=0\text{ for all $s\in S$}\,\}.
$$
Then $S^\perp$ is a closed subspace of $X$. We obviously have
$S\cap S^\perp = 0$ and $S\subset S^{\perp\perp}$.
Claim: If $S$ is a closed subspace of $X$, $x\in X$, and $P_S x$ the
projection of $x$ onto $S$, then $x-P_S x\in S^\perp$. Indeed,
if $s\in S$ is arbitrary and $t\in\R$, then
$$
\|x-P_S x\|^2 \le \|x-P_S x - t s\|^2=\|x-P_S x\|^2-2t(x-P_S x,s)+t^2\|s\|^2,
$$
so the quadratic polynomial on the right hand side has a minimum at $t=0$.
Setting the derivative there to 0 gives $(x-P_S x,s)=0$.
Thus we can write any $x\in X$ as $s+s^\perp$ with $s\in S$ and $s^\perp\in
S^\perp$ (namely $s=P_S x$, $s^\perp=x-P_S x$). Such a decomposition
is certainly unique (if $\bar s+\bar s^\perp$ were another one
we would have $s-\bar s=\bar s^\perp-s^\perp\in S\cap S^\perp=0$.)
We clearly have $\|x\|^2=\|s\|^2+\|s^\perp\|^2$.
An immediate corollary is that $S^{\perp\perp}=S$ for $S$ a closed subspace,
since if $x\in S^{\perp\perp}$ we can write it as $s+s^\perp$, whence
$s^\perp\in S^\perp\cap S^{\perp\perp}=0$, i.e., $x\in S$. We thus
see that the decomposition
$$
x=(I-P_S)x+P_S x
$$
is the (unique) decomposition of $x$ into elements of $S^\perp$ and
$S^{\perp\perp}$. Thus $P_{S^\perp}=I-P_S$. For any subset $S$ of $X$,
$S^{\perp\perp}$ is the smallest closed subspace containing $S$.
\subsubhead Orthonormal sets and bases in Hilbert space \endsubsubhead
Let $e_1$, $e_2$, \dots, $e_N$ be orthonormal elements of a Hilbert space
$X$, and let $S$ be their span. Then $\sum_n\ $. This map is a linear isometry of $X$ onto $X^*$.
For a complex Hilbert space it is a conjugate linear isometry
(it satisfies $j_{\alpha y}=\bar\alpha j_y$).
\endproclaim
\demo{Proof} It is easy to see that $j$ is an isometry of $X$ into $X^*$ and
the main issue is to show that any $f\in X^*$ can be written as $j_y$ for
some $y$. We may assume that $f\ne0$, so $\Nu(f)$ is a proper closed subspace
of $X$. Let $y_0\in[\Nu(f)]^\perp$ be of norm $1$ and set $y=(fy_0)y_0$.
For all $x\in X$, we clearly have that $(fy_0)x-(fx)y_0\in \Nu(f)$, so
$$
j_y(x)=\ 0$, then $E(0_Y,r)\subset T(B(0_X,2))$.
\endproclaim
\demo{Proof} Let $U=T(E(0_X,1))$. Let $y\in Y$, $\|y\| w*>> \delta_0$.
\proclaim{Theorem (Alaoglu)} The unit ball in $X^*$ is weak* compact.
\endproclaim
\demo{Proof} For $x\in X$, let $I_x=\{\,t\in\R\,:\, |t|\le\|x\|\,\}$,
and set $\Omega=\Pi_{x\in X} I_x$. Recall that this Cartesian product
is nothing but the set of all functions $f$ on $X$ with $f(x)\in I_x$
for all $x$. This set is endowed with the Cartesian product topology,
namely the weakest topology such that for all $x\in X$,
the functions $f\mapsto f(x)$ (from $\Omega$ to $I_x$) are continuous.
Tychonoff's Theorem states that $\Omega$ is compact with this topology.
Now let $E$ be the unit ball in $X^*$. Then $E\subset\Omega$ and
the topology thereby induced on $E$ is precisely the weak* topology.
Now for each pair $x,y\in X$ and each $c\in R$, define
$F_{x,y}(f)=f(x)+f(y)-f(x+y)$, $G_{x,c}=f(c x)-cf(x)$. These are
continuous functions on $\Omega$ and
$$
E=\bigcap_{x,y\in X}F_{x,y}^{-1}(0)\cap\bigcap\Sb x\in X\\c\in\R\endSb
G_{c,x}^{-1}(0).
$$
Thus $E$ is a closed subset of a compact set, and therefore compact itself.
\qed
\enddemo
\proclaim{Corollary} If $f_n@>w^*>>f$ in $X^*$, then
$\|f\|\le\displaystyle\liminf_{n\to\infty} \|f_n\|_{X^*}$.
\endproclaim
\demo{Proof} Let $C=\liminf\|f_n\|$ and let $\epsilon>0$ be arbitrary.
Then there exists a subsequence (also denoted $f_n$) with
$\|f_n\|\le C+\epsilon$. The ball of radius $C+\epsilon$ being
weak* compact, and so weak* closed, $\|f\|