Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 6 down vote accepted

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.

share|improve this answer
    
Perfect, thanks again Davide, you've managed to clear up the two main questions I've had so far in my revision. –  James Nov 5 '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 5 '12 at 17:15
1  
1  
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 5 '12 at 20:16
1  
@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. Nov 5 '12 at 20:33

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.