Problem. Assume that $f:[1,2]\to \Bbb R$ is absolutely continuous, with $f(2) = 0$. Prove that $$\left|\int_1^2 f'(x) \log x\, dx \right| \le \int_1^2 |f(x)|\, dx \tag{1}$$ Source: Problem $2$, Pg. $2$.
The proof is quite straightforward; we use integration by parts to get $$\int_1^2 f'(x) \log x\, dx = f(2)\log 2 - f(1)\log 1 - \int_1^2 \frac{f(x)}{x}\, dx = -\int_1^2 \frac{f(x)}{x}\, dx$$ so that $$\left|\int_1^2 f'(x) \log x\, dx \right| = \left|\int_1^2 \frac{f(x)}{x}\, dx \right|\le \int_1^2 \left|\frac{f(x)}{x} \right|\, dx \le \int_1^2 |f(x)|\, dx$$ as required.
My Question: Do we really need absolute continuity of $f$ to get $(1)$? What if $f$ is just (i) differentiable almost everywhere on $[1,2]$? or (ii) of bounded variation? Essentially, I would like to know if a weaker hypothesis on $f$ could also lead to $(1)$.
Note that functions of bounded variation can be decomposed as a sum of two monotone functions, and hence, are differentiable almost everywhere. We interpret $\int f'(x)\log x\, dx$ in the sense of the Lebesgue measure on $\Bbb R$.
"Differentiable almost everywhere on $[1,2]$". No, using Cantor function you can construct $f$ such that $f'=1$ almost everywhere and that $f(2)=0$, yet $\| f\|_{\infty}$ is close to $0$ however you want.
"Of continuous bounded variation".
Yes, but you have to specify what you mean by the integrand with $f'(x)$.. As OP has changed his question, the answer for this case is now NO. Counterexamples are as above.