For interval $[0,1]$ define $f_n = n\chi_{[0,1/n]}$.Show that the squence of $f_n$ does
not have weak convergence subsequence in $L^1[0,1]$
This is an example in Royden fourth edition page 173.The proof as follows:
Suppose exist a weak convergence subsequence
for $[c,d]\subset(0,1)$ ,$\chi_{[c,d]} \in L^\infty[0,1]$,we have $\int^d_c f = \lim_{k\to \infty}\int^d_cf_{n_k}$ since for sufficient large $k$,we have $1/n_k<c$ hence $\int^d_cf = 0$.
Here is the question:
- I have a err that change $0<c<d\le1$ to $0<c<d<1$.I don't understand
why this change is necessary.Since it may not influence the integral. - The book uses the lemma that if for all $c,d$ such that $0<c<d<1$ we have $\int^d_cf = 0$ then $f = 0$ a.e. I find a point here in the lemma13 page 126.It define the integral on a open interval $(x_1,x_2)\subset [a,b] $!. This open interval seems necessary here.I don't understand why define on the open interval here and why the err change in the book is necessary?
Best Answer
I don't know such a book, but I would argue so. I hope this somehow helps to figure it out. Suppose that a (sub)sequence converges weakly to some function $f\in L_1[0,1]$. Then for any $g\in L_\infty[0,1]$ we have $\displaystyle\int\limits_0^1 f_n(x)g(x)dx\to\int\limits_0^1 f(x)g(x)dx$. Choosing $g=1$ we have $\displaystyle\int\limits_0^1 f(x)dx=1$, so $f$ cannot be zero almost everywhere. Denote $M\subset(0,1)$ such a set of positive measure on which $f$ is not equal to zero (the extreme points of the segment for obvious reasons can not be considered).
Now fix the $0<\varepsilon<1$ so that $M$ falls into $[\varepsilon,1]$ and consider the function $h_\varepsilon=\chi_{E_1}-\chi_{E_2}\in L_\infty[0,1]$, where $E_1=\{x\in[\varepsilon,1]:\;f(x)\geq0\}$, $E_2=\{x\in[\varepsilon,1]:\;f(x)<0\}$. Then, considering large enough $n$ you can easily check that $\displaystyle\int\limits_\varepsilon^1 |f(x)|dx=0$, and $f=0$ on $[\varepsilon,1]$, which is the contradiction.