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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 ?

Community
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teshile
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  • How do you show that $D_{n+1}\subset D_n$? Actually, if we wouldn't be able to find such a sequence, a theorem of James would give us that $L^1$ is reflexive (which is not true). So I think finding such a sequence will use this fact. – Davide Giraudo Oct 23 '12 at 14:22
  • What do you mean by contradiction? If $\phi$ is a linear functional which doesn't take its norm (i.e. we can't find a $f$ in the closed unit ball such that $\phi(f)=\lVert\phi\rVert)$, then define $C_n:={f\in L^1,\lVert f\rVert\leq 1,\phi(f)\geq\lVert \phi\rVert-n^{-1}}$. So the problem is to find such a linear functional. – Davide Giraudo Oct 23 '12 at 15:29
  • Closed in what topology? – Nate Eldredge Oct 23 '12 at 17:28
  • @NateEldredge At least, as the set are convex there can be closed either in the weak topology or the strong one. Do you have an other topology in mind? – Davide Giraudo Oct 23 '12 at 19:31
  • @DavideGiraudo: Well, it doesn't actually say we are working in $L^1$. We could be using the uniform topology, or the product topology... – Nate Eldredge Oct 23 '12 at 22:35
  • @NateEldredge Yes, at least the space on which we are working has to be clarified (I assumed it was $L^1$, but indeed, we aren't sure about that). – Davide Giraudo Oct 24 '12 at 08:00
  • @DavideGiraudo I've edited to let the space be L1 – teshile Oct 24 '12 at 13:29

2 Answers2

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(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.

fgp
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  • this seems not nested to me . How about change the last part to -e^x ≤fx≤e^x so that it is nested ? But is this still convex ? – teshile Oct 23 '12 at 15:15
  • @teshile Uh, right, I initially had it bounded only from above, then added the bound from below to exclude the function $f(x)=0$, and didn't realized that'd break the nesting :-( – fgp Oct 23 '12 at 18:35
  • @fgp How do you know $D_n$ is nonempty for all $n$? I can't seem to show a function that is in every $D_n$. – anegligibleperson Oct 30 '12 at 18:59
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    @anegligibleperson You don't need (or have!) one function which is in every $D_n$. If there was such a function, the intersection of all the $D_n$ wouldn't be empty. But if you pick a particular $n$, you can easily find a function which is in $D_n$, take for example $\mathbf{1}_{[n,n+1]} \in D_n$. There's a difference between $\exists f : \forall n : f \in D_n$ (which is false), and $\forall n : \exists f : f \in D_n$ (which is true). – fgp Nov 05 '12 at 10:34
  • @fgp thanks for clarifying that point! – anegligibleperson Nov 05 '12 at 13:36
  • I think that for your sets to be convex, you need to add $f\geq 0$. So your $D_n$ would be the set of all $f\geq 0$ such that $\int_{-\infty}^{+\infty} f\leq 2$ and $\int_n^{+\infty}f\geq 1$. It works well then. – Julien May 17 '13 at 01:23
  • @julien: You are right. The $f$ should be a.e. non-negative. – Travis Wang Feb 17 '14 at 16:24
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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 $D_n = \{f : \|f\| \le 1, \ell(f) \ge 1-\frac{1}{n}\}$.

Nate Eldredge
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