I asked this question on MO but haven't received any answers, so I thought I'd try here. Denote by $L^1(0,1)$ the space of Lebesgue integrable functions on the interval $(0,1)$.
Question: Does there exist a function $F:(0,1)\rightarrow\mathbb{R}$ such that:
- $\frac{F(x)}{x}\in L^1(0,1)$,
- $\frac{F'(x)}{x}\in L^1(0,1)$,
- $\frac{F(x)}{x^2}\notin L^1(0,1)$?
My guess is that the answer is yes and the point is to construct $F$ such that $F$ and $F'$ behave suitably near zero. It seems quite delicate. I checked that $F$ cannot be a polynomial or a power function (since then $F'\simeq \frac{F}x$, thus conditions 2 and 3 cannot hold simultaneously).
I would appreciate any hints!
No such $F$ exists. From 2. we obtain that $$F(x) = \int_0^x t\cdot g(t)\,dt$$ for some $g \in L^1(0,1)$ (perhaps with additional regularity if e.g. $F$ should be continuously differentiable rather than merely absolutely continuous). A nonzero integration constant would violate 1. Since $$\lvert F(x)\rvert \leqslant x \int_0^x \lvert g(t)\rvert\,dt$$ the function $x \mapsto F(x)/x$ is bounded (and continuous), hence belongs to $L^1(0,1)$.
By Tonelli's theorem we have \begin{align} \int_0^1 \frac{\lvert F(x)\rvert}{x^2}\,dx &\leqslant \int_0^1 \frac{1}{x^2} \int_0^x t\cdot \lvert g(t)\rvert\,dt\,dx \\ &= \int_0^1 t\cdot \lvert g(t)\rvert \int_t^1 \frac{dx}{x^2}\,dt \\ &= \int_0^1 t\cdot\lvert g(t)\rvert \biggl(\frac{1}{t} - 1\biggr)\,dt \\ &\leqslant \int_0^1 \lvert g(t)\rvert\,dt + \int_0^1 t\cdot \lvert g(t)\rvert\,dt \\ &< +\infty\,, \end{align} hence also $\frac{F(x)}{x^2} \in L^1(0,1)$.