Problem 3

Exercise 3.3 (Product of absolutely continuous functions)   Let $f,g:[0,1]\to\mathbb{R}$ be absolutely continuous. Then their product is absolutely continuous.

Proof. Select $\delta_1,\delta_2>0$ via absolute continuity so that

$$\sum_{i=1}^n \vert y_i - x_i\vert < \delta_1 \implies \sum_{i=1}^n \vert f(y_i) - f(x_i)\vert < \epsilon/\Vert g\Vert$$ (3.19)

$$\sum_{i=1}^n \vert y_i - x_i\vert < \delta_2 \implies \sum_{i=1}^n \vert g(y_i) - g(x_i)\vert < \epsilon/\Vert f\Vert$$ (3.20)

and set $\delta=\min\{\delta_1,\delta_2\}$. Then we have

$$\sum_{i=1}^n \vert(fg)(y_i) - (fg)(x_i)\vert = \sum_{i=1}^n \vert f(y_i)g(y_i) - f(x_i)g(x_i)\vert$$ (3.21)

$$= \sum_{i=1}^n \vert f(y_i)g(y_i) - f(y_i)g(x_i) + f(y_i)g(x_i) - f(x_i)g(x_i)\vert$$ (3.22)

$$\leq \sum_{i=1}^n \vert f(y_i)\vert\vert g(y_i)-g(x_i)\vert + \vert g(x_i)\vert\vert f(y_i)-f(x_i)\vert$$ (3.23)

$$\leq \sum_{i=1}^n \Vert f\Vert\vert g(y_i)-g(x_i)\vert + \Vert g\Vert\vert f(y_i)-f(x_i)\vert$$ (3.24)

$$\leq \Vert f\Vert\sum_{i=1}^n \vert g(y_i)-g(x_i)\vert + \Vert g\Vert\sum_{i=1}^n\vert f(y_i)-f(x_i)\vert$$ (3.25)

These sums are both bounded by $\epsilon$ if $\sum_i \vert y_i-x_i\vert<\delta$, indicating $fg : [0,1]\to\mathbb{R}$ is absolutely continuous. $\qedsymbol$