The first: Suppose $g$ is increasing and differentiable on $[0,1]$. For every $f\in L^2(0,1)$ define $f^*(x)$, for $x\in [0,1]$, by: $$f^*(x)=g'(x)\int_0^x f(t)\,dt .$$ If $f_n\to f$ in $L^2(0,1)$, then prove that $f_n^*\to f^*$ in $L^1(0,1)$.
I started like this: First notice that since $g$ increasing, $g'(x)\geq 0$. Now let $\epsilon>0$. Let $$f_n^*(x)=g'(x)\int_0^x f_n(t)\,dt .$$ We need to prove: $$\int_0^1|f_n^*-f^*|\,dx<\epsilon, $$ provided $$\left(\int_0^1 |f_n-f|^2 \,dx\right)^{1/2}<\epsilon_1. $$ \begin{align} \int_0^1|f_n^*-f^*|\,dx&\leq\int_0^1 g'(x)\int_0^x\left|f_n-f\right|\,dt\,dx\\ &\leq \int_0^1 g'(x)\int_0^1 \left|f_n-f\right|\chi_{\{x\ge t \ge 0\}}(t)\,dt\,dx\\ &\leq \int_0^1 g'(x)\left(\int_0^1 |f_n-f|^2\,dt\right)^{1/2}\left(\int_0^x 1 \,dt\right)^{1/2}\,dx\qquad\text{Holder's}\\ &=\epsilon_1 \int_0^1 \sqrt{x}\,\,g'(x) \,dx \end{align} If that was right, then what after?
You're almost done. Note that $g'(x)$ dominates $\sqrt{x} g'(x)$ on this interval, and the integral of $g'(x)$ is less than or equal to $g(1)-g(0)$. This doesn't require absolute continuity for $g$. See Proposition 22 here.