= Solution
The <Lp norm> of the forcing at the critical exponent is constant:
$$
\|f_m\|_{L^{6/5}(B)}^{6/5}
=|E_m|^{-1}\int_B\chi_{E_m}\,dx=1.
$$
To prove <weak convergence> in $L^{6/5}(B)$, the <duality of Lp spaces> says that we must test against an arbitrary $g\in L^6(B)$. <Holder inequality>, restricted to the shrinking <measurable set> $E_m$, gives
$$
\begin{aligned}
\left|\int_B f_mg\,dx\right|
&\leq |E_m|^{-5/6}\int_{E_m}|g|\,dx\\
&\leq |E_m|^{-5/6}|E_m|^{5/6}
\left(\int_{E_m}|g|^6\,dx\right)^{1/6}
=\left(\int_{E_m}|g|^6\,dx\right)^{1/6}\longrightarrow0.
\end{aligned}
$$
The last step is <absolute continuity of the Lebesgue integral>, since $|E_m|=r_m^3|E|\to0$ and $|g|^6$ is integrable. Thus
$$
\boxed{f_m\rightharpoonup0\text{ in }L^{6/5}(B),\qquad\|f_m\|_{6/5}=1.}
$$
Testing only bounded or smooth functions would not by itself establish this <weak convergence>; the restricted <Holder inequality> handles every member of the full dual space.
Back to article page
