$|f|$ is of bounded variation $\implies f$ is of bounded variation

Question

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.

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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?

Answer 1

Hint for the counterexample. Consider the map $f$ such that $f(x)=1$ for $x\in \mathbb{Q}$ and $f(x)=-1$ otherwise.

Answer 2

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 $$ TV(f, [a,b]) \leq TV(|f|, [a,b]). $$