Let $x=0.a_1a_2\dots$ be the decimal expansion of a number $x$, $0<x<1$. If two decimal expansions of $x$ exist, the one that ends with $0$’s is taken. For what values of $q > 1$ is the function of bounded variation? \begin{align*} f_q(x) = \sum\limits_{k=1}^\infty q^{-k} a_k \end{align*}
I am not sure how to approach this problem. I know that if I want to show a function if not of bounded variation I must find an appropriate partition, but I don't know what intuition will help me find one for this function. I will appreciate any hints. Thanks!
For $1<q<10$, $f$ is not of bounded variation: observe that $f$ jumps at every $x$ with ambiguous decimal expansion. Say $x=0.a_1\ldots a_{n-1}10000\ldots$. Compare $f(x)$ with $f(y)$, where $$ y=0.a_1\ldots a_{n-1}099\ldots 9000\ldots , $$ with a long string of $9$'s. Then $$ f(x) - f(y) \to q^{-n}\left( 1 - \frac{9}{q-1} \right) $$ as we increase the length of this segment. Since there $10^{n-1}$ initial segments $a_1\ldots a_{n-1}$ that can go here, we see that these jumps contribute $\gtrsim (10/q)^n$ to the total variation, and $f\notin BV$.
For $q\ge 10$, $f$ is indeed increasing and thus BV. If $x>y$ and $n$ is the first digit where the decimal expansions of $x,y$ differ, then it suffices to observe that $$ \sum_{j>n} 9q^{-j} = \frac{9}{q-1} q^{-n} \le q^{-n} . $$