# fail of cantor intersection property on closed , bounded , convex sets of integrable functions

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 ?

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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

(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.

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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
@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 5 '12 at 10:34
@fgp thanks for clarifying that point! – anegligibleperson Nov 5 '12 at 13:36

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 $D_n = \{f : \|f\| \le 1, \ell(f) \ge 1-\frac{1}{n}\}$.

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