If $f\colon\Bbb R\to\Bbb R$ then $\operatorname{var}(f, [a,b]):=\sup \{\sum_{k=1}^n |f(x_k)-f(x_{k-1})| \}$,
where supremum is taken over all finite sequences $(x_k)$ such that $a=x_0 Let $f:R \rightarrow R$ be $1$-periodic and $c \in R$. Is it then $\operatorname{var}(f, [0,1])=\operatorname{var}(f, [c, c+1])$ ?
Variation of periodic function
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0I want to know how to prove it in case when $c$ is not integer. – 2012-05-08
4 Answers
Consider $c\in (p-1,p)$ with $p$ integer Then
$$V[c,c+1]= V[c,p]+V[p,c+1]=V[p-1,c]+V[c,p]+V[p,c+1]-V[p-1,c]$$
Where we used that $V[a,c]+V[c,b]=V[a,b]$,
Noticing that $V[a+1,b+1]=V[a,b]$ then
$$V[c,c+1]=V[p-1,p]+V[p-1,c]-V[p-1,c]=V[p,p+1]=V[0,1].$$
PS:That is the same solution for the integral of a 1-periodic function of an interval with length 1.
Yes.
Claim 1 Let $(x_k)$ be a finite sequence such that $a = x_0 < x_1<\ldots < x_n = b$. Let $(y_k)$ be a refinement of $(x_k)$ (meaning that every point of $x_k$ is some point in $y_k$) with total of $m+1$ points. Then $\sum_{k = 1}^n |f(x_k) - f(x_{k-1})| \leq \sum_{k = 1}^{m} |f(y_k) - f(y_{k-1})|$. That is, refinements increases the sum.
(Hint this follows by triangle inequality.)
Claim 2 Every $(x_k)$ with $x_0 = 0$ and $x_n = 1$ can be refined to a sequence $(y_k)$ such that $[c]$ (the fractional part of the number $c$) is a point in $(y_k)$.
Claim 3 For $(y_k)$ as above, a cyclic permutation of the indices + a translation allows you to identify it with some sequence $(z_k)$ with $z_0 = c$ and $z_{m} = c+1$. Furthermore the associated sums $\sum_{k=1}^m |f(y_k) - f(y_{k-1})| = \sum_{k=1}^m |f(z_k) - f(z_{k-1})|$ are equal.
Similarly any sequence on $[c,c+1]$ can be refined to a sequence that can be identified with one on $[0,1]$. Use this to argue that the supremum must be equal.
The proposed statement is true. Suppose $c$ is not an integer and $c I think most of the work is in proving the first bulleted statement above; your question makes it appear that you know how to do the rest.
Is this homework? What have you tried so far?
If you want to show that the claim is true, here is the strategy:
You might want to show $$\operatorname{var}(f,[0,1])\leq \operatorname{var}(f,[c,c+1])$$ and vice versa.
Choose a sequence $0=x_0<\cdots