Does $L^p(E)$ with $m(E)<\infty$ with smaller norm preserve the Banach

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

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| is a Cauchy sequence which is not convergent.
• 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? – Ameryr Apr 28 at 14:40
• @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?. – Kavi Rama Murthy Apr 28 at 23:16
• 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. – Kavi Rama Murthy Apr 29 at 0:17