Let $f:[a,b]\to \mathbb{R}$ and let $D=\{x_o,x_1,...,x_n\}$ be a division of $[a,b]$. We say that $f$ is of bounded variation on $[a,b]$ if $\displaystyle \sup_{D\in \mathscr{D}} \sum_{i=1}^n |f(x_i)-f(x_{i-1})|<\infty$, where $\mathscr{D}$ is a collection of divisions of $[a,b]$.
My question is: Does Bounded Variation implies that a function is Bounded? Thanks.
$\endgroup$3 Answers
$\begingroup$Look $BV(\mathbb{R}) \subset L^\infty(\mathbb{R})$ in fact $$|f(x)| = |f(x)-f(a)+f(a)|\le|f(x)-f(a)|+|f(a)|\le|f(a)|+M$$
Where M is the total variation of the function.
$\endgroup$ $\begingroup$Yes, because for arbitrary $x$ you can easily find $D$ showing that $|f(x)-f(a)|\le $ the given bound.
$\endgroup$ 3 $\begingroup$For any division $D \in \mathfrak{D}$, the inequality $\sum_{i=1}^n |f(x_i)-f(x_{i-1})| \leq \sup_{D\in\mathfrak{D}} \sum_{i=1}^n|f(x_i)-f(x_{i-1})| =: C < \infty$ holds. For any $x \in [a,b]$, let $D = \{a,x\}$ so you get $|f(x)-f(a)| \leq C \Rightarrow |f(x)| = |f(x)-f(a)+f(a)| \leq |f(x)-f(a)|+|f(a)| \leq C+|f(a)|$.
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