Analysis Problem of the Day 17

Today’s problem comes as Problem 6 on the UCLA Fall 2016 Analysis Qual:

Problem 6: a) Show that there is no sequence $\{u_n\} \in l^1$ such that $\|u_n\| \geq 1$ for all $n$ and $u_n \rightharpoonup 0.$

b) Show that $l^1$ has the Schur property, i.e. if $u_n$ converges weakly to $u$ in $l^1,$ then $u_n$ converges to $u$ in $l^1$ norm.

Solution: a) For sake of contradiction, suppose there is such a sequence $\{u_n\}.$ By a standard diagonalization argument, one may choose a subsequence (which we will still call $\{u_n\}$ by abuse of notation) such that after the signs of the first $k$ elements of $u_n$ are the same for all $k \geq n.$ Our goal now will be to show that the mass of elements in this subsequence cannot concentrate on any finite set.

Fix a small $\epsilon>0,$ and pick a finite set of indices $S_1$ and an element $u_{n_1}$ such that $\|u_{n_1}\|_{l^1(S_1^c)} < \epsilon.$ We claim that there exist infinitely many $u_j$ such that $\|u_j\|_{l^1(S_1)} < \epsilon.$ Indeed, if not, $\|u_j\|_{l^1(S_1)}>1-\epsilon$ with at most finitely many exceptions. For large enough $j,$ elements of $u_j$ have the same pattern of signs on $S_1.$ Thus, defining $v \in l^\infty$ as $v = \lim_{j \to \infty} \text{sign}(u_j) \cdot 1_{S_1},$ we see that $\liminf_{j \to \infty}\langle u_j,v \rangle \geq \|u_j\|_{l^1(S^1)} \geq 1-\epsilon,$ which contradicts the fact that $u_n$ converges weakly to zero. Thus, one many pick $u_{n_2}$ and a finite set $S_2$ disjoint from $S_1$ such that $\|u_{n_2}\|_{l^1(S_2^c)}<\epsilon.$

Recursively, we may apply the exact same argument to conclude that given $u_{n_1},...,u_{n_k},$ there exists $u_{n_{k+1}}$ and a set $S_{k+1}$ disjoint from $S_1 \cup ... \cup S_k$ such that $\|u_{n_k}\|_{S_{k+1}^c} <\epsilon.$ Finally, let $v = \lim_{j \to \infty} \text{sign}(u_{n_k}) \in l^\infty$ be the sequence of (eventually constant) signs of $u_{n_k}.$ Then, the inner product between each $u_{n_k}$ and $v$ will contribute $1-\epsilon$ on $S_k$ and at most $\epsilon$ on $S_k^c,$ i.e.

$\langle u_{n_k}, v \rangle \geq 1-\epsilon - \epsilon > 0$

for $\epsilon < \frac12.$ But this contradicts the claim that $u_{n_k}$ converges weakly to zero, and we are done.

b) By subtracting $u,$ it suffices to show that if $v_n$ converges to zero weakly in $l^1,$ then it converges in $l^1$ norm. If not, there exists a subsequence $v_{n_k}$ such that $\|v_{n_k}\| \geq \epsilon,$ i.e. $\|\frac{v_{n_k}}{\epsilon}\| \geq 1.$ But $\frac{v_{n_k}}{\epsilon}$ converges to zero weakly, contradicting a). Thus, every weakly convergent sequence in $l^1$ is norm convergent.

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