0
$\begingroup$

We proved in class today that if $f ,g : [a,b] \rightarrow \mathbb{R}$ are functions of bounded variation then so if $fg$. It was done directly using definitions and it was rather lengthy in my opinion. We then discussed Jordan's theorem. i.e A function $f$ is of bounded variation on $[a, b] $ if and only if it is a difference of 2 increasing functions on $[a, b].$ My question is does any have a proof of if $f ,g : [a,b] \rightarrow \mathbb{R}$ are functions of bounded variation then so if $fg$ using Jordan's theorem. I feel like using Jordan's theorem might lead to a cleaner proof.

$\endgroup$

1 Answer 1

2
$\begingroup$

First of all, note that any function of bounded variation is automatically bouded (for instance, as every increasing function $f\colon [a, b] \to \mathbb{R}$ is bounded by $f(a)$ and $f(b)$).

Also, note that if $f$ is of bounded variation, we can write it as the difference of two positive increasing functions. In fact, if $f=f_1-f_2$, and $f_1, f_2$ are increasing, then $f=(f_1+c)-(f_2+c)$ for any $c \in \mathbb{R}$; choosing a large enough value of $c$ ensures us that $f_1+c, f_2+c$ are positive and increasing. Keeping this in mind, we can prove your required result one step at a time.

  1. If $f, g$ are of bounded variation, then $f+g$ is also of bounded variation. In fact, write them as $f=f_1-f_2, g=g_1-g_2$, where each $f_i, g_i$ is increasing. Then $f+g=(f_1+f_2)-(g_1+g_2)$ is of bounded variation, as the sum of increasing functions is increasing. A similar argument shows that $cf$ is also of bounded variation if $c$ is a constant.

  2. If $f$ is of bounded variation, then $f^2$ is also. In fact, we can write $f=f_1-f_2$, where $f_1, f_2$ are positive and increasing. Then $f_1^2, f_2^2$ and $f_1f_2$ are positive and increasing as well, and $f^2=(f_1^2+f_2^2)-(2f_1f_2)$.

  3. If $f$ and $g$ are of bounded variation, we can write $fg=((f+g)^2-(f-g)^2)/4$. Here $f+g$ and $f-g$ are of bounded variation by 1., and then we conclude by 2. and 1. again.

$\endgroup$
1
  • $\begingroup$ This is much cleaner than the proof i was shown in class today. very nice $\endgroup$ Jan 17, 2022 at 22:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .