Answer to Question #164578 in Real Analysis for Ebenezer

Question

Show that the linear combination of two functions of bounded variation is also of bounded variation. Is the product of two such functions also of bounded variation? 

Answer 1

Let $f(x)$ and $g(x)$ be two functions of bounded variations.

So, the sum i.e., $f(x)+g(x)$

We can get $h(x)=f(x)+g(x)$

Using the definition of a function of continuous variation we know that the total variation is atmost the variation of $f(x)+g(x)$

So, $h(x)=f(x)+g(x)$ is a function of continuous variation.

Now, let $c(x)=f(x)\cdot g(x)$

We can re-write the function as:-

$|(fg)(x)-(fg)(y)|\leq |f(x)||g(x)-g(y)|+|g(y)||f(x)-f(y)|\>\> [triangle\>inequality]$

So, this is a transposition of the previous function i.e., the sum of the functions.

So, we conclude that the product of two functions is also of bounded variation.