Need Hint; Show that the Limit Exists when $f\in C^1(0,1)$ and...

Question

The problem is as follows:

Assume that $f\in C^1(0,1)$ and $$ \int_{(0,1)}x|f'|^p\,dx<+\infty\qquad\text{for some }p>2. $$ Show that $\lim_{x\rightarrow 0^+}f(x)$ exists.

Note: $C^1(0,1)$ is the space of continuously differentiable functions on $(0,1)$.

What I've considered so far;

• I know that $x\in C^1(0,1)$.
• I know that $f$ is differentiable.
• I know that the definition of the right-hand limit here will be important (which I have written down on my scratch work).

However, I am having difficulty in figuring out where to continue off from here. I think I might be missing some important Theorem.

Answer 1

We can find that for $0<x<y<1$, $$\begin{align*} |f(x)-f(y)| &\le\int_x^y |f'(t)|\mathrm dt\\ &\le\left(\int_x^y t|f'(t)|^p\mathrm dt\right)^{1/p}\left(\int_x^y t^{\frac1{1-p}}\mathrm dt\right)^{1-1/p}\\ &\le \left(\int_0^1 t|f'(t)|^p\mathrm dt\right)^{1/p}\left(\frac{p-1}{p-2}\cdot\left(y^{\frac{p-2}{p-1}}-x^{\frac{p-2}{p-1}}\right)\right)^{1-1/p} \end{align*}$$ by Holder's inequality, which implies that $f$ is uniformly continuous on $(0,1)$. Hence, $f$ can be continuously extended to $[0,1]$ and $\lim_{x\to 0^+}f(x)$ exists.