Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
I want to know how to prove it in case when $c$ is not integer. –  L.T May 8 '12 at 12:38

4 Answers 4

up vote 2 down vote accepted

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.

share|improve this answer

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.

share|improve this answer

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.

share|improve this answer

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?

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.