Denote by $L^1(0,1)$ the space of Lebesgue integrable functions on the interval $(0,1)$.


$\textbf{Question:}$ Does there exist a function $F:(0,1)\rightarrow\mathbb{R}$ such that:


1. $\frac{F(x)}{x}\in L^1(0,1)$,
2. $\frac{F'(x)}{x}\in L^1(0,1)$,
3. $\frac{F(x)}{x^2}\notin L^1(0,1)$?


I'm guessing that the answer is positive 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!



Example of a function with a curious property
There is no such function. First of all, $|F(a)-F(b)|\leqslant \int\_a^b |F'(x)|dx\to 0$ when $a,b\to 0$. So $F$ has a limit $c$ at point 0. If $c\ne 0$, then 1) fails. So $\lim\_{x\to 0} F(x)=0$.


Next, $$|F(a)|\leqslant \int\_{0}^a|F'(x)|dx\leqslant a\int\_{0}^a\frac{|F'(x)|}x dx=o(a),\quad\text{when}\quad a\to 0.\quad (1)$$
Now
$$
\int\_a^b \frac{F(x)}{x^2}dx=\frac{F(a)}a-\frac{F(b)}b+\int\_a^b \frac{F'(x)}xdx. \quad(2)
$$
Consider two cases:


1. $F$ has fixed sign near 0. Then choosing $a,b$ close to 0 we conclude from (1) and (2) that $\int \frac{F(x)}{x^2}dx$ converges at 0, but this is equivalent to the convergence of $\int \frac{|F(x)|}{x^2}dx$ which we need.
2. $F$ has infinitely many zeroes in any neighborhood of 0. Then choosing $(a\_k,b\_k)$ being inclusion-maximal intervals of the open set $\{x:F(x)\ne 0\}$ and applying (2) for $a=a\_k,b=b\_k$ we bound $\int\_0^c \frac{|F(x)|}{x^2}dx$ via
$\int\_0^c \frac{|F'(x)|}{x}dx$. Here $c=b\_1$, for example.


