Does bounded variation imply boundedness

Question

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

1.When the domain is a bounded interval $[a,b]$, the statement holds.

2.When the domain is $\mathbb{R}$ and the function is monotone, the statement holds both ways (if and only if).

But what about in general? My guess is true, and here is my arguement:

If $||f||_{TV} < \infty$, then $f$ only has jump discontinuities, so we can bound $|f|\leq g$ where $g$ is monotone. By this construction, $||g||_{TV}\leq ||f||_{TV}<\infty$. By (2), we know $g$ is bounded, thus $f$ is bounded.

Edit: I forgot, I probably need to impose a limit behavior at $\pm \infty$ for the function $g$. would $\limsup_{x\rightarrow \infty} g(x) - |f(x)| = 0$ be enough?

is this correct and thank you for your help!

Answer 1

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 $x$.

Answer 2

If $f$ is a real valued function of bounded variation on an interval [a,b] then $f$ is bounded because

$$ |f(x)| \leq |f(a)| + |f(x) - f(a)| \leq |f(a)| + \sup_{x_0 < \cdots < x_n}\sum_{i = 1}^{n} |f(x_i) - f(x_{i - 1})|. $$ for all $x \in [a,b]$. I believe the argument by Hagen was for the closed interval $[0,1]$.

Answer 3

We can also do it by contradiction: assuming $f$ is not bounded, then using definition of unbounded functions, choose such a partition and hence prove that our function can not be of bounded variation.