# Is Lp space complete with this norm? [duplicate]

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...

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

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.