7

In a recent post, a question arose as to the requirements on $f:[0,1] \to \mathbb{R}$ in order to ensure that

$$\tag{1}\lim_{x \to 0+}\log(x) \int_0^x f(t) \, dt = 0$$

I showed that Riemann integrability of $f$ is a sufficient condition. Assuming $f$ is bounded (and all Riemann integrable functions must be bounded) we have $f(t) \leqslant M$ for some $M \geqslant 0$ and, hence,

$$\tag{2}0 \leqslant \left|\log(x) \int_0^x f(t) \, dt \right|= -\log(x)\left|\int_0^x f(t) \, dt\right|\leqslant -\log(x)\int_0^x|f(t)| \, dt\leqslant -Mx\log(x)\underset{x \to 0+}\longrightarrow 0$$

My questions is what is the most general class of functions for which (1) holds?

It clearly holds for some unbounded functions like $x \mapsto x^{-1/2}$ and I believe it is even true for $x \mapsto x^{-1}\sin x^{-1}$ which is not Lebesgue integrable.

Furthermore, since $x \log(x) \to 0$ as $x \to 0+$, it really is a question of the convergence or boundedness of $\frac{1}{x}\int_0^x f(t) \, dt$ and perhaps an application of the Lebesgue differentiation theorem.

RRL
  • 92,835
  • 7
  • 70
  • 142
  • 3
    You probably know this already, but by Holder, a slight adaptation of your argument shows that $f\in L^p([0,1])$, with $1<p\leq \infty$, suffices; but this is, of course, a very crude estimate. – peek-a-boo Apr 29 '23 at 01:29
  • 4
    A sufficient condition is that $$\lim_{x\to0^+} x(\log x)^2 f(x) = 0. $$ – Sangchul Lee Apr 29 '23 at 01:41
  • 2
    Another sufficient condition is that if $F(x) \in o\left(\dfrac{1}{- \log(x}\right)$ and $F(0)=0$ then $f(x) = \dfrac{d}{dx} F(1/x)$ satisfies the property. – Sandipan Samanta Apr 29 '23 at 06:26

0 Answers0