$J_{\delta}$ is a standard mollifier. $f'\in L^2[\delta', 1-\delta']$ is the weak derivative of some $f\in L^2[\delta', 1-\delta']$, with $\delta < \delta'$.
Then we know $(J_{\delta}*f)'=J_{\delta}*(f')$.
We know $(J_{\delta}*f)' \to f'$ almost everywhere on $[\delta', 1-\delta']$.
I also know that $(J_{\delta}*f)' \rightharpoonup f'$.
I want to use Radon Riesz property to show that $(J_{\delta}*f)' \to f'$ strongly.
So I'm left with showing $\lim_{\delta \to 0} ||(J_{\delta}*f)'||_{L^2} = ||f'||_{L^2}$.
I think this must follow from the pointwise convergence, since if it would follow from the weak convergence, then this additional condition in Radon Riesz property wouldn't be necessary.
We want $$ \lim_{\delta\to 0} \int ((J_{\delta}*f)'(x))^2\;dx = \int (f'(x))^2 dx $$
I don't know where to start. I have also noticed that $||(J_\delta * f)'||_{L^2}=||J_\delta * (f')||_{L^2} \leq ||J_\delta||_{L^1} * ||f'||_{L^2} = ||f'||_{L^2}$ by Young's convolution inequality, so
$$ \lim_{\delta\to 0} \int ((J_{\delta}*f)'(x))^2\;dx \leq \int (f'(x))^2 dx, $$
but I don't know if that brings us any closer to getting that $\leq$ to an $=$.