Lebesgue measure #2

Thu 17 July 2014

Filed under math

Tags math

This is a proof of basic closure-type properties with respect to functions of bounded variation. I had the chance to study these when I took a class by Dr. Ovchinnikov, who was in the process of writing a book on the subject at the time.

Show that the sum, difference and product of two BV-functions is a BV-function.

Let $f$, $g$, be two BV functions over $[a, b]$. Define a partition, $P = {x_i = 1 \leq i \leq n }.$ Then,

$ \begin{align*} \sum_{i = 1}^{n} | (f + g )(x_i) - (f + g)(x_{i-1}) | &= \sum_{i = 1}^{n} | { f(x_i) + g(x_i) } - { f(x_{i - 1}) + g(x_{i-1}) } | \\ &\leq \sum_{i = 1}^{n} | f(x_i) - f(x_{i - 1}) | + \sum_{i = 1}^{n} | g(x_i) - g(x_{i - 1})| \\ &\leq V(f, P) + V(g, P) \end{align*}$

$\Rightarrow V(f + g, P) \leq V(f, P) + V(g, P) \leq V(f, P) + V(g, P) \Rightarrow V(f + g, P) \leq V(f, P) + V(g, P). $

Thus, $(f + g)$ is a function of bounded variation.

To show that $(f - g)$ is of bounded variation, the proof is the same and $V(f - g) \leq V(f) + V(g)$

To show that the product of two functions of bounded variation is also of bounded variation, notice that both $f$ and $g$ are bounded, and thus there exists $K \in \mathbb{N}$ such that $|f(x)| \leq k, |g(x) | \leq k, \forall x \in P$.

Thus, $$\begin{align*} \sum_{i = 1}^{n} | (fg)(x_i) - (fg)(x_{i - 1}|&= \sum_{i = 1}^{n} | f(x_i)g(x_i) - f(x_{i -1})g(x{i - 1})| \\ &= \sum_{i = 1}^{n} |f(x_i) {g(x_i) - g(x_{i - 1}) } + g(x_{i-1}) { f(x_i) - f(x_{i - 1}) } | \\ &\leq \sum_{i = 1}^{n} { | f(x_i) | |g(x_i) - g(x_{i - 1}) | + | g(x_{i - 1}) | | f(x_i) - f(x_{i - 1}) | } \\ &\leq k \sum_{i = 1}^{n} |g(x_i) - g(x_{i-1}) | + k \sum_{i = 1}^{n} | f(x_i) - f(x_{i -1})| \\ &\leq k V(g) + kV(f). \end{align*}$$ and so the product of two functions of bounded variation is also of bounded variation.


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