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I was wondering if $L^p$ is complete whenever we have $0<p<1$ when we define the metric $d(f,g) = \int |f-g|^p$? I think the answer would be yes, but I cannot come up with a proof. It seems to me that it would be in the same spirit of proving the case for $p \geq 1$ but I am unsure.

Jose Avilez
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GnarlySquid
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    Did you follow the proof of the $p \ge 1$ case? If so, at what point were you unable to proceed? – Umberto P. Jun 16 '21 at 19:53
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    @Ian $d(0,1)+d(1,2)=2$ and $d(0,2)=2^p<2$ because $p<1$. –  Jun 16 '21 at 20:02
  • @UmbertoP. I am unsure how to go about the argument. When I try to mirror steps used in $p \geq 1$ I run into difficulties because $d$ is only a metric. – GnarlySquid Jun 16 '21 at 20:07
  • @Krull -- Can you prove that $d(x,y)=|x-y|^p$ is a complete metric on $(-\infty,\infty)$ for $0<p<1$? – uniquesolution Jun 16 '21 at 20:10
  • @Ian Thank you for your comment. Isn't $d(0,2) \leq d(0,1)+d(1,2)$ what we want? As you said $2^p < 2$, so that example does not work. – GnarlySquid Jun 16 '21 at 20:16
  • @uniquesolution: I have not tried proving this, but will this help with the overall proof? – GnarlySquid Jun 16 '21 at 20:17
  • Gah I'm forgetting what the property that breaks is. There is an important property that goes awry for $p<1$. Maybe it was that the balls aren't convex? – Ian Jun 16 '21 at 20:58
  • @Krull -- I believe it will. Doing mathematics means doing, not asking what do. – uniquesolution Jun 16 '21 at 21:03
  • @Ian the balls aren't convex, and the metric is not a norm (because it is not homogeneous). – supinf Jun 16 '21 at 21:07
  • @WolfgangKrull if you already tried to mirror the steps, you should include these steps in your question and point out exactly where it goes wrong. Otherwise people have to redo your work again. – supinf Jun 16 '21 at 21:09

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This problem may have already several answers. Here is yet another one. Some details (How Fatou's is used and so on are left for the OP).

For $0<p<1$, $a^p+b^p\geq (a+b)^p$ for all $a,b\geq0$. From this, it follows that $$d(f, g)=\int|f-g|^p\leq \int(|f-h|+|h-g|)^p\leq d(f, h) + d(h, g)$$ Completeness follows a similar proof than that of $L_p$, $p\geq1$. Suppose $\{f_n: n\in\mathbb{N}\}$ is Cauchy in $L_p$ and choose subsequence so that $d(f_n,f_{n_k})<2^{-k}$ for all $n\geq n_k$. Define $$\begin{align}G_K&=|f_{n_1}|+\sum^K_{j=1}|f_{n_{j+1}}-f_{n_j}|\\ G&=|f_{n_1}|+\sum^\infty_{j=1}|f_{n_{j+1}}-f_{n_j}|\end{align}$$ Then $$\begin{align} \int G^p_K&=\int(|f_{n_1}|+\sum^K_{j=1}|f_{n_{j+1}}-f_{n_j}|)^p\leq \int|f_{h_1}|^p+\sum^n_{j=1}\int|f_{n_{j+1}}-f_{n_j}|^p\\ &\leq d(0,f_{n_1})+\sum^\infty_{j=1}2^j<\infty \end{align}$$ From that and monotone convergence, $G$ converges a.s. Hence the series $g= f_{n_1}+\sum^\infty_{j=1}(f_{n_{j+1}}-f_{n_j})$ converges absolutely almost surely and so, $f_{n_j}$ converges almost surely to say $f$. The rest is and application of Fatou's to check that $f\in L_p$ and that $d(f_{n_j},f)\xrightarrow{j\rightarrow\infty}0$.

Mittens
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