Behavior at $0$ of a function that is absolutely continuous on $[\epsilon, 1]$

Question

The function $f$ on $[0,1]$ is absolutely continuous on $[\epsilon,1]$ for $0<\epsilon<1.$ I further have that $$\int_0^1x|f'(x)|^pdx<\infty.$$

I'm trying to show that $$ \lim_{x\to 0}f(x)\ \text{exists and is finite}\qquad \text{if}\ p>2, $$ $$ \frac{f(x)}{|\log x|^{1/2}}\to 0\ \text{as}\ x\to 0\qquad \text{if}\ p=2,\ \text{and} $$ $$ \frac{f(x)}{x^{1-\frac{2}{p}}}\to 0\ \text{as}\ x\to 0\qquad \text{if}\ p<2. $$

(Of course, for $0<\epsilon\leq x\leq 1$ we have that $$ f(\epsilon)=f(x)-\int_{\epsilon}^x f'(x)dx, $$ and, for $s>-1$, $r\geq1$, Hölder's inequality gives $$ \int_0^1x^{s-(s-1)/r}|f'(x)|^{p/r} dx<\infty; $$ in particular, $$ \int_0^1x^a|f'(x)|^{p/2} dx<\infty\qquad \text{for}\ a>0.) $$

I'd greatly appreciate any suggestions!

Answer 1

Hölder's inequality should be used to estimate $\int_\epsilon^1 |f'(x)|\,dx$. $$ \int_\epsilon^1 |f'(x)|\,dx = \int_\epsilon^1 \left(x^{1/p} |f'(x)|\right) x^{-1/p} \,dx \le \left(\int_\epsilon^1 x|f'(x)|^p\,dx\right)^{1/p} \left(\int_\epsilon^1 x^{1/(1-p ) }\,dx\right)^{1-1/p} \tag{1} $$ Since the first factor in the product on the right of (1) is bounded by $C = \left(\int_0^1 x|f'(x)|^p\,dx\right)^{1/p} $, we have $$ \int_\epsilon^1 |f'(x)|\,dx \le C \left(\int_\epsilon^1 x^{1/(1-p) }\,dx\right)^{1-1/p} \tag{2} $$

• When $p>2$, the integral on the right of (2) converges.
• When $p=2$ it behaves like $|\log \epsilon|$, which is then raised to the power of $1-1/p = 1/2$.
• When $1< p<2$ it behaves like $x^{1+1/(1-p)} = x^{(2-p)/(1-p)}$ which is then raised to the power of $1-1/p$, producing $x^{(p-2)/p}$.
• When $p=1$, the above calculations need an adjustment: $$ \int_\epsilon^1 |f'(x)|\,dx = \int_\epsilon^1 \left(x |f'(x)|\right) x^{-1} \,dx \le \epsilon^{-1} \int_\epsilon^1 x|f'(x)| \,dx $$ hence $f(\epsilon)\le C/\epsilon$.

This yields the desired conclusions about the behaviour of $f(\epsilon)$ as $\epsilon\to 0$.