# To show a sequence of functions converges almost everywhere

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!

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.
• @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! Commented Aug 17, 2021 at 20:21
• @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. Commented Aug 17, 2021 at 20:42