Some questions about functions of bounded variation: Jordan's theorem

Question

I was trying to do some of these questions to check my understanding about the topic, but I'm not sure if they're correct. Here are my answers.

1) Suppose $f$ is continuous on $[0,1]$. Must there be a nondegenerate closed subinterval $[a,b]$ of $[0,1]$ for which the restriction of $f$ to $[a,b]$ is of bounded variation?

No, because $f(x)$ might be a constant function. In this case, $f(x)$ will not be the difference of two increasing functions...and so will not be of bounded variation.

2) Let $f$ be the Dirichlet function, the characteristic function of the rationals in $[0,1]$. Is $f$ of bounded variation on $[0,1]$?

No, because we cannot express it as the difference of two increasing functions.

3) Let $f$ be a step function on $[a,b]$. Find a formula for its total variation.

If we partition $[a,b]$ in such a way that for each $[x_{i-1}, x_i]$ (which is part of the partition), $x_{i-1}$ and $x_i$ each correspond to the midpoint of a step. I know this is not a formal answer, but I'm just trying to check my understanding.

Answer 1

1. $f(x)$ might be a constant function

   It is true, but it does not contradict the claim, as a constant function is of bounded variation.

   I think the following can give a counter-example: for each integer $n$, let $f_n$ be the polygonal interpolation of the points $(k2^{-n},(-1)^k)$, $0\leqslant k\leqslant2^n$ and $f:=\sum_{n\geqslant 1}n^{-2}f_n$. This define a continuous function which is not of bounded variation on any interval $[a,b]$.

2. No, because we cannot express it as the difference of two increasing functions.

   That is true, but this need justifications (it is like if you answered "No, because $f$ is not of bounded variation").

   To get a justification, consider an integer $N$, and consider a subdivision $0=t_1<t_2<\dots<t_{2N}=1$, where $t_k$ is rational for $k$ odd and irrational for $k$ even.

3. Call $c_i$ the value of $f$ on $(x_{i-1},x_i)$. There is a subdivision such that the associated sum is $\sum_{j=0}^{n-1}\left|c_{j+1}-c_j\right|$. Now take an arbitrary partition and show we cannot do better. There are case we need to consider, like there are several points on $[x_{i-1},x_i)$, or at the contrary this interval is avoided.