# Find a sequence $f_n$ so that $\int_0 ^1 |f_n(x)| = 2$ and $\lim_{n \to \infty} f_n(x) = 1$.

(a) Find a sequence $f_n: [0,1] \rightarrow \mathbb{R}$ such that $\int_0 ^1 |f_n(x)| = 2$ for all $n \in \mathbb{N}$ and $\lim_{n \to \infty} f_n(x) = 1$ for all $x \in [0,1]$. (b) If the $f_n$ are as in part (a), then prove $\lim_{n \to \infty} \int_0 ^1 |f_n(x) -1| dx = 1$.

I'm having trouble finding the proper $f_n$ to fit this situation. Any help is appreciated. Once I find $f_n$, I may need tips for part (b) as well, but I think I could figure that part out given (a). Thanks in advance.

• Try $f(x)=1+(n+2)(n+1)(1-x)x^n$. – Professor Vector Dec 11 '17 at 18:52

Define $f_n$ as a triangle on $[0,1/n]$ with $f(x) = 1$ for $x = 0$ and $x \in [1/n,1]$ (the triangle top is choosen such that the total integral is $2$).
For part b). By Egorov Theorem, for every $\epsilon>0$, there exists some measurable set $M_{\epsilon}\subseteq[0,1]$ such that $M_{\epsilon}^{c}:=[0,1]-M_{\epsilon}$ satisfies $|M_{\epsilon}^{c}|<\epsilon$ and $f_{n}\rightarrow 1$ uniformly on $M_{\epsilon}$.
We see that \begin{align*} \int_{0}^{1}|f_{n}-1|&=\int_{M_{\epsilon}}|f_{n}-1|+\int_{M_{\epsilon}^{c}}|f_{n}-1|\\ &\leq\int_{M_{\epsilon}}|f_{n}-1|+\int_{M_{\epsilon}^{c}}|f_{n}|+|M_{\epsilon}^{c}|\\ &=\int_{M_{\epsilon}}|f_{n}-1|+\int_{0}^{1}|f_{n}|-\int_{M_{\epsilon}}|f_{n}|+|M_{\epsilon}^{c}|\\ &=\int_{M_{\epsilon}}|f_{n}-1|-\int_{M_{\epsilon}}|f_{n}|+2+|M_{\epsilon}^{c}|\\ &<\int_{M_{\epsilon}}|f_{n}-1|-\int_{M_{\epsilon}}|f_{n}|+2+\epsilon, \end{align*} taking $n\rightarrow\infty$, we get \begin{align*} \limsup_{n}\int_{0}^{1}|f_{n}-1|\leq 0-1+2+\epsilon=1+\epsilon, \end{align*} so \begin{align*} \limsup_{n}\int_{0}^{1}|f_{n}-1|\leq 1. \end{align*} On the other hand, \begin{align*} \int_{0}^{1}|f_{n}-1|\geq\int_{0}^{1}|f_{n}|-1=2-1=1. \end{align*}
• This makes total sense. I hadn't realized that it doesn't necessarily matters what $f_n$ is, only how it is defined in part (a). Thanks! – obewanjacobi Dec 11 '17 at 21:21
Hint for (b): Apply Fatou's lemma to $|f_n| + 1- |f_n-1|.$
Choose $$f(x):=4n^2x+1 \text{ for }x\in \left[0,\frac{1}{2n}\right]\\ f(x):=-4n^2x+1+4n \text{ for }x\in \left[\frac{1}{2n},\frac{1}{n}\right]\text{and }f(x)=1\text{ otherwise}.$$