# Question about the $L^p$ being complete.

I was wondering if $$L^p$$ is complete whenever we have $$0 when we define the metric $$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.

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

For $$0, $$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$$.
• Thank you for your answer. I think the first line you wrote should read $(a+b)^p \leq a^p + b^p$ no? Jun 17, 2021 at 1:16