Variation of periodic function

Question

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 $a=x_0<x_1<\cdots<x_{n-1}<x_n=b$, is called a variation of $f$ on $[a,b]$.

Let $f:R \rightarrow R$ be $1$-periodic and $c \in R$.

Is it then $\operatorname{var}(f, [0,1])=\operatorname{var}(f, [c, c+1])$ ?

Answer 1

The proposed statement is true. Suppose $c$ is not an integer and $c<n<c+1$, and $n$ is an integer.

• The variation of $f$ on $[c,c+1]$ is the sum of the variations of $f$ on $[c,n]$ and on $[n,c+1]$.
• The variation of $f$ on $[n,c+1]$ is the same as that on $[n-1,c]$.
• Therefore, the variation of $f$ on $[c,c+1]$ is the same as that on $[n,n+1]$.

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.

Answer 2

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.

Answer 3

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 $0=x_0<\cdots<x_n=1$. It suffices to show that $$\sum |f(x_{i+1})-f(x_i)|\leq \operatorname{var}(f,[c,c+1]).$$ How can you relate the sequences $c=x_0+c<\cdots<x_n+c=1+c$ and $c=x_0+c<\cdots<x_j+c<z<x_{j+1}+c<x_n+c=1+c$ for some integer $z$? How does the triangle inequality help? What is the value of $$\sum |f(x_{i+1}+c)-f(x_i+c)|+|f(x_{j+1}+c)-f(z)|+|f(z)-f(x_j+c)|$$ compared to the above?

Answer 4

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.