Does that exponent of (absolute value of derivative) is constrained implies Lipschitz continuity?

Question

Given $C^1([a, b])$ functions $f_n$ that converge to a continuous real-valued function $f_n \to f$ on a closed interval $[a, b] \subset \mathbb R$, suppose $$ \int_a^b |f_n'(x)|^{1 + \epsilon} dx < M $$ for all $n > 0$ and fixed $M < \infty$. For a small $\epsilon$, does this implies

1. $f$ is absolutely continuous
2. $f$ is Lipschitz

Can we tell more about this function?

I have $\epsilon = 0$ does not hold for Cantor function and some approximation $f_n$.

Answer 1

Q1: Yes. Since $f'_n$ is bounded in $L^p$, a suitable subsequence will converge $f_n'\to g$ weakly. Then also $$ f(x) =\lim f_n(x) = \lim \left( f_n(a)+\int_a^x f'_n(t)\, dt \right) = f(a) + \int_a^x g(t)\, dt , $$ so $f$ is absolutely continuous (with derivative in $L^p$).

Q2: No. For example $f(x)=|x|^{\alpha}$ has a derivative $f'\in L^p$ for $p<1/(1-\alpha)$, so can easily be approximated by functions with bounded $L^p$ derivatives.