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Suppose $f_n$ is a sequence of non-negative functions in $L_2[0, 1]$, satisfying $\|f_n\|_1=1$ for all $n\in \mathbb{N}$. Assume, further, that

$$|\|f_n\|_2-1|\leq 2^{-n}$$

I need to show that $f_n → 1$ almost everywhere.

What I've tried/know so far: $|\|f_n\|_2-1|\leq 2^{-n}$ implies that $\sum |\|f_n\|_2-1|< \infty$ so that $\lim_{n\to\infty}\|f_n\|_2=1$. My idea is to consider the following

$$\int\sum |f_n-1|$$

and show that it is finite. That would imply that $f_n\to1$, as required. All I have so far though is:

$$\int\sum |f_n-1|=\sum \int|f_n-1|\leq \sum (\int|f_n-1|^2)^{1/2}$$

I'm not sure how to proceed from here. If I can show that $(\int|f_n-1|^2)^{1/2}\leq 2^{-n}$, then I'm through but I'm unable to prove this. Hints are greatly appreciated. Thanks!

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The idea is to control the weaker sum: $$ \int\sum_n (f_n-1)^2\, dx. $$ First step is to note that $$ \int(f_n-1)^2\, dx = \int (f_n^2-2f_n +1)\, dx = \int f_n^2\, dx -1=\| f_n\|_2^2-1, $$ owing to the fact that $f_n\geq 0$ and $\| f_n\|_1=1$. Now it remains to control this last quantity; for this note that by the bound in the hypotheses we get $\| f_n\|_2\leq 2$, and so multiplying by the conjugate we have $$ \dfrac{\|f_n\|_2^2-1}{3} \leq \dfrac{(\| f_n\|_2-1)(\| f_n\|_2+1)}{\| f_n\|_2 +1} = \| f_n\|_2-1 \leq 2^{-n}. $$ This gives the required summability and so $f_n\to 1$ a.e.

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  • $\begingroup$ Beautiful solution. What was the insight/intuition into considering that weaker sum? (Also, why’d you call it weaker? Weaker in what sense?) $\endgroup$
    – tangentbundle
    Aug 17, 2021 at 20:07
  • $\begingroup$ Also, minor but there’s a sign error in the first step. Second integral negative 1 should be positive 1. I can’t edit. $\endgroup$
    – tangentbundle
    Aug 17, 2021 at 20:13
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    $\begingroup$ @tangentbundle: Weaker in the sense that $\sum a_n^2\leq\left(\sum a_n\right)^2$ for any sequence of nonnegative numbers. Thanks for spotting the typo! $\endgroup$
    – Jose27
    Aug 17, 2021 at 20:21
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    $\begingroup$ @tangentbundle: As far as intuition for picking this weaker version, I actually got the bound for $\| f\|_2^2-1$ first, so it seemed natural to work with an $L^2$ quantity; the naive choice $(f_n-1)^2$ worked out once you expanded. $\endgroup$
    – Jose27
    Aug 17, 2021 at 20:42

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