2
$\begingroup$

This is from my recent homework. I am asked to find a descending nested sequence of closed , bounded , nonempty convex sets $\{D_n\}$ in $L^1(\mathbb{R})$ such that the intersection is empty , where elements in $D_n$ should be integrable functions defined on R.

There is a discussion on mathoverflow which says we could replace unit ball part in James theorem by convex closed set . As suggested in the comments , possibly this is needed for the question.

Could anyone help me with this ?

  • 0
    @DavideGiraudo I've edited to let the space be L12012-10-24

2 Answers 2

2

(My previous idea turned out to be wrong, here's a new one)

I think $ D_n = \{f \in L_1 \::\: ||f||_1 \leq 2,\: ||\mathbf{1}_{[n,\infty)}f||_1 \geq 1\} $ could work. Since $||\mathbf{1}_{[a,\infty)}f||_1 \geq ||\mathbf{1}_{[b,\infty)}f||_1$ if $a \leq b$, the sets are nested. They are bounded by $||f||_1 \leq 2$. The fact that $||\lambda f||_1 = \lambda||f||_1$ makes them convex. For every specific $f$, $||\mathbf{1}_{[n,\infty)}||_1 \to 0$ as $n \to \infty$, which shows that the intersection of all the $D_n$ is empty. They are also closed, because if $f_n \to f$ in $L_1$, then $||\mathbf{1}_{[n,\infty)}(f-f_n)||_1$ must go to zero.

  • 0
    @julien: You are right. The $f$ should be a.e. non-negative.2014-02-17
2

Suggestion: find a linear functional $\ell$ on $L^1$ which does not attain its norm. That is, $\|\ell\| = 1$ but $|\ell(f)| < 1$ for all $\|f\| \le 1$. (You can write one down explicitly; no need to invoke James's theorem.) Then let $D_n = \{f : \|f\| \le 1, \ell(f) \ge 1-\frac{1}{n}\}$.