Question

net & SIR" be open shown that for any LP62) ¢ La (20. such that miel=0. 15p<q{o, then [m is lebesque measure I

Answer 1

2. Let 2 CRM be open sit min) claim for any 15Þ<q 300, L(O) 01960). where mis lebesgue measure to prove theorem, we prove the theorem by contradiction Let La CLP Ket {fn} be a sequerice in L2 converges to & from 22 norm, and to a for the th norm. We extract a subsequence converge's almost everywhere to f and . hence, a by closed graph theorem then, LP, 22 both are Bramach space! .: we can find ero st. 1 fl sciflq Since can be written as increasing union of finite measure Sets An. then m( Anys sem(A) * => m(An) et sc. since p&q. = m(An) sch Now, we take limit as n o, we get m(2) < Change m(-)< . - contradiction. Similarly, if we take L" CL? again we can find m(2) 200 contradiction So, L"() € L? (12) for min)=00 (Proved)