Abstract Hilbert Space Results
We have learned a little about the Hilbert spaces L 2 ÝUÞ and § 2 and we have at least defined
H 1 ÝUÞ and the scale of Hilbert spaces H p ÝUÞ. Now we are going to develop additional facts
that are true about any Hilbert space. Later these facts will be helpful in formulating and
solving partial differential equations in a Hilbert space setting.
1. Subspaces
A subset M of Hilbert space H is a subspace of it is closed under the operation of forming
linear combinations;i.e.,
-x, y 5 M, C 1 x + C 2 y 5 M,
- C 1 , C 2 5 R.
The subspace M is said to be closed if it contains all its limit points; i.e., every sequence of
elements of M that is Cauchy for the H-norm, converges to an element of M. In a Euclidean
space every subspace is closed but in a Hilbert space this is not the case.
Example 1.11. If U is a bounded open set in R N then H = L 2 ÝUÞ is a Hilbert space containing M = CÝUÞ
as a subspace. It is easy to find a sequence of functions in M that is Cauchy for the H norm
but the sequence converges to a function in H that is discontinuous and hence not in M. For
example, if N = 1, and U = Ý?1, 1Þ the sequence
nx
u n ÝxÞ =
if
|x| ² 1/n
1
if
1/n ² x ² 1
?1
if
? 1 ² x ² ?1/n
can be shown to converge in the L 2 ? norm to uÝxÞ = sgnÝxÞ. Here sgnÝxÞ 5 L 2 ß?1, 1à but
sgnÝxÞ 6 Cß?1, 1à; i.e., the limit does not belong to the subspace of continuous functions.
This proves that M = Cß?1, 1à is not closed in H = L 2 ß?1, 1à.
2. Every finite dimensional subspace of a Hilbert space H is closed. For example, if M
denotes the span of finitely many elements áx 1 , ....x N â in H, then the set M of all possible
linear combinations of these elements is finite dimensional (of dimension N), hence it is
closed in H.
3. Let M denote a subspace of Hilbert space H and let M e denote the orthogonal
complement of M.
def
M e = áx 5 H : Ýx, yÞ H = 0, -y 5 Mâ.
Then M e is easily seen to be a subspace and it is closed, whether or not M itself is closed.
To see this, suppose áx n â is a Cauchy sequence in M e converging to a limit x 5 H. For
arbitrary y 5 M, Ýx n , yÞ H = 0 for every n, and hence
Ýx, yÞ H = Ýx ? x n , yÞ H + Ýx n , yÞ H = Ýx ? x n , yÞ H + 0 í 0, as n tends to infinity.
1
Then the limit point x is orthogonal to every y in M which is to say, x is in M e , and M e is
closed.
#
If M is a subspace of H that is not closed, then M is contained in a closed subspace M
#
of H, consisting of M together with all its limit points. M is called the closure or completion
# . This means that for every x in M
# there is a sequence
of M and M is said to be dense in M
of elements of M that converge to x in the norm of H. Equivalently, to say M is dense in
# means that for every x in M
# and every O > 0, there is a y in M such that qx ? yq H < O.
M
Then we can show,
#.
Lemma 1.1 Let M denote a subspace of Hilbert space H. Then ÝM e Þ e = M
Lemma 1.2 A subspace M of Hilbert space H is dense in H if and only if M e = á0â.
Example 1.2
1.Recall the Hilbert-Sobolev space of order one,
H 1 ÝUÞ =
uÝxÞ 5 L 2 ÝUÞ : / x i uÝxÞ 5 L 2 ÝUÞ for i = 1, ..., n
where / x i uÝxÞ denotes the distributional derivative of uÝxÞ with respect to x i . We defined an
inner product on H 1 ÝUÞ as follows,
Ýu, vÞ 1 = XU ßuÝxÞvÝxÞ + 4u 6 4và dx
for u, v 5 H 1 ÝUÞ.
and showed that H = H 1 ÝUÞ is complete for the norm induced by this inner product. The
linear space C K ÝUÞ, of infinitely differentiable functions is contained in H 1 ÝUÞ and is
therefore a subspace of H 1 ÝUÞ. We can show, using a technique called regularization, that
M = C K ÝUÞ is dense in H; i.e., the completion of C K ÝUÞ in the norm of H 1 ÝUÞ is the whole
Hilbert space. Then, according to the previous lemma, only the zero function is orthogonal,
in the H 1 ÝUÞ ? inner product, to every function in C K ÝUÞ.
2. The linear space of test functions M = C Kc ÝUÞ is also a subspace of H 1 ÝUÞ The closed
# obtained by completing the test functions in the norm of H 1 ÝUÞ is denoted by
subspace M
H 10 ÝUÞ and we refer to this closed subspace as the H 1 ÝUÞ functions that ”vanish on the
boundary of U”. In fact, what we can show is that for every g 5 H 10 ÝUÞ, and for every point
p 5 /U, the integral of g over the ball, B P ÝpÞ, tends to zero as P tends to zero. This says that
if g 5 H 10 ÝUÞ, then g has zero mean value near every point on /U.
We will show now that the closed subspace H 10 ÝUÞ is not equal to H 1 ÝUÞ. To do this, we
need an inequality known as the Poincaré inequality. This inequality asserts that for any
bounded subset U Ð R n there exists a constant, C depending only on U, such that
||d|| 20 ² C || 4d|| 20
-d 5 C Kc ÝUÞ
||d|| 21 ² C || 4d|| 20
-d 5 C Kc ÝUÞ
||d|| 0 =
XU d 2 dx
1/2
or, equivalently
2
The proof of this inequality will be given later. We will use it to show there are functions in
H 1 ÝUÞ that are not in H 10 ÝUÞ. For example, let 1ÝxÞ 5 H 1 ÝUÞ denote the constant function
having value one at every point. Then
||1 ? d|| 21 = ||1 ? d|| 20 + ||4Ý1 ? dÞ|| 20 = ||1 ? d|| 20 + ||4d|| 20
Clearly then
||1 ? d|| 21 ³ ||1 ? d|| 20
The first of these results implies
-d 5 C Kc ÝUÞ.
||1 ? d|| 21 ³ || 4d|| 20 .
and
||1 ? d|| 21 ³ ||1|| 20 ? ||d|| 20
which leads to
||1 ? d|| 21 + ||d|| 20 ³ ||1|| 20 = XU 1 dx = |U| = area of U
Then the Poincaré inequality implies ||1 ? d|| 21 + C ||4d|| 20 ³ |U|,
Ý1 + CÞ||1 ? d|| 21 ³ |U|.
and finally,
It is clear from this inequality that there is no sequence of test functions d n which can
converge to 1ÝxÞ in the norm of H 1 ÝUÞ. Then 1ÝxÞ does not belong to H 10 ÝUÞ although it
does belong to H 1 ÝUÞ.
2. Projections
A Hilbert space H is said to be separable if H contains a countable dense subset áh n â. In
this case, for every x in H and every O > 0 there exists an integer N O and scalars áa n â such
that
N
x ? > n=1 a n h n
H
<O
for N > N O
If H is a separable Hilbert space, then the Gram-Schmidt procedure can be used to
construct an orthonormal basis for H out of a countable dense subset. Recall that an
orthonormal basis for H is a set of mutually orthogonal unit vectors, ád n â in H with the
following property:
1) For f 5 H, Ýd n , fÞ H = 0 -n é
f=0
When the orthonormal set ád n â has property 1, then it is said to be dense or complete in
H. Of course, not every orthonormal set in H is complete. Recall that other equivalent ways
of characterizing completeness for orthonormal sets can be stated as follows:
2) For all f in H and every O > 0, there exists an integer N O such that
N
f ? > n=1 Ýf, d n Þ H d n
H
<O
for N > N O
3
K
3) For every f in H, > n=1 f 2n = qfq 2H where f n = Ýf, d n Þ H
Hilbert Space Projection Theorem
In a Euclidean space,E, where all subspaces M are closed, it is a fact that for each y 5 E
there is a unique z 5 M such that qy ? zq E is minimal. This element z, which is just the
orthogonal projection of y onto M, is the ”best approximation to y from within M”. In an
infinite dimensional Hilbert space, a similar result is true for closed subspaces but for
subspaces that are not closed there may fail to be a ”best” approximation in M.
Hilbert Space Projection Theorem
let y in H be given. Then
Let M be a closed subspace of Hilbert space H and
(i) there exists a unique x y in M such that qy ? x y q H ² qy ? zq H for all z in M
(x y is the unique point of M that is closest to y, the best approximation to y in M )
(ii) Ýy ? x y , zÞ H = 0 for all z in M; i.e., y ? x y ó M
(iii) every y in H can be uniquely expressed as y = x y + z y
where
Qy = z y 5 M e
Py = x y 5 M,
and
qyq 2H = qPyq 2H + qQyq 2H
i.e., H = M ã M e .
The proof of this result will be given later.
3. Linear Functionals and Bilinear Forms
A real valued function defined on H, is said to be a functional on H. The functional, L, is
said to be:
(a) Linear if, for all x and y in H, LÝC 1 x + C 2 yÞ = C 1 Lx + C 2 Ly, for all scalars C 1 ,C 2 .
(b) Bounded if there exists a constant C such that |Lx| ² Cqxq H for all x in H
(c) Continuous if qx n ? xq H ¸ 0 implies that
|Lx n ? Lx| í 0
It is not difficult to show that the only example of a linear functional on a Euclidean space E
is Lx = Ýx, zÞ E for some z in E, fixed. For example, if F is a linear functional on E, then for
arbitrary x in E,
n
FÝxÞ = F > i=1 x i ē i
n
n
= > i=1 x i FÝē i Þ = > i=1 x i F i = Ýx, z F Þ E = x f z F
where áe i â denotes the standard basis in E and z# F denotes the n-tuple whose i-th
4
component is F i = FÝē i Þ. This displays the isomorphism between functionals F and
elements, z F , in E. This isomorphism also exists in an abstract Hilbert space.
Riesz Representation Theorem For every continuous linear functional f on Hilbert
space H there exists a unique element, z f in H such that fÝxÞ = Ýx, z f Þ H for all x in H.
Proof- Let N f = áx 5 H : fÝxÞ = 0â.Then N f is easily seen to be a closed subspace of H. If
N f = H then z f = 0 and we are done. If N f ® H then H = N f ã N ef by the Hilbert space
projection theorem. Since N f is not all of H, N ef must contain nonzero vectors, and we
denote by z 0 an element of N ef such that qz 0 q H = 1. Then for any x in H,
w = fÝxÞz 0 ? fÝz 0 Þx,
belongs to N f hence w ó z 0 . But in that case,
ÝfÝxÞz 0 ? fÝz 0 Þx, z 0 Þ H = fÝxÞÝz 0 , z 0 Þ H ? fÝz 0 ÞÝx, z 0 Þ H = 0.
This leads to, fÝxÞ = fÝz 0 ÞÝx, z 0 Þ H = Ýx, fÝz 0 Þz 0 Þ H which is to say z f = fÝz 0 Þz 0 .
To see that z f is unique, suppose that
-x 5 H
fÝxÞ = Ýz f , xÞ H = Ýw f , xÞ H
Subtracting leads to the result that
-x 5 H.
Ýz f ? w f , xÞ H = 0
In particular, choosing x = z f ? w f leads to qÝz f ? w f q H = 0.n
A real valued function aÝx, yÞ defined on H × H is said to be:
(a) Bilinear if, for all x 1 , x 2 , y 1 , y 2 5 H and all scalars C 1 , C 2
aÝC 1 x 1 + C 2 x 2 , y 1 Þ = C 1 aÝx 1 , y 1 Þ + C 2 aÝx 2 , y 1 Þ
aÝx 1 , C 1 y 1 + C 2 y 2 Þ = C 1 aÝx 1 , y 1 Þ + C 2 aÝx 1 , y 2 Þ
(b) Bounded if there exists a constant b > 0 such that,
|aÝx, yÞ| ² bqxq H qyq H for all x,y in H
(c) Continuous if x n í x, and y n í y in H, implies aÝx n , y n Þ í aÝx, yÞ in R
(d) Symmetric if aÝx, yÞ = aÝy, xÞ for all x, y 5 H
(e) Positive or coercive if there exists a constant a 0 > 0 such that
aÝx, xÞ ³ a 0 qxq 2H for all x in H
It is not hard to show that for both linear functionals and bilinear forms, boundedness is
5
equivalent to continuity. If aÝx, yÞ is a bilinear form on H × H, and FÝxÞ is a linear functional
on H, then ®ÝxÞ = 12 aÝx, xÞ ? FÝxÞ + Const is called a quadratic functional on H. In a
Euclidean space a quadratic functional has a unique extreme point located at the point
where the gradient of the functional vanishes. This result generalizes to the infinite
dimensional situation.
Lemma 3.1 Suppose aÝx, yÞ is a positive, bounded and symmetric bilinear form on Hilbert
space H, FÝxÞ is a bounded linear functional on H and C is a constant. Consider the
following problems
(a) minimize ®ÝxÞ =
1
2
aÝx, xÞ ? FÝxÞ + C
over H
(b) find x in H satisfying aÝx, yÞ = FÝyÞ for all y in H.
Then
i) x in H solves (a) if and only if x solves (b)
ii) there is at most on x in H solving (b)
iii) there is at least one x in H solving (a)
Proof- For t in R and x, y fixed in H, let fÝtÞ = ®Ýx + tyÞ . Then fÝtÞ is a real valued function
of the real variable t and it follows from the symmetry of aÝx, yÞ that
fÝtÞ =
1 2
t
2
aÝy, yÞ + tßaÝx, yÞ ? FÝyÞà +
1
2
aÝx, xÞ ? FÝxÞ + C
and
f v ÝtÞ = t aÝy, yÞ + ßaÝx, yÞ ? FÝyÞà
It follows that ®ÝxÞ has a global minimum at x in H if and only if fÝtÞ has a global minimum at
t = 0; i.e.,
®Ýx + tyÞ = ®ÝxÞ + tf v Ý0Þ + t 2 /2 aÝx, xÞ ³ ®ÝxÞ, -t 5 R and -y 5 H
if and only if
f v Ý0Þ = aÝx, yÞ ? FÝyÞ = 0.
-y 5 H.
This establishes the equivalence of (a) and (b).
To show that ÝbÞ has at most one solution in H, suppose
aÝx 1 , yÞ = FÝyÞ
and
aÝx 2 , yÞ = FÝyÞ for all y in H.
Then aÝx 1 , yÞ ? aÝx 2 , yÞ = aÝx 1 ? x 2 , yÞ = 0 for al l y in H. In particular, for y = x 1 ? x 2
6
0 = aÝx 1 ? x 2 , x 1 ? x 2 Þ ³ a 0 qx 1 ? x 2 q 2H ; i.e., x 1 = x 2 .
To show that ®ÝxÞ has at least one minimum in H, let J = inf x5H ®ÝxÞ. Now
®ÝxÞ =
1
2
aÝx, xÞ ? FÝxÞ ³
1
2
a 0 qxq 2H ? bqxq H
and it is evident that ®ÝxÞ tends to infinity as qxq H tends to infinity. This means
J > ?K (i.e.,” the parabola opens upward rather than downward”). Moreover since J is an
infimum, there exists a sequence x n in H such that ®Ýx n Þ í J as n tends to infinity. Note
that
2ßaÝx n , x n Þ + aÝx m , x m Þà = aÝx n ? x m , x n ? x m Þ + aÝx m + x n , x m + x n Þ
which leads to the result,
®Ýx m Þ + ®Ýx n Þ = 1 aÝx m ? x n , x m ? x n Þ + 2 ®ßÝx m + x n Þ/2à ³ 1 Cqx m ? x n q 2H + 2J.
4
4
But ®Ýx m Þ + ®Ýx n Þ tends to 2J as n tends to infinity and in view of the previous line, the
minimizing sequence áx n â must be a Cauchy sequence with limit x in the Hilbert space H.
Finally, since ®ÝxÞ is continuous, ®Ýx n Þ í ®ÝxÞ = J.n
Applications of the lemma1. Lemma 3.1 can now be used to prove the Hilbert space projection theorem.
For M a closed subspace in H it follows that M is itself a Hilbert space for the norm and
inner product inherited from H.
For y a fixed but arbitrary element in H, we can define
aÝz, xÞ = Ýz, xÞ H
FÝzÞ = Ýy, zÞ H
and
®ÝzÞ = 12 aÝz, zÞ ? FÝzÞ +
Note that -x 5 M
1
2
qx ? yq 2H =
=
1
2
1
2
1
2
qyq 2H
-x, z 5 M
-z 5 M,
-z 5 M.
Ýx ? y, x ? yÞ H
áqxq 2H ? 2Ýx, yÞ H + qyq 2H â = ®ÝxÞ
Clearly aÝz, xÞ is a positive, bounded and symmetric bilinear form on M, F is a bounded
linear functional on M. Then it follows from the lemma that there exists a unique element
x y 5 M which minimizes ®ÝzÞ over M. It follows also form the equivalence of problems (a)
and (b) that x y satisfies
Ýx y , zÞ H = Ýy, zÞ H -z 5 M.
aÝx y , zÞ = FÝzÞ, -z 5 M; i.e.,
7
But this is just the assertion that
Ýx y ? y, zÞ H = 0
-z 5 M; i.e., x y ? y ó M.
Finally, for y in H, fixed, let the unique element x y in M be denoted by Py = x y 5 M. Then
y ? Py ó M, and z = y ? Py 5 M e .
To see that this decomposition of elements of H is unique, suppose
and
y = x y + z,
y = X y + Z,
x y 5 M,
X y 5 M,
z 5 Me,
Z 5 Me,
Then
x y + z = X y + Z, and x y ? X y = Z ? z.
Z ? z 5 Me,
M V M e = á0â,
But
x y ? X y 5 M,
and it follows that
x y ? X y = Z ? z = 0.n
2. Recall that for U open and bounded in R n , the Hilbert Sobolev space of order one,
H 1 ÝUÞ = H is a Hilbert space which contains C K ÝUÞ as a dense subspace and also contains
the closed subspace H 10 ÝUÞ, obtained by completing the subspace of test functions in the
H 1 -norm. We showed in an earlier example, that H 10 ÝUÞ is not equal to H 1 ÝUÞ. Then by the
projection theorem, every y in H can be uniquely expressed as a sum, y = x y + z, with
e
e
x y 5 H 10 ÝUÞ, and z 5 ÝH 10 ÝUÞÞ . To characterize the subspace ÝH 10 ÝUÞÞ , choose arbitrary
d 5 C K0 ÝUÞ and f 5 C K ÝUÞ and write
Ýd, fÞ H = X ßdf + 4d 6 4fà dx = X dßf ? 4 2 fàdx + X d / N fdS
U
U
/U
= Ýd, f ? 4 2 fÞ 0 + 0. (Here Ýu, vÞ 0 denotes the H 0 ÝUÞ = L 2 ÝUÞ inner product).
e
Now suppose f 5 C K ÝUÞ V ÝH 10 ÝUÞÞ . Then Ýd, fÞ H = 0, for all d 5 C K0 ÝUÞ, and since C K0 ÝUÞ is
dense in H 10 ÝUÞ, Ýu, fÞ H = 0, for all u 5 H 10 ÝUÞ. That is, Ýu, f ? 4 2 fÞ 0 = 0 - u 5 H 10 ÝUÞ. But
e
this implies that f 5 C K ÝUÞ V ÝH 10 ÝUÞÞ satisfies f ? 4 2 f = 0, in H 0 ÝUÞ. Then, since C K ÝUÞ is
dense in H = H 1 ÝUÞ it follows that
e
ÝH 10 ÝUÞÞ = áz 5 H 1 ÝUÞ : z ? 4 2 z 5 H 0 ÝUÞ, and z ? 4 2 z = 0 â.
The Lax-Milgram Lemma
Lemma 3.1 requires that the bilinear form aÝx, yÞ be symmetric. For application to existence
theorems for partial differential equations, this is an unacceptable restriction. Fortunately,
the most important part of the result remains true even when the form is not symmetric.
8
Lax-Milgram LemmaSuppose aÝu, vÞ is a bounded and positive bilinear form on Hilbert
space H; i.e., for positive constants a 0 , a 1
i)
ii)
-u, v 5 H
| aÝu, vÞ| ² a 1 quq H qvq H
2
aÝu, uÞ ³ a 0 quq H
-u 5 H.
Suppose also that FÝvÞ is a bounded linear functional on H. Then there exists a unique
u F 5 H such that
aÝu F , vÞ = FÝvÞ -v 5 H.
Proof- For each fixed u 5 H, the mapping v $ aÝu, vÞ is a bounded linear functional on H.
It follows that there exists a unique z u 5 H such that
aÝu, vÞ = Ýz u , vÞ H -v 5 H.
Let Au = z u ; i.e., aÝu, vÞ = ÝAu, vÞ H -u 5 H. Clearly A is a linear mapping of H into H, and
since
qAuq 2H = |ÝAu, AuÞ H | = | aÝu, AuÞ| ² a 1 quq H qAuq H
it is evident that A is also bounded. Note further, that
a 0 quq 2H ² aÝu, uÞ = ÝAu, uÞ H ² qAuq H quq H
i.e.,
a 0 quq H ² qAuq H -u 5 H.
This estimate implies that A is one-to one and that R A , the range of A, is closed in H.
Finally, we will show that R A = H. Since the range is closed, we can use the projection
theorem to write, H = R A ã R eA . If u 5 R eA , then
0 = ÝAu, uÞ H = aÝu, uÞ ³ a 0 quq 2H ;
i.e., R eA = á0â.
Since FÝvÞ is a bounded linear functional on H, it follows from the Riesz theorem that there
is a unique z F 5 H such that FÝvÞ = Ýz F , vÞ H for all v 5 H. Then the equation
aÝu, vÞ = FÝvÞ can be expressed as
ÝAu, vÞ H = Ýz F , vÞ H -v 5 H; i.e., Au = z F .
But A has been seen to be one-to-one and onto and it follows that there exists a unique
u F 5 H such that Au F = z F. n
9
Convergence in H
In R N convergence of x n to x means
N
qx n ? xq R N = > i=1 ßÝx n ? xÞ 6 e i à 2
1/2
í 0 as n í K.
Here e i denotes the i-th vector in the standard basis.This is equivalent to,
Ýx n ? xÞ 6 e i í 0 as n í K, for i = 1, ..., N,
and to
Ýx n ? xÞ 6 z í 0 as n í K, for every z 5 R N
In an infinite dimensional Hilbert space H, convergence of x n to x in H means
qx n ? xq H í 0 as n í K.
This is called strong convergence in H and it implies that
Ýx n ? x, vÞ H í 0 as n í K -v 5 H.
This last mode of convergence is referred to as weak convergence and, in a general
Hilbert space, weak convergence does not imply strong convergence. Thus while there is
no distinction between weak and strong convergence in a finite dimensional space,
the two notions of convergence are not the same in a space of infinite dimensions.
In R N the so called Bolzano-Weierstrass theorem asserts that every bounded
sequence áx n â contains a convergent subsequence. The theorem is proved by noting that
áx n 6 e 1 â is a bounded sequence of real numbers and hence contains a subsequence
áx n,1 6 e 1 â that is convergent. Similarly, áx n,1 6 e 2 â is also a bounded sequence of real
numbers and thus contains a subsequence áx n,2 6 e 2 â that is convergent. Proceeding in this
way, we can generate a sequence of subsequences, áx n,k â Ð áx n â such that áx n,k 6 e j â is
convergent for j ² k. Then the diagonal sequence áx n,n â is such that á² x n,n 6 ejâ is
convergent for 1 ² j ² N, which is to say, áx n,n â is convergent. In a general Hilbert space
we have a weaker result,
In H every bounded sequence áx n â contains a subsequence that is weakly convergent.
To see this, suppose that qx n q ² M for all n and let ád j â denote a complete orthonormal
family in H. Proceeding as we did in R N , let áx n,k â Ð áx n â denote a subsequence such
that áÝx n,k , d j Þ H â is convergent (in R) for j ² k. Then for each j, Ýx n,j , d j Þ H converges to a
real limit a j as n tends to infinity. It follows that the diagonal subsequence áx n,n â is such that
Ýx n,n , d j Þ H converges to a j for j ³ 1. Now define
FÝvÞ = Lim n Ýx n,n , vÞ H
Then
for v 5 H.
|FÝvÞ| ² | Lim n Ýx n,n , vÞ H | ² M qvq H
from which it follows that F is a continuous linear functional on H. By the Riesz theorem,
10
there exists an element, z F in H such that
But
FÝvÞ = Ýz F , vÞ H for all v in H.
FÝvÞ = F > i Ýv, d i Þ H d i = lim n Ýx n,n , > i Ýv, d i Þ H d i Þ H
= > i lim n Ýx n,n , d i Þ H Ýv, d i Þ H = > i a i Ýv, d i Þ H ;
for all v in H.
That is,
FÝvÞ = Ýz F , vÞ H = > i a i Ýv, d i Þ H
Then by the Parseval-Plancherel identity, it follows that
zF = >i aidi
and
Ýx n,n , vÞ H í Ýz F , vÞ H
for all v in H.
11