1

Claim

$f$ is Of Bounded Variation $\Rightarrow$ $f$ is Bounded

Proof

To prove above claim, I had seen the proof like below:

"That $\|f\|_{TV}<\infty$ implies that $f$ is bounded is quite straightforward: $|f(x)|\le|f(0)|+|f(x)-f(0)|\le |f(0)|+\|f\|_{TV}$ holds for all $x$"

EDIT $$||f||_{TV} := \sup_{x_0<\cdots<x_n}\sum_{i=1}^{n} |f(x_i) - f(x_{i-1})|.$$

How does one can derive the fact that $\mid f(x)-f(0) \le \|f\|_{TV}$?

Daschin
  • 695

2 Answers2

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Let $f$ be a function of bounded variation on a interval $I\subseteq\mathbb{R}$. Let us fix a point $a\in I$. Let $x\in I$; assume, w.l.o.g., $x > a$.

Using the partition $x_0=a$, $x_1=x$ in the definition of total variation, you clearly have $$ |f(x) - f(a)| \leq TV(f, [a,x]), $$ hence $$ |f(x)| \leq |f(x) - f(a)| + |f(a)| \leq TV(f, [a,x]) + |f(a)| \leq TV(f) + |f(a)|. $$

Rigel
  • 14,634
2

$f$ is of Bounded variation on $[a, b]$ means that

$sup_P \sum_{i = 0}^{N_p} |f(x_{i+1}) - f(x_{i})| = L < \infty$, where sup over all partitions P of $[a b]$.

Then you can find some partition, which contains $x$ (Actually for all $x$ you can find partition with $x_i = x$ for some $i$)

$|f(x)| = |f(x_i)| = |f(x_i) - f(x_{i-1}) + f(x_{i-1})| \leq |f(x_i) - f(x_{i-1})| + |f(x_{i-1})| \leq ... \leq \sum_{j = 0}^{i-1} |f(x_{j+1}) - f(x_{j})| + |f(a)| \leq |f(a)| + L$

duncan
  • 415