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Construct a continuous $f$ such that $f$ on $[0, 1]$ is not of bounded variation on any interval. Hint on book for this question is modified Cantor lebesgue function.

The things I thought is I can prove that it is not lipitchitz function, then can I just say that it is not bounded variation on any interval by moving term? If not, how can I modified and get the results?

I know if without hint, we can consider a function like $xsin(1/x)$. But I want to know some examples that make by hint.

Edit: The full hint in book is as below enter image description here

Apple
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  • A better way than using the hint: Any continuous nowhere differentiable function is not of bounded variation on any interval. – geetha290krm Oct 12 '23 at 08:50
  • $x \sin (\frac 1 x)$ is of bounded varaition on $[c,1]$ for any $c>0$. – geetha290krm Oct 12 '23 at 08:52
  • The book is Measure and Integral by Richard L. Wheeden and Antoni Zygmund. It is an exercise on it. Because this section is talk about Cantor lebesgue function so this hint give like that. Do you have any idea about it? – Apple Oct 12 '23 at 10:32
  • @geetha290krm: To me the hint looks pretty bad. --- For more context about this problem, see Exercise 26 on p. 49 [= .pdf page 31] of this copy of the 1977 edition. In particular, note that the exercise follows that of various Cantor set constructions and is much simpler than constructing (with proof) a nowhere differentiable continuous function (to say nothing of the yet-to-be-proved nontrivial fact that BV functions are differentiable almost everywhere, or even somewhere). Note also the follow-up Ex. # 27. – Dave L. Renfro Oct 12 '23 at 10:52
  • I don't have time to work out and write-up an answer, but it's possible that some of the references I posted here could help in getting an idea of what is intended. – Dave L. Renfro Oct 12 '23 at 10:59
  • I think it want to make a polygonal function something like Weierstrass function, then try to prove it nowhere differentiable function. Finally prove it is not of bounded variation on any interval, right? How I can prove that it nowhere differentiable function? – Apple Oct 12 '23 at 11:05
  • Nowhere differentiability is much more elaborate than what you want or need, and besides, the proof that a monotone function has even a single point of differentiability is not trivial (done much later in WZ's text). The type of stuff required for the WZ exercise involves much more basic mathematical reasoning, the type I use in my answer to Curve In a Closed Interval with an Infinite Length. See also my comment to Curve with non finite arc length (understanding the calculations). – Dave L. Renfro Oct 12 '23 at 11:34
  • Sorry, I make a mistake that the second step form the hint won't change the value of $(1/3,2/3)$ right? So that is won't be a polygonal function that I say above. – Apple Oct 12 '23 at 11:45

1 Answers1

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Define $f:\Bbb R \to \Bbb R$ by $f(x)=x$, for $0\leqslant x\leqslant\frac12$, with $f(x)=\frac12-x$ for $\frac12<x\leqslant1$ and $f(x)=0$ otherwise. This function is continuous and piecewise affine: It comprises two non-zero affine functions respectively on two adjacent intervals, and is zero elsewhere.

The basic idea is to build a function in stages. Starting with $f_1:=f$, at stage $n$ $(n=1,2,...)$ we construct a function $f_n$. At stage $n+1$, we make smaller, steeper, and shifted copies of $f$ and add them to $f_n$ to build a more jagged function $f_{n+1}$. These little copies form a relative uptick and downtick in each affine segment of $f_n$, while preserving continuity and piecewise affinity.

At each stage $n=2,3,...\,$, the number of affine segments is doubled. Thus, suppose that we have just built $f_n$, which has $2^n$ affine segments of equal length. To make $f_{n+1}$, to each segment is added a small scaled shifted copy of $f$ to fit. Thus, say that $f_n$ has an affine segment between $k/2^n$ and $(k+1)/2^n$, where $k\in\{0,...,2^n-1\}$; what we add is $$f_{n,k}:x\mapsto\frac1{n^2}f(2^nx-k)\quad(x\in\Bbb R),$$ so that $$f_{n+1}:=f_n+\sum_{k=0}^{2^n-1}f_{n,k}\quad(n=1,2,...).$$ Then $f_\infty:=\lim_{n\to\infty}f_n$ is continuous but of unbounded variation on any subinterval of $[0\,\pmb,\,1]$.

John Bentin
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  • I edit the whole hint above the text. Do you have some idea about the hint? I see you answer but not really understand your construction, can you write $f_1$,$f_2$ what they really look like? And we the last sentence you get unbounded variation, from where we can conclude? – Apple Oct 12 '23 at 13:12