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For a sequence $\{f_{n}\}$ which converges in $\mathcal{L}^{p}$ space, can we extract a subsequence which is dominated by a function $g \in \mathcal{L}^{p}$?

Can anyone help with this? I thought about using a rapidly Cauchy subsequence but can't get any further...

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Actually the existence of such $g$ is contained in the standard proof of Riesz-Fischer theorem. Indeed, assuming $f_{0} = 0$ for notational simplicity, we can find a constant $C$ and a subsequence $(f_{n_j})$ with $n_{1} = 0$ such that $\| f_{n_{j+1}} - f_{n_{j}} \| \leq C 2^{-j}$. Then $g = \sum_{j=1}^{\infty} |f_{n_{j+1}} - f_{n_{j}}|$ is a dominating function for this subsequence. –  sos440 Nov 20 '11 at 5:58
    
@sos440: Any hints why the $g$ defined as you described belongs to $L^p$? –  mathreader Jul 11 '13 at 9:16
    
Ok, I think I know how to show that $g$ belongs to $L^p$. Define $g_k=\sum_{j=1}^k|f_{n_{j+1}}-f_{n_j}|$. Then $g_k\to g$ pointwise, $g_k\in L^p$, and $\|g_{k+m}-g_k\|_p\le\sum_{j=k+1}^{k+m}\|f_{n_{j+1}}-f_{n_j}\|_p\le2^{-k-1}$ by the Minkowski's inequality and the choice of $f_{n_k}$. Thus $\int|g_{k+m}-g_k|^p\le2^{(-k-1)p}$, and by Fatou's lemma applied to $k$ fixed and $m\to\infty$, we get $\int|g-g_k|^p\le2^{(-k-1)p}$, and since $\|g\|_p\le\|g-g_k\|_p+\|g_k\|_p$, we get $g\in L^p$. –  mathreader Jul 11 '13 at 12:20

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up vote 2 down vote accepted

If $f_n\to f$ in $L^p$, then there is a subsequence $(f_{n_k})$ such that $\|f_{n_k} - f\|_p < 2^{-k}$. Take $g=|f|+\sum\limits_{k=1}^\infty|f_{n_k}-f|$.

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How to show $g$ is in $L^{p}$? –  Red Rover Nov 20 '11 at 8:16
    
NVM. Got it. Thanks a ton. –  Red Rover Nov 20 '11 at 8:44
    
@Jonas: Indeed, why is $g\in L^p$? If the sum in $g$ were finite, then Minkowski's inequality would do the trick. But it is not the case. –  mathreader Jul 11 '13 at 9:15
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@mathreader: In any Banach space $X$, if $\sum_k\|x_k\|$ converges to a finite number, then $\sum_k x_k$ converges to an element of $X$. –  Jonas Meyer Jul 11 '13 at 13:25
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@JonasMeyer: Thanks! That observation makes it much easier than what I wrote in the comment to the original question above. Basically, if we define $g_n=\sum_{k=1}^nx_k$, then $$\|g_{n+m}-g_n\|=\|\sum_{k=n+1}^{n+m}x_k\|\le\sum_{k=n+1}^{n+m}\|x_k\|$$ can be made smaller than arbitrary $\epsilon>0$ for big enough $n$ and all $m$, and thus $\{g_k\}$ is a Cauchy sequence in a Banach space, hence is convergent to $g=\sum_{k=1}^\infty x_k$. –  mathreader Jul 11 '13 at 14:13

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