Record a problem

Let $H$ be a Hilbert space. It is well-known that if $Z$ is a convex and closed subset of $H$, then for every $x\in H$, there is a unique element $z\in Z$ such that \[
\|x-z\|=d(x,Z).
\] Let's denote this unique element $z$ by $P_Zx$.

Problem. Let $Z_1\supseteq Z_2\supseteq Z_3\supseteq \cdots$ be a chain of closed convex subsets of a Hilbert space $H$, show that:
(a) If $\capp_{n=1}^\infty Z_n \neq\emptyset$, then $\|x-P_{Z_n}x\|\to \|x-P_Zx\|$.
(b) If $\capp_{n=1}^\infty Z_n=\emptyset$, then $\|x-P_{Z_n}x\|\to \infty$.

Since $Z_n$'s are not subspace of $H$, $P_{Z_n}$ itself is not a linear map, and therefore standard tools from functional analysis cannot be used. Nevertheless, by assuming $Z_n$'s are further a closed subspace, it is instructive to see a weak-convergence argument to conclude the result in (a) --- a slightly complicated solution.

Some also asked me can $\capp_{n=1}^\infty Z_n=\emptyset$ happen? Below is an example to show there is a chain of closed convex subsets with empty intersection:

Example. Let $Z_n=\{(1,\frac{1}{\sqrt{2}},\dots,\frac{1}{\sqrt{n}},a_1,a_2,\dots):(a_1,a_2,\dots)\in \ell^2\}$, then $Z_n$ is convex, a closed subset of $\ell^2$, and descending in $n$. However, if there is $(x_1,x_2,\dots)\in \capp_{n=1}^\infty Z_n$, then necessarily $x_n=1/\sqrt{n}$ for every $n$, and this element is not in $\ell^2$, a contradiction. So $\capp_{n=1}^\infty Z_n=\emptyset$.
　　Next what is $d(x,Z_n)$ in this particular example? By definition for $z\in Z_n$ we have \[
\|x-z\|^2 = \sum_{k=1}^n\left|x_k-\frac{1}{\sqrt{k}}\right|^2+\sum_{k>n} |x_k-z_k|^2,
\] hence for fixed $x\in \ell^2$ the smallest possible $\|x-z\|$ is \[
d(x,Z_n)=\sqrt{\sum_{k=1}^n \left|x_k-\frac{1}{\sqrt{k}}\right|^2},
\] of course this diverges to infinity as $n\to\infty$ for any fixed $x\in \ell^2$. What's more, \[P_{Z_n}x =\textstyle (1,\frac{1}{\sqrt{2}},\dots,\frac{1}{\sqrt{n}}, x_{n+1},x_{n+2},\dots).\qed\]