# Variation of periodic function

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

• I want to know how to prove it in case when $c$ is not integer. – A.B May 8 '12 at 12:38

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.

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.

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.

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?