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This question already has an answer here:

Let $E$ be a measurable set of finite measure and $1\leq a<b<\infty$. Consider the $L^b(E)$ space normed by $L^a$ norm. Is this space a Banach space?

I think this is wrong, so I tried to find a counterexample but it is hard to find one...

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marked as duplicate by Nate Eldredge, JonMark Perry, user91500, mrp, Claude Leibovici Mar 31 '17 at 10:56

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Suppose that $L^b(E)$ is complete under the $L^a$ norm. Then by the Banach isomorphism theorem, there would be a constant $C$ such that for any $f\in L^b(E)$, $\lVert f\rVert_b\leqslant C\lVert f\Vert_a$. In particular, if $f=\chi_A$ for $A\subset E$ measurable, we would get $\mu(A)^{1/b-1/a}\leqslant C$, hence $$\inf_{A:\mu(A)>0}\mu(A)>0.$$ This condition implies that, together with finiteness of the measure space, that $E$ is a finite union of atoms and in this case, $L^b(E)$ is finite dimensional.

Conclusion: $L^b(E)$ is complete under the $L^a$ norm if and only if it is finite-dimensional.

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  • $\begingroup$ Umm...I learned only real analysis in undergraduate level.. Can you explain more easily?? $\endgroup$ – user122794 May 14 '14 at 16:56
  • $\begingroup$ Also I think inequality is wrong $\endgroup$ – user122794 May 14 '14 at 16:56
  • $\begingroup$ If you expect an answer in undergraduate level, no problem. Maybe you could state it in the question.// Which inequality are you talking about? $\endgroup$ – Davide Giraudo May 14 '14 at 18:25
  • $\begingroup$ I think 4th line is wrong.. b norm is larger than a norm multiply c $\endgroup$ – user122794 May 15 '14 at 1:27
  • $\begingroup$ I find counter example. x^(-0.5+1/n) on(0 , 1] $\endgroup$ – user122794 May 15 '14 at 1:28

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