f(x_{i-1})|=\sum_{i=1}^{n-1}|f(x_i)-f(x_{i-1})|+|f(x_{n-1})|$$

By using the definition of variation and the fact that $|f(y)|\leq\|f\|_\infty$, we conclude from the last inequality that $$\sum_{i=1}^n|f(x_i)-f(x_{i-1})|\leq Var_{[a,x_{n-1}]}f+\|f\|_\infty$$

If we take the limit of the last inequality when $x_{n-1}\rightarrow b^-$, we conclude that $$\sum_{i=1}^{n-1}|f(y_i)-f(y_{i-1})|\leq Var_{[a,b)}f+\|f\|_\infty$$

for the new partition $y_0=a<y_1<...<y_{n-2}<y_{n-1}=b$ of $[a,b]$, where $y_0=x_0,y_1=x_1,...,y_{n-2}=x_{n-2}$. By taking the sup, we conclude that $$Var_{[a,b]} f\leq Var_{[a,b)}f+\|f\|_\infty$$