A question I came across.
Let $(X,\mathcal{F},\mu)$ be a $\sigma$ -finite measure space. Let $f_1,f_2,\dotsc:X\to\mathbb R$ be measurable functions such that $n^2\cdot\lVert f_n\rVert_2\leq1$ for all $n\geq1$ . Is it necessarily true that $f_{n}\rightarrow0$ a.e. for $n\to\infty$?
I'm pretty sure the answer is yes. My attempt at proving it:
It is given that: $\parallel f_{n}\parallel_{2}\leq\frac{1}{n^{2}}$
So we know $\underset{n\rightarrow\infty}{lim}\parallel f_{n}\parallel_{2}=\underset{n\rightarrow\infty}{lim}(\intop_{X}\mid f_{n}\mid^{2}d\mu)^{1/2}=0$
Since square root is a continuous function - $(\underset{n\rightarrow\infty}{lim}\intop_{X}\mid f_{n}\mid^{2}d\mu)^{1/2}=0$
We would like to take the limit inside the integral. Fatou's Lemma tells us that for non-negative functions $\underset{n\rightarrow\infty}{liminf}\intop_{X}\, f_{n}d\mu\geq\intop_{X}\,\underset{n\rightarrow\infty}{liminf}f_{n}d\mu$ so we can deduce that $\underset{n\rightarrow\infty}{liminf}\mid f_{n}\mid^{2}=0$ . I am however stuck here since I want the same conclusion for the $lim$ and not $liminf$.
Any suggestions?
Thanks for the help!
Your proof is incorrect. Note that when you claim $$ \sum \int_X |f_n|^2 ~d\mu<\infty$$ this is not the same as $$\sum ||f_n||_2 <\infty.$$ In fact, in worst case the former sum can be like $\sum \frac{1}{n} = \infty$.
Also, let us note that it doesn't take all that effort to conclude $\lim \int |f_n|^2~d\mu = 0$, since this number is bounded by $\frac{1}{n}$ anyway.
The easiest way is probably to directly apply Fatou's lemma to the sequence of $L^1$ functions $|f_n|^2$.