Prove that $ \frac {1}{f}$ is a function of bounded variation on $[a,b]$.

Question

I am studying for a test in measure theory. Please help with the following question:

Let $f:[a,b]\to R$ a continuous function of bounded variation, when $f(x)\ne 0$ for every $x \in [a,b]$.
Prove that $ \frac {1}{f}$ is a function of bounded variation on $[a,b]$.

Ofcourse, since $f$ is a continuous function and $f(x)\ne 0$ for every $x \in [a,b]$, then $f(x)>0$ or $f(x)<0$ for every $x \in [a,b]$. Please help to go on.

Answer 1

This is the question's solution following the good hinted answer you got in the other answer:

Since $\;f(x);$ is continuous on $\;[a,b]\;$ , it is bounded there, and since $\;f(x)\neq 0\;\;\forall\,x\in [a,b]\;$ also $\;\dfrac1{f(x)}\;$ is continuous and thus also bounded on $\;[a,b]\;$ , say $\;(*)\;\;\dfrac1{|f(x)|}\le N\;\;\forall\,x\in [a,b]\;$ .

For any partition $\;\{x_k\}_{k=1}^n\;$ of $\;[a,b]\;$ we thus get

$$\sum_{k=1}^{n-1}\left|\frac1{f(x_{k+1})}-\frac1{f(x_k)}\right|=\sum_{k=1}^n\left|\frac{f(x_{k+1})-f(x_k)}{f(x_k)f(x_{k+1})}\right|\stackrel{(*)}\le\sum_{k=1}^nN^2\,|f(x_{k+1})-f(x_k)|\le N^2V(f,[a,b])$$

and we're done as we're given $\;f\;$ is of bounded variation on $\;[a,b]\;$ .

The hint was, imo, enough to deduce the solution to the question, yet some time already passed and there were doubts whether the hint really helped. I think it did.

Answer 2

Hint

$f$ continuous on $[a,b]$ and $\forall x \in [a,b] , \,\, f(x) \neq 0$ implies that either $f > 0$ and it reaches a minimum $v_0 > 0$ or $f < 0$ and it reaches a maximum $v_0 < 0$. Which means $|f| > 0$ and it reaches a minimum $v_0 > 0$.

So for $a \le y_1 < y_2 \le b$, $|\frac{1}{f(y_2)} - \frac{1}{f(y_1)}| = \frac{|f(y_2) - f(y_1)|}{|f(y_1) \cdot f(y_2)|} \le \frac{|f(y_2) - f(y_1)|}{v_0^2} $