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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

wfw
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2 Answers2

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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 $L^p$.

Christopher A. Wong
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  • The negation would be that there exists an $f \in L^p$ s.t. $\forall C > 0$ we have $|f|_{p} > C |f|_1$ If we take the sequence given by $f_n(x) = f/n$ then this sequence converges to zero in $L^1$ but the norm is unbounded in $L^p$? – Monolite Apr 05 '17 at 00:47
  • I am having a hard time understanding what $f$ could there exist s.t. $\forall C > 0$ we have $|f|_{p} > C |f|_1$ . – Monolite Apr 05 '17 at 00:50
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    @Monolite Your quantifiers are reversed: For every $C > 0$, you can find $f$ such that $|f|_p > C |f|_1$. – Christopher A. Wong Apr 06 '17 at 22:04
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If for all $k \in \mathbb{N}$ there is an $f_k$ s.t $\Vert f_k \Vert_1 < 2^{-k}\Vert f_k \Vert_p$, then on dividing both sides by $\Vert f_k \Vert_p$, we may assume that there exists $g_k$ s.t $\Vert g_k \Vert_p = 1$ but $\Vert g_k \Vert_1 < 2^{-k}$. Thus $$ \sum_{1}^{\infty} g_k $$ Is in $L^1$ but not in $L^p$

Vatsa Srinivas
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