The Unapologetic Mathematician

Mathematics for the interested outsider

Functions of Bounded Variation II

Let’s consider the collection of functions of bounded variation on \left[a,b\right] a little more deeply. It turns out that they form a subring of the ring of all real-valued functions on \left[a,b\right]. Just to be clear, the collection of all real-valued functions on an interval becomes a ring by defining addition and multiplication pointwise.

Okay, so to check that we’ve got a subring we just have to check that the sum, difference, and product of two functions of bounded variation is again of bounded variation. Let’s take f and g to be two functions of bounded variation on \left[a,b\right], and let (x_0,...,x_n) be a partition of \left[a,b\right]. Then we calculate

|f(x_i)g(x_i)-f(x_{i-1})g(x_{i-1})|=
|f(x_i)g(x_i)-f(x_{i-1})g(x_i)+f(x_{i-1})g(x_i)-f(x_{i-1})g(x_{i-1})|\leq
|f(x_i)g(x_i)-f(x_{i-1})g(x_i)|+|f(x_{i-1})g(x_i)-f(x_{i-1})g(x_{i-1})|=
|g(x_i)||f(x_i)-f(x_{i-1})|+|f(x_{i-1})||g(x_i)-g(x_{i-1})|\leq
A|f(x_i)-f(x_{i-1})|+B|g(x_i)-g(x_{i-1})|

where A is the least upper bound of |g(x)| on \left[a,b\right], and B is the least upper bound of |f(x)|. Then we find AV_f+BV_g is an upper bound for the sum over the partition. In fact, this not only shows that the product fg is of bounded variation, it establishes the inequality V_{fg}\leq AV_f+BV_g.

The proofs for the sum and difference are similar. You should be able to work them out, and to establish the inequality V_{f\pm g}\leq V_f+V_g.

We can’t manage to get quotients of functions because we can’t generally divide functions. The denominator might be {0} at some point, after all. But if f(x) is bounded away from {0} — if there is an m with 0<m\leq|f(x)| — then g(x)=\frac{1}{f(x)} is of bounded variation, and V_g\leq\frac{V_f}{m^2}. Indeed, we can check that

\displaystyle\left|g(x_i)-g(x_{i-1})\right|=\left|\frac{1}{f(x_i)}-\frac{1}{f(x_{i-1})}\right|=
\displaystyle\left|\frac{f(x_i)-f(x_{i-1})}{f(x_i)f(x_{i-1})}\right|\leq\frac{\left|f(x_i)-f(x_{i-1})\right|}{m^2}

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March 6, 2008 - Posted by | Analysis

2 Comments »

  1. Very minor typo: A is the lub of |g(x)|, and similarly for B. (Feel free to delete this comment.)

    Comment by Todd Trimble | March 6, 2008 | Reply

  2. good work.

    Comment by simiyu | December 19, 2013 | Reply


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