# Bounded sequence in $L^1$ with no weakly convergent subsequence and bounded sequence in $L^\infty$ with no weakly convergent subsequence

Can you give two elementary examples that show:

1. a bounded sequence in $L^1$ with no weakly convergent subsequence;
2. a bounded sequence in $L^\infty$ with no weakly convergent subsequence??

For the $L^1$ case, I was thinking about an approximation of the Dirac delta, say rectangles of area 1 that get higher and smaller around the origin, but I cannot see if I can make it work. I'm totally lost about the $L^\infty$ case.

• @Ravi I've been thinking about it, but got nowhere. – user428573 Jul 18 '17 at 18:24
• Why don't you tell us some of your thoughts? What might be a strategy to find such a sequence in $L^1$? – Jason Jul 18 '17 at 18:26
• @Jason For the $L^1$ case I was thinking about an approximation of the Dirac delta, say rectangles of area 1 that get higher and smaller arond the origin, but I cannot see if I can make it work. I'm totally lost about the $L^\infty$ case. – user428573 Jul 18 '17 at 18:38
• Your idea for $L^1$ will work. Prepare a sequence of rectangles based at the origin of area one with width $\frac{1}{n}$ and height $n$. This is clearly a bounded sequence. Now try the evaluation at $0$ functional. – Ravi Jul 18 '17 at 19:00
• @Ravi I doubt that that functional is continuous on $L^1$ – user428573 Jul 18 '17 at 19:12

For the case of $L^1$ your example does work. Let $f_n= n\chi_{(0,1/n)}.$ Suppose, to reach a contradiction, that $f_{n_k}$ converges weakly in $L^1.$ Passing to a further subsequence, which I'll still denote by $f_{n_k},$ we can assume $n_{k+1}/n_k \to \infty.$ Define $g\in L^\infty$ by setting $g=(-1)^k$ on $(1/n_k, 1/n_{k+1}).$ Then $\int_0^1 f_{n_k}g \to -1$ as $k\to \infty$ through odd integers, while $\int_0^1 f_{n_k}g \to 1$ through even integers. Thus $\int_0^1 f_{n_k}g$ fails to have a limit as $k\to \infty,$ contradiction.

Added later: Why do we need $n_k/n_{k+1} \to \infty?$ Hopefully this will help:

$$\int_0^1 f_{n_k}g = n_k\int_0^{1/n_k} g = n_k\sum_{j=k}^{\infty}(-1)^j(1/n_j- 1/n_{j+1}) = (-1)^k(1- n_k/n_{k+1}) + r_k.$$

Verify that $|r_k| \le n_k/n_{k+1}.$ Thus the above integrals have the behavior described.

• Why do you need $n_{k+1}/n_k \to \infty$ as $k \to \infty$? – user193319 May 15 '18 at 17:46
• @user193319 I added to my answer. – zhw. May 15 '18 at 18:32

Let me first prove that your example in the $$L^1$$ case works. Let

$$f_n=n 1_{(-1/(2n), 1/(2n))}.$$

Assume by contradiction that your sequence has a weakly convergent subsequence and let $$f\in L^1$$ be the weak limit. Let

$$M_{+,\epsilon}=\{x\in \mathbb{R}\setminus (-\epsilon, \epsilon): f(x)>0\}$$

and

$$M_{-,\epsilon}=\{x\in \mathbb{R}\setminus (-\epsilon, \epsilon): f(x)<0\}.$$

Then we define the functionals $$T_{\pm,\epsilon}(g)=\int_{M_{\pm ,\epsilon}} g$$. Using weak convergence, one has $$T_{\pm , \epsilon}(f)=0$$. From this we obtain by the monotone convergence theorem that

$$\Vert f \Vert = \lim_{\epsilon \rightarrow 0} T_{+, \epsilon}(f) - T_{-, \epsilon}(f) =0.$$

This implies $$f=0$$. Define $$T(g)=\int_{-1}^1 g$$. For every member of our sequence, this functional equal 1, however, $$T(f)=0$$ which gives you a contradiction.

For the $$L^\infty$$ case set

$$f_n=(1-n\vert x \vert )1_{(-1/n,1/n)}$$

Similarly as above, one proves that $$f=0$$ (if it wasn't zero, there would be $$C,R>0$$ such that $$M_+=\{f\geq C\}\cap (-R, R)$$ or $$M_-=\{f<-C\}\cap (-R, R)$$ is not a null set. Consider then the functionals $$L_\pm(g)=\int_{M_\pm} g$$. By dominated convergence, we get the contradiction $$L_\pm (f)=0$$). Now define

$$T: C^0(\mathbb{R}, \mathbb{R}) \rightarrow \mathbb{R}, \ T(g)=g(0)$$

Using Hahn-Banach, we can extend this functional to all of $$L^\infty$$. Call the extension $$S$$. Then we get the contradiction (note that the $$f_n$$ are continuous and $$f_n(0)=1$$)

$$0=S(f)= \lim_{n\rightarrow \infty} S(f_n)=1.$$

• For your first example: Why is this sufficient to show the rising towers sequence has no weakly convergent subsequence? In particular, it might be possible to find a subset of the original sequence (ie subsequence) which does weakly converge. – yoshi May 22 at 17:28
• @yoshi I don't understand the question. This is exactly what the OP is asking for. A bounded sequence in $L^1$ with no weakly converging subsequence. My proof is by contradiction. I assume by contradiction that there exists a weakly converging subsequence, then I show that it must converge weakly to zero and finally show that this yields a contradiction. – Severin Schraven May 22 at 19:41
• ah okay, I was getting mixed up with the logic. if you wish, you can add your comment as an answer in the other post and I'll accept it. – yoshi May 22 at 20:11
• @yoshi I added "by contradiciton" in my answer, I hope it is clear this way. – Severin Schraven May 22 at 20:54