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Let $f : [a, b] \to R$ be continuous and suppose that $|f|$ is of bounded variation over $[a, b]$.

Show that then $f$ is of bounded variation over $[a, b]$ and give a counterexample to the above statement in case $f$ is not continuous.

To prove above claim, I had tried to used the reverse triangular inequality, but only succeeded to prove "$f$ is of bounded variation $\implies |f|$ is of bounded variation".

Any adivce/hint to prove above claim?

Daniele Tampieri
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Daschin
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2 Answers2

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Let $a = x_0 < x_1 < \cdots < x_n = b$ be a partition of $[a,b]$.

Let $i\in\{1,\ldots,n\}$. If $f(x_{i-1}) \cdot f(x_i) < 0$, then by the continuity of $f$ there exists a point $t\in (x_{i-1}, x_i)$ such that $f(t) = 0$. We clearly have $$ |f(x_i) - f(x_{i-1})| \leq |f(x_i) - f(t)| + |f(t) - f(x_{i-1})| = ||f(x_i)| - |f(t)|| + ||f(t)| - |f(x_{i-1})|| $$ In this case, let us add the point $t$ to the partition.

In this way we construct a new partition $a=y_0 < y_1 < \cdots y_m = b$ such that $$ \sum_{i=1}^n |f(x_i) - f(x_{i-1})| \leq \sum_{j=1}^m ||f(y_j)| - |f(y_{j-1})|| \leq TV(|f|, [a,b]). $$ This shows that $$ TV(f, [a,b]) \leq TV(|f|, [a,b]). $$

Rigel
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  • Very nice proof +1. The last inequality also holds in reverse (because of triangle inequality) so that total variation of $f$ is same as that of $|f|$. – Paramanand Singh May 28 '17 at 06:01
  • How does this equality -$|f(x_i) - f(t)| + |f(t) - f(x_{i-1})| = ||f(x_i)| - |f(t)|| + ||f(t)| - |f(x_{i-1})|| $ hold? – Daschin May 28 '17 at 06:08
  • @Daschin: Use the fact that $f(t) =0$. – Paramanand Singh May 28 '17 at 06:11
  • @ParamanandSingh got it. – Daschin May 28 '17 at 06:31
  • $$ \sum_{i=1}^n |f(x_i) - f(x_{i-1})| \leq \sum_{j=1}^m ||f(y_j)| - |f(y_{j-1})||$$ from this part, does the inequality still hold even though we changed the partition to the new one? – Daschin May 28 '17 at 06:37
  • Indeed in general the inequality holds only if we change the partition. Each term at the lhs is estimated through the first inequality in the answer. Thus at the rhs you need one or two terms for each term at the lhs, according with the sign of $f$ at $x_{i-1}$ and $x_i$. – Rigel May 28 '17 at 07:53
  • You have to put in the new partition only one point where $f=0$ for any interval where $f$ changes sign. – Rigel May 28 '17 at 10:36
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Hint for the counterexample. Consider the map $f$ such that $f(x)=1$ for $x\in \mathbb{Q}$ and $f(x)=-1$ otherwise.

Robert Z
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  • for your recommended $f$ want to construct partition to makke the BV of f diverge, I would like to alternatively put rational number and irrational number. Any nice pattern of this kind to put infinitely many terms in finite range of $\Bbb R$? – Daschin May 28 '17 at 06:52
  • Yes, this is the idea. – Robert Z May 28 '17 at 06:54