Analysis Problem of the Day 32

After a brief hiatus, today’s problem is Problem 2 from the Spring 2009 UCLA Analysis Qual:

Let $H$ be an infinite-dimensional real Hilbert space.

a) Prove that the unit sphere of $H$ is weakly dense in the unit ball of $H.$

b) Prove that there exists a sequence of bounded linear operators $T_n$ on $H$ such that $\|T_n\|=1$ for all $n$ but $T_n(x) \to 0$ for all $x \in H.$

Solution: a) I claim that in any infinite-dimensional normed space $X,$ the weak closure of the unit sphere $\overline{S^w}$ is the closed unit ball $B$ in $H.$ Indeed, one inclusion is clear – since a convex set is closed if and only if it is weakly closed, $B$ is weakly closed, and so $\overline{S^w} \subseteq B.$ Conversely, recall that the weak topology is the coarsest topology on $H,$ such that the linear functional evaluation maps $x \to \phi(x) = \langle x, \phi \rangle$ are continuous. Thus, the basic open neighborhoods in the weak topology of some $x \in H$ are the sets $U=\{y \in H: \langle y-x, \phi_i \rangle < \epsilon, i=1,...,n\}.$ For $x \in B,$ note that $y-x \in \bigcap_{i=1}^n \ker \phi_i,$ and since kernels of linear functionals have finite codimension and $X$ is infinite-dimensional, the intersection of the kernels is infinite-dimensional and therefore contains a line $L=\{tv: t \in \mathbb{R}, v \in X\}$ through the origin. In particular, since $\|x\| \leq 1,$ if $y-x \in L,$ so $y = x+L$ intersects $S$ (since for $t=0$ one has $\|y\| \leq 1$ and for $t$ large $\|y\| \to \infty.$ ) Thus, any basic open neighborhood of $x \in B$ intersects $S,$ i.e. $S$ is weakly dense in $B.$ Along with the other inclusion, it follows that $\overline{S^w}=B.$

b) Note that since $H$ is reflexive, the weak topology on the unit ball is metrizable. In particular, let $x_n \in S$ be a sequence of unit vectors converging weakly to 0, and define $T_n: H \to H$ by $T_n x= \langle x,x_n\rangle x_n.$ Then, $T_n x \to 0$ for all $x \in H$ since $x_n \rightharpoonup 0,$ but $\|T_n x \| \leq \|x\|$ and $\|T_n x_n\| = \|x_n\|=1,$ i.e. $\|T_n\|=1.$ Thus, $T_n$ satisfies the necessary properties.

Remark: A general topological fact is that a closed set in a second-countable space is sequentially closed. Conversely, a convex weakly sequentially closed set in a Banach space is weakly closed. One can also show that if $X^*$ is separable, then $0$ is in the weak sequential closure of the unit sphere. On the other hand, if $X^*$ is not separable, as in the case of $(l^1)^*=l^\infty,$ since the weak and norm topology on $l^1$ coincide, the unit sphere is closed and therefore weakly sequentially closed, yet not weakly closed.

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