1. Prove that if fn converges to 0 in L2 and ‖fn‖L2≤1 for each n, then fn(x)n converges pointwise to 0 a.e.
Proof: Pick ϵ>0. Let An={x:|fn(x)n|≥ϵ}. Then we have
1n2≥∫|fnn|2≥∫An|fnn|2≥∫Anϵ2=ϵ2m(An)
so that m(An)≤1ϵ2n2. Then ∑An<∞ and so by Borel-Cantelli, lim supAn=0, so for almost all x, lim sup|fn(x)/n|<ϵ. Since ϵ was arbitrary, we’re done.