Weak convergence under nonlinearities
In this paper, we prove that if a Nemytskii operator maps Lp(omega, E) into Lq(omega, F), for p, q greater than 1, E, F separable Banach spaces and F reflexive, then a sequence that converge weakly and a.e. is sent to a weakly convergent sequence. We give a counterexample proving that if q = 1 and p is greater than 1 we may not have weak sequential continuity of such operator. However, we prove that if p = q = 1, then a weakly convergent sequence that converges a.e. is mapped into a weakly convergent sequence by a Nemytskii operator. We show an application of the weak continuity of the Nemytskii operators by solving a nonlinear functional equation on W1,p(omega), providing the weak continuity of some kind of resolvent operator associated to it and getting a regularity result for such solution.
Main Authors: | , |
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Format: | Digital revista |
Language: | English |
Published: |
Academia Brasileira de Ciências
2003
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Online Access: | http://old.scielo.br/scielo.php?script=sci_arttext&pid=S0001-37652003000100002 |
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Summary: | In this paper, we prove that if a Nemytskii operator maps Lp(omega, E) into Lq(omega, F), for p, q greater than 1, E, F separable Banach spaces and F reflexive, then a sequence that converge weakly and a.e. is sent to a weakly convergent sequence. We give a counterexample proving that if q = 1 and p is greater than 1 we may not have weak sequential continuity of such operator. However, we prove that if p = q = 1, then a weakly convergent sequence that converges a.e. is mapped into a weakly convergent sequence by a Nemytskii operator. We show an application of the weak continuity of the Nemytskii operators by solving a nonlinear functional equation on W1,p(omega), providing the weak continuity of some kind of resolvent operator associated to it and getting a regularity result for such solution. |
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