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Let $(\Omega,\mathscr{F},\mu)$ be a probability space. It is well known that $(L_1,\|\cdot\|_1)$ and $(L_\infty,\|\cdot\|_\infty)$ are complete.

Question. Is it true that $(L^\infty,\|\cdot\|_1)$ is complete?

My attempt: A subspace $S$ of a complete metric space $(X,d)$ is complete if and only $S$ is $d$-closed, see e.g. here. Hence it would be enough to prove that $L^\infty$ is $\|\cdot\|_1$-closed. To this aim, let $(f_n)$ be a sequence of functions in $L^\infty$ converging to $f \in L^0$, i.e., $$ \int_\Omega|f_n-f|\,\mathrm{d}\mu=\|f_n-f\|_1 \to 0 \,\,\text{ as }n\to \infty. $$ How can we conclude that $f \in L^\infty$?

Community
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    You are attempting to prove the answer is "yes". Let's think first about whether or not we believe the answer is "yes". After all, unbounded $L^1$ functions usually exist... – Thompson Apr 06 '17 at 15:02

2 Answers2

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Consider the sequence of functions $f_n(x)=\frac{1}{\sqrt x}\mathbf{1}_{(\tfrac{1}{n},1)}(x)$. It is Cauchy in $L_\infty(0,1)$ endowed with the integral norm. Is it convergent?

Tomasz Kania
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Hint: you can have an unbounded $f$ with $\|f\|_1<\infty$ that is $L^1$-limit of $L^\infty$ functions.

Martín-Blas Pérez Pinilla
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