# Is $L^{p_2}$ complete under the $L^{p_1}$ norm?

Given $p_1,p_2$ such that $1 \leqslant p_1 < p_2 < \infty$ and a measurable set $E$ of finite measure, I'm trying to determine whether the space $L^{p_2}(E)$, which I know to be contained in $L^{p_1}(E)$, is complete under the $L^{p_1}$ norm.

I think that it is not, and am trying to come up with a counter-example as follows: let $E = (0,1]$, then if we let $f(x) = x^k$ where $k = -\frac{1}{2}( \frac{1}{p_1} + \frac{1}{p_2} )$, $f$ is in $L^{p_1}$ but not $L^{p_2}$. I would then like to produce a sequence of functions in $L^{p_2}$ converging to $f$ under the $L^{p_1}$ norm, but am having trouble doing so.

Any help is appreciated!

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Take $f_n=f\cdot\chi_{[1/n,1]}$. –  David Mitra Dec 7 '12 at 23:44
@DavidMitra Thanks, that was exactly what I needed. –  Jonas Dec 7 '12 at 23:58

Assume that $L^{p_2}$ is complete for the $p_1$-norm. Then $$\iota\colon (L^{p_2},\lVert\cdot\rVert_{p_2})\to (L^{p_2},\lVert\cdot\rVert_{p_1})$$ defined by $\iota(f)=f$ is a surjective linear continuous map. By Banach isomorphism theorem, we can find a constant $C>0$ such that for all $f\in L^{p_2}$, we have $$\lVert f\rVert_{p_2}\leqslant C\lVert f\rVert_{p_1}.$$ In particular, for all $A\subset E$ measurable of non-zero measure, we would have $$\mu(A)^{1/p_2}\leqslant C\mu(A)^{1/p_1},$$ so $\mu(A)\geqslant K$ where $K$ depends on $C,p_1$ and $p_2$.

So we can't have the claim when we can find sets of arbitrary small positive measure.

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I have not covered the Banach isomorphism theorem yet, so most of this answer goes over my head. Regardless, thank you for the help! –  Jonas Dec 7 '12 at 23:57
No problem. David Mitra already gave you a sequence which shows that the result doesn't hold in general. Here, I give the reason why it may not work generally. Maybe you can read the answer when you will have studied Baire's theorem. –  Davide Giraudo Dec 8 '12 at 9:48