# a product of functions of bounded variation is a function of bounded variation using Jordan's theorem

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.

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.