Assume that $f$ is of bounded variation on $[a,b]$ and $f(x)\geq \epsilon>0$ $\forall x \in [a,b]$. Prove that $\ln f$ is of bounded variation on $[a,b]$.
Definition: Let $f$ be defined on $[a,b]$. If $P = \{ x_0, \ldots, x_n\}$ is a partition of $[a,b]$, write $\Delta f_k= f(x_k)-f(x_{k-1})$, for $k=1,\ldots,n$. If there exists $ M\in\mathbb{R^+}$ such that $$\sum_{k=1}^{n}|\Delta f_k| \leq M$$ for all partitions of $[a,b]$, then $f$ is of bounded variation on $[a,b]$.
Attempt: Suppose $f$ is of bounded variation on $[a,b]$ and $f(x)\geq \epsilon>0$ $\forall x \in [a,b]$. Define $g(x) = \ln f(x)$. Then, $$\Delta g_k = \ln f(x_k) - \ln f(x_{k-1})$$
Then, \begin{align} \sum_{k=1}^{n}|\Delta g_k| &= \sum_{k=1}^{n}|\ln f(x_k) - \ln f(x_{k-1})|\\ &= |\ln f(x_1) - \ln f(x_{0})| +\cdots+ |\ln f(x_n) - \ln f(x_{n-1})| \\ &\leq |\ln f(x_1)|+|\ln f(x_0)|+\cdots+ |\ln f(x_n)|+|\ln f(x_{n-1})|, \text{ by triangle inequality} \\ &= |\ln f(x_0)|+|\ln f(x_n)|+2\sum_{k=1}^{n-1}|\ln f(x_k)|, \text{ some terms show up twice}\\ &\leq 2(|\ln f(x_0)|+|\ln f(x_n)|)+2\sum_{k=1}^{n-1}|\ln f(x_k)| \\ &= 2\sum_{k=0}^{n}|\ln f(x_k)| \\ &= M_1, \text{ not sure how to justify} \end{align}
I'm actually not convinced that what I am to prove is actually true. I can see that it's true for certain types of functions, like constant and monotonic functions. Outside of those types, I'm not convince. Any help would be appreciated.
EDIT: Using hints provided below, \begin{align} \sum_{k=1}^{n}|\Delta g_k| &= \sum_{k=1}^{n}|\ln f(x_k) - \ln f(x_{k-1})| \\ &= \sum_{k=1}^{n}\big|\ln\left( \frac{f(x_k)}{f(x_{k-1})}\right)\big| \\ &= \sum_{k=1}^{n}\big|\ln\left( 1+ \frac{f(x_k)-f(x_{k-1})}{f(x_{k-1})}\right)\big| \\ &\leq \sum_{k=1}^{n} \frac{|f(x_k)-f(x_{k-1})|}{\min\{f(x_k),f(x_{k-1})\}}, \text{ from hint below} \\ &\leq \sum_{k=1}^{n} \frac{|f(x_k)-f(x_{k-1})|}{\epsilon}, \text{ since $f(x) \geq \epsilon >0$ for all $x\in [a,b]$}\\ &= \frac{1}{\epsilon}\sum_{k=1}^{n} |f(x_k)-f(x_{k-1})|\\ &\leq \frac{M}{\epsilon} \text{ since $f$ is of bounded variation} \end{align}
If $f(x_k) \geqslant f(x_{k-1})$, then
$$|\ln f(x_k) - \ln f(x_{k-1})| = \ln \frac{f(x_k)}{f(x_{k-1}) } =\ln \left(1 + \frac{f(x_k) - f(x_{k-1})}{ f(x_{k-1})} \right)\leqslant \frac{f(x_k) - f(x_{k-1})}{f(x_{k-1})} = \frac{|f(x_k) - f(x_{k-1})|}{|f(x_{k-1})|}.$$
If $f(x_k) \leqslant f(x_{k-1})$, then
$$|\ln f(x_k) - \ln f(x_{k-1})| = \ln \frac{f(x_{k-1})}{f(x_{k}) } =\ln \left(1 + \frac{f(x_{k-1}) - f(x_{k})}{ f(x_{k})} \right)\leqslant \frac{f(x_{k-1}) - f(x_{k})}{f(x_{k})} = \frac{|f(x_k) - f(x_{k-1})|}{|f(x_{k})|}.$$
Since $f(x_k), f(x_{k-1)} \geqslant \epsilon > 0$ we have
$$|\ln f(x_k) - \ln f(x_{k-1})| \leqslant \frac{|f(x_k) - f(x_{k-1})|}{\epsilon}.$$
This should enable you to finish.