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Folland Real Analysis Exercise 6.34

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The question for reference

Folland 6.34 - If $f$ is absolutely continuous on $[\epsilon,1]$ for $0<\epsilon<1$and $\int_0^1 x|f'(x)|^p dx< \infty$, then $\lim_{x\to 0}f(x)$ exists(and is finite) if $p>2$, $\frac{|f(x)|}{|\ln(x)|^\frac{1}{2}}\to 0$as $x\to 0$ if $p=2$, $\frac{|f(x)|}{x^{1-\frac{2}{p}}}\to 0$ as $x> \to 0$ if $p<2$.

Now I've gotten the cases for $p>2$ and $p<2$. However I literally am unable to figure out $p=2$. Proceeding similary to how I did the others, let $M:=\int_0^1 x|f'(x)|^2dx$ and $e^{-1}\geq x > 0$.$$\begin{align*}|f(x)| &\leq \int_x^1 |f'(t)|dt + |f(1)|\\&\leq \int_0^1 \left(\chi_{(0,\infty)}(t-x)t^{-\frac{1}{2}}\right)\left(t^{\frac{1}{2}}|f'(t)|\right)dt+|f(1)|\\&\leq \left(\int_0^1 t|f'(t)|^2 dt\right)^{\frac{1}{2}}\left(\int_0^1 \chi_{(0,\infty)}(t-x)t^{-1} dt\right)^\frac{1}{2}+|f(1)| \tag{Holder}\\&= M\biggl(\ln(t)\Bigr|_x^1\biggr)^\frac{1}{2}+|f(1)|\\\frac{|f(x)|}{|\ln(x)|^\frac{1}{2}} &\leq M + |f(1)|\end{align*}$$But this bound only shows it's finite. I tried some weird stuff and found that the ratio is $> 1$ but that's pretty useless. L'Hopital's didn't go anywhere when I tried. Fairly certain I just need to try something else but I'm out of ideas.


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