Let (X, M, $\mu$) be a finite measure space. Show the whole of $L^p(d \mu )$ is closed in $L^1(d\mu)$ iff $\exists C>0$ such that $||f||_{L^p}\le C\cdot||f||_{L^1}$ for all $f\in L^p(d\mu)$
$\Leftarrow$ is easy to show, as any sequence $\{f_n\}\subset L^p\cap L^1$ is cauchy in $L^1$ implies it being cauchy in $L^p$
This is a question on an analysis qualifying exam I've been looking at. I am a little uncomfortable showing the other direction. Does anyone have any hints? I understand that this would come about iff (X,M,$d\mu$) is finite dimensional, but this question is supposed to be answered without looking into that information.
$L^p(d\mu)\hookrightarrow L^1(d\mu)$ since this is a finite measure space.
Convergence in $L^1$ is equivalent to convergence $\mu$-a.e. and convergence of the norm values (i.e. $f_n\rightarrow f$ in $L^1$ iff $f_n\rightarrow f$ $\mu$-a.e. and $||f_n||_{L^1}\rightarrow||f||_{L^1}$) I feel like this can be useful for showing the other direction, but am curious if someone can show a good clean way to solve this problem.
Thanks
What you want to do is show that if it is not true that $\|f\|_{p} \le C \|f\|_1$, then you could form a sequence that converges in $L^1$ but not $L^p$.