$xf''(x) , xf', f \in L^{2}$ is $f' \in L^{1}$? I am stuck on the following problem. I have a function $f$ such that 


*

*$f$ is bounded on $(0,1)$,

*$xf'(x)$ is bounded on $(0,1)$,

*$f \in L^{2}(0,1)$,

*$xf' \in L^{2}(0,1)$, and

*$xf'' \in L^{2}(0,1)$.


I wish to show that $f' \in L^{1}(0,1)$ or find a counter example that shows this to be untrue. 
So far, the only functions that I think could present possibly work as counter examples are functions for which $\lim\limits_{x \to 0^+} f(x)$ does not exist. But the common examples ($\sin(1/x)$ and $\sin(\ln(x))$) don't work. 
Thanks.
 A: The following is not a complete answer. In the following let $f$ be some candidate for a counterexample.
First observation: $f$ cannot be monotone. If it were, say, increasing, then we would have $$\int_0^x |f'| dx \leq f(x)-f(0)$$ by the Lebesgue decomposition theorem. Similarly $\int_0^x |f'(x)| dx \leq f(0) - f(x)$ if $f$ is decreasing. 
Second observation: the only way you could have $f' \not \in L^1$ is if it has a non-integrable singularity at $0$, because elsewhere it is bounded by condition $2$. Between condition $2$ and the non-integrable singularity requirement, this means it must diverge only slightly slower than or exactly as fast as $1/x$. Combining these first two requirements also means that $f$ cannot be monotone on $[0,\varepsilon]$ for any $\varepsilon > 0$.
Third observation: condition $3$ is redundant to condition $1$, and condition $4$ is redundant to condition $2$. So you can drop those.
My intuition, then, is to choose $f'$ so that $|f'(x)|=1/x$ but $\int_0^x f'(y) dy$ stays within, say, $\pm 1$. I think we can do this by defining:
$$A_k = \left [e^{-k},e^{-k+1} \right ] \\
A = \bigcup_{k=1}^\infty A_{2k-1} \\
B = \bigcup_{k=1}^\infty A_{2k} \\
g'(x) = \frac{\chi_A(x) - \chi_B(x)}{x}$$
As it stands, this doesn't provide a counterexample; this satisfies condition $1$, $2$, $3$, and $4$, but not condition $5$, since $|xg''(x)|=1/x \not \in L^2$. 
My intuition is to now precompose with a certain singular function, i.e. to define $f'(x) = g'(s(x))$ where $s'=0$ a.e. Then $f''$ would be zero a.e. and we would have $x f'' \in L^2$. But we can't have a function with $s'=0$ a.e. such that $s(x)$ grows like $x$; all of them grow like $x^\alpha$ for some $\alpha < 1$. (The Cantor function, for example, grows like $x^{\log_3(2)}$.)
