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.

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    $\begingroup$ Did you follow the proof of the $p \ge 1$ case? If so, at what point were you unable to proceed? $\endgroup$
    – Umberto P.
    Jun 16, 2021 at 19:53
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    $\begingroup$ @Ian $d(0,1)+d(1,2)=2$ and $d(0,2)=2^p<2$ because $p<1$. $\endgroup$
    – user239203
    Jun 16, 2021 at 20:02
  • $\begingroup$ @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. $\endgroup$ Jun 16, 2021 at 20:07
  • $\begingroup$ @Krull -- Can you prove that $d(x,y)=|x-y|^p$ is a complete metric on $(-\infty,\infty)$ for $0<p<1$? $\endgroup$ Jun 16, 2021 at 20:10
  • $\begingroup$ @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. $\endgroup$ Jun 16, 2021 at 20:16

1 Answer 1


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 $d(f_{n_j},f)\xrightarrow{j\rightarrow\infty}0$.

  • $\begingroup$ Thank you for your answer. I think the first line you wrote should read $(a+b)^p \leq a^p + b^p$ no? $\endgroup$ Jun 17, 2021 at 1:16
  • $\begingroup$ @WolfgangKrull: That's right. I just corrected the direction of the inequality. Thanks! $\endgroup$ Jun 17, 2021 at 1:18

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