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If $1\leq p < q <\infty$ and $E$ a subset of $\mathbb{R}$ with finite measure if we consider the space $L^q(E)$ is it a Banach space with the norm $||.||_p$. I know that $L^p$ space is a Banach space with respect to the $p$ norm.

And I know using Holder inequality that $||f||_p \leq ||f||_q (m(E))^{\frac{q-p}{pq}}$ so $E$ has finite measure then $L^q(E) \subset L^p(E)$. I was trying to construct a sequence $f_n \in L^q$ which is Cauchy with respect to the norm $||.||_p$ which does not converge but I could not? Or is that space a Banach space with smaller norm. I found similar questions but I did not find any counter example the solvers talk about atoms and the open map theorem, I am searching for a Cauchy sequence which diverge. The questions

1) Is Lp space complete with this norm?

2) Is $L^{p}$ space with alternate norm Banach?

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If it is Banach then the identity map from $L^{q}$ with $L^{p}$ norm to $L^{q}$ with $L^{q}$ would be continuous by Open Mapping Theorem. So there would be a constant $C$ such that $\|f\|q \leq C \|f\|p$. Take the example $f_n=n^{1/q}I_{(0,1/n)}$ to get a contradiction. Aliter: if $f\in L^{p}\setminus L^{q}$ then $(fI_{1/n <|f|<n)})$ is a Cauchy sequence which is not convergent.

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  • $\begingroup$ The sequence $ f_n $ is converge in $p$ norm, $\int |n^{p/q}| I_{(0,1/n)}=n^{p/q-1} $ which clearly goes to zero? $\endgroup$ – Ameryr Apr 28 at 14:40
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    $\begingroup$ @Ameryr Each $f_n$ is in $L^{r}$ for every $r>0$. $\|f_n\|_p \to 0$ but $\|f_n\|_q $ does not tend to $0$ and that is how you get a contradiction. When you have clearly said $1\leq p <q <\infty$ why are you talking about negative exponents?. $\endgroup$ – Kavi Rama Murthy Apr 28 at 23:16
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    $\begingroup$ To show that $L^{q}$ is not a Banach space w.r.t. $p-$norm I am not constructing Cauchy sequence which does not converge. I am using a theorem to prove that this space cannot be complete. Have been asked to construct a Cauchy sequence explicitly? If not you can change your approach to the one I have suggested. $\endgroup$ – Kavi Rama Murthy Apr 29 at 0:17
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    $\begingroup$ Did I not mention Open Mapping Theorem in my answer? Please read my answer with some patience. $\endgroup$ – Kavi Rama Murthy Apr 29 at 0:25
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    $\begingroup$ @Ameryr I have included an example of a Cauchy sequence which is not convergent. $\endgroup$ – Kavi Rama Murthy Apr 29 at 5:20

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