# A sequence of functions such that..

I am wondering if one can find a sequence of functions $f_{n}(x)$ such that uniformly in $n$, we have $f_n (x) \in L^1(\mathbb R)$ and $\lim_{n \rightarrow \infty} f_n (x)=1$ and $\lim_{n \rightarrow \infty}\Vert f_n \Vert_{L^{1}}=1$ or equal to another finite constante $C>0$.

Thanks.

• @John $\mathbb R$ – Lee-Jin Jun 8 '15 at 2:22
• Do you mean $n \to \infty$ instead of $n \to 0$? – Robert Israel Jun 8 '15 at 2:23
• @RobertIsrael yes !! sorry – Lee-Jin Jun 8 '15 at 2:27
• If it's going pointwise or a.s. to 1 over all of $\mathbb{R}$, why would you think the limit of the norms would be finite? – Batman Jun 8 '15 at 2:34
• @Batman Just a guess with a Gaussian function, probably the limit could be infinite to – Lee-Jin Jun 8 '15 at 2:38

## 1 Answer

By Fatou's lemma, if $\lim_{n \to \infty} |f_n(x)| = 1$ almost everywhere then $$\lim \inf_{n \to \infty} \|f_n \|_1 \ge \int_{\mathbb R} 1\; dx = \infty$$

• Thanks for your nice answer ! – Lee-Jin Jun 8 '15 at 2:40