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It is rather straight forward to show that $L_p$ is complete for $p\geqslant 1$, but I am having trouble showing the same thing when $p<1$. For the former case I have shown that every absolutely convergent sequence converges by constructing a a function in $L_p$ but bigger than the series and used the dominated convergence theorem (I can also do a similar thing using Cauchy sequences rather than absolutely convergent series). The problem is that in showing that my upper bound function is an element of $L_p$ I have used the triangle inequality which I cannot do for $p<1$. Does anyone have any ideas of a way around this?

I noticed there are similar questions to this already, but they have either not been answered or closed.

Davide Giraudo
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James
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1 Answers1

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Let $0<p<1$ fixed. Let $\{f_n\}$ a Cauchy sequence for $$d(f,g):=\int |f-g|^pd\mu.$$ It's a metric, as concavity of $t\mapsto t^p$ on $\Bbb R_{\geqslant 0}$ helps us to show that $(a+b)^p\leq a^p+b^p$ when $a$ and $b$ are non-negative real numbers.

We extract a subsequence $\{g_k\}:=\{f_{n_k}\}$ such that $d(g_{k+1},g_k) \leqslant 2^{-k}$. Let $A_k:=\{x,|g_{k+1}(x)-g_k(x)|\geqslant 2^{-k}\}$. Then $$2^{-kp}\mu(A_k)\leqslant \int |g_{k+1}-g_k|^pd\mu\leqslant 2^{-k},$$ so $\mu(A_k)\leqslant 2^{-k(1-p)}$. As $1-p>0$, a Borel-Cantelli's like argument shows that $\mu(\limsup_{k\to+\infty} A_k)=0$. So, for almost all $x$, we can find $K(x)$ such that for $k\geq K(x)$, $|g_{k+1}(x)-g_k(x)|\leqslant 2^{-k}$. Let $g(x):=\lim_{k\to +\infty}g_k(x)$.

Using Fatou's lemma, we can see that $g\in L^p$ and $d(g,g_n)\to 0$.

Now fix $\varepsilon>0$, $N$ such that $d(f_n,f_m)\leqslant\varepsilon$ and $d(g_n,g)\leqslant\varepsilon$ whenever $m,n\geqslant N$. If $k\geqslant N$, $n_k\geqslant N$ so $$d(f_k,g)\leqslant d(f_k,g_k)+d(g_k,g)\leqslant 2\varepsilon.$$

Note that this works without assumptions on the measure space we are working on, except that the measure is non-negative.

Davide Giraudo
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  • Perfect, thanks again Davide, you've managed to clear up the two main questions I've had so far in my revision. – James Nov 05 '12 at 17:08
  • You are welcome. Could you give the link(s) of unanswered questions about the completeness of $L^p$ you talk about in the OP? – Davide Giraudo Nov 05 '12 at 17:15
  • http://math.stackexchange.com/questions/223182/verify-that-l-p-space-is-complete – James Nov 05 '12 at 17:41
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    http://math.stackexchange.com/questions/223140/let-l-p-be-the-complete-separable-space-with-p0 – James Nov 05 '12 at 17:42
  • Davide, just rereading it and I have one further question. In the last line you use the triangle inequality for d. But, if $p<1$ then the triangle inequality does not hold (I am right in saying this right?). Have you assumed something else I am not seeing? – James Nov 05 '12 at 17:55
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    Actually, you can check that $(a+b)^p\leq a^p+b^p$ when $0<p<1$, $a,b\geq 0$, which ensures us that $d$ satisfies triangular inequality (use concavity of $t\mapsto t^p$). – Davide Giraudo Nov 05 '12 at 20:16
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    @James Note that the distance is not defined as in the case $p>1$, that is look at $$d(f,g)= \int|f-g|^p dx$$ and not at $$\tilde{d}(f,g):=\left(\int|f-g|^p dx\right)^{1/p}.$$ It is $\tilde{d}$ that does not satisfy the triangle inequality (when $0<p<1$). – AD - Stop Putin - Nov 05 '12 at 20:33
  • Ahh, I see, thanks again – James Nov 05 '12 at 23:01