as a Hilbert space. This is
another interesting application of Theorem 17.1.8. First here is a definition which
involves abominable and atrociously misleading notation which nevertheless seems to be
well accepted.
Definition 17.2.1Let L ∈ℒ
(U,H )
, the bounded linear maps from U to H for U,HHilbert spaces. For y ∈ L
(U )
, let L^{−1}y denote the unique vector in
{x : Lx = y} ≡ My
which is closest in U to 0.
PICT
Note this is a good definition because
{x : Lx = y}
is closed thanks to the continuityof L and it is obviously convex. Thus Theorem 17.1.8applies. With this definition definean inner product on L
(U)
as follows. For y,z ∈ L
(U )
,
( −1 − 1)
(y,z)L(U) ≡ L y,L z U
The notation is abominable because L^{−1}
(y)
is the normal notation for M_{y}.
In terms of linear algebra, this L^{−1} is the Moore Penrose inverse. There you obtain
the least squares solution x to Lx = y which has smallest norm. Here there is an actual
solution and among those solutions you get the one which has least norm. Of course a
real honest solution is also a least squares solution so this is the Moore Penrose inverse
restricted to L(U).
First I want to understand L^{−1} better. It is actually fairly easy to understand in
terms of geometry. Here is a picture of L^{−1}
(y)
for y ∈ L
(U )
.
PICT
As indicated in the picture, here is a lemma which gives a description of the
situation.
Lemma 17.2.2In the context of the above definition, L^{−1}
(y)
is characterized by
( −1 )
L (y),x U = 0 for all x ∈ ker(L)
L (L−1 (y)) = y, (L −1(y) ∈ My )
In addition to this, L^{−1}is linear and the above definition does define an innerproduct.
Proof: The point L^{−1}
(y)
is well defined as noted above. I claim it is characterized
by the following for y ∈ L
(U)
( −1 )
L ((y),x U) = 0 for( all x ∈ ker(L ))
L L−1 (y) = y, L −1(y) ∈ My
Consider the last claim. If L is one to one, then for y ∈ L
(U)
, there is only one vector
which maps to y. Therefore,
L −1(L(x)) = x.
Hence for y ∈ L
(U)
,
∥y∥ ≡ ∥∥L−1(y)∥∥
L(U ) U
Also,
∥ ∥
∥Lu∥L(U) ≡ ∥L− 1(L(u))∥ ≡ ∥u∥U
U
Now here is another argument for various continuity claims.
∥ −1 ∥
∥Lx∥L(U) ≡ ∥L (Lx)∥U ≤ ∥x∥U
because L^{−1}
(Lx)
is the smallest thing in U which maps to Lx and x is something which
maps to Lx so it follows that the inequality holds. Hence L ∈ℒ
(U,L (U))
and in fact,
∥L∥
_{ℒ(U,L(U))
} = 1. Next, letting y ∈ L
(U)
,
∥∥L−1y∥∥ ≡ ∥y∥
U L(U)
and so
∥∥L−1∥∥
_{ℒ(L(U),U)
} = 1 and this shows that L ∈ℒ
(U,L(U ))
while L^{−1}∈ℒ
(L(U ),U )
and both have norm equal to 1.
Now
∥∥ ( −1 )∥∥ ∥∥ − 1 ∥∥
∥Lx∥H = L L (Lx ) H ≤ ∥L ∥ℒ(U,H) L (Lx ) U ≡ ∥L ∥ℒ(U,H)∥Lx∥L(U) ■
Now here are some other very interesting results. I am following [?].
Lemma 17.2.4Let L ∈ℒ
(U,H )
. Then L
(B-(0,r))
is closed and convex.
Proof: It is clear this is convex since L is linear. Why is it closed? B
(0,r)
is compact in the weak topology by the Banach Alaoglu theorem, Theorem
15.5.4 on Page 1326. Furthermore, L is continuous with respect to the weak
topologies on U and H. Here is why this is so. Suppose u_{n}→ u weakly in U. Then if
h ∈ H,
(Lu ,h ) = (u ,L∗h) → (u,L ∗h ) = (Lu,h)
n n
which shows Lu_{n}→ Lu weakly. Therefore, L
(------)
B (0,r)
is weakly compact because it is
the continuous image of a compact set. Therefore, it must also be weakly closed because
the weak topology is a Hausdorff space. (See Lemma 16.3.2 on Page 1436, and so you can
apply the separation theorem, Theorem 16.2.5 on Page 1408 to obtain a separating
functional. Thus if x≠y, there exists f ∈ H^{′} such that Ref
(y)
> c >Ref
(x)
and so
taking
2r < min(c− Re f (x),Re f (y)− c),
Bf (x,r)∩ Bf (y,r) = ∅
where
Bf (x,r) ≡ {y ∈ H : |f (x− y)| < r}
is an example of a basic open set in the weak topology.)
Now suppose p
∕∈
L
( )
B-(0,r)-
. Since the set is weakly closed and convex, it follows by
Theorem 16.2.5 and the Riesz representation theorem for Hilbert space that there exists
z ∈ H such that
Re (p,z) > c > Re (Lx,z)
for all x ∈B
(0,r)
. Therefore, p cannot be a strong limit point because if it
were, there would exist x_{n}∈B
(0,r)
such that Lx_{n}→ p which would require
Re
(Lxn,z)
→Re
(p,z)
which is prevented by the above inequality. This proves the
lemma.
Now here is a very interesting result about showing that T_{1}
(U1)
= T_{2}
(U2)
where
U_{i} is a Hilbert space and T_{i}∈ℒ
(Ui,H)
. The situation is as indicated in the
diagram.
H
T1 ↗ ↖ T2
U1 U2
The question is whether T_{1}U_{1} = T_{2}U_{2}.
Theorem 17.2.5Let U_{i},i = 1,2 and H be Hilbert spaces and let T_{i}∈ℒ
(Ui,H )
. Ifthere exists c ≥ 0 such that for all x ∈ H
∗ ∗
||T1x||1 ≤ c||T2x||2 (17.2.12)
(17.2.12)
then
(------) (------)
T1 B (0,1) ⊆ T2 B (0,c) (17.2.13)
(17.2.13)
and so T_{1}
(U1)
⊆ T_{2}
(U2 )
. If
∗
||T1x||
_{1} =
∗
||T x||
_{2}for all x ∈ H, then T_{1}
(U1)
= T_{2}
(U2)
and in addition to this,
|| −1 || || −1 ||
||T1 x||1 = ||T2 x||2 (17.2.14)
(17.2.14)
for all x ∈ T_{1}
(U1)
= T_{2}
(U2 )
. In this theorem, T_{i}^{−1}refers to Definition 17.2.1.
Proof: Consider the first claim. If it is not so, then there exists u_{0},
||u0||
_{1}≤ 1
but
(------)
T1(u0) ∕∈ T2 B (0,c)
the latter set being a closed convex nonempty set thanks to Lemma 17.2.4. Then by the
separation theorem, Theorem 16.2.5 there exists z ∈ H such that
Re (T (u ),z) > 1 > Re(T (v),z)
1 0 H 2 H
for all
||v||
_{2}≤ c. Therefore, replacing v with vθ where θ is a suitable complex number
having modulus 1, it follows
| |
||T∗1z|| > 1 > |(v,T∗2z)U2| (17.2.15)
(17.2.15)
for all
||v||
_{2}≤ c. If c = 0, 17.2.15 gives a contradiction immediately because of 17.2.12.
Assume then that c > 0. From 17.2.15, if
| | 1 1
||T2∗z||U2 = sup |(v,T∗2z)U2| ≤ c < c ||T∗1z||
||v||≤1
which contradicts 17.2.12. Therefore, it is clear that T_{1}
(U1 )
⊆ T_{2}
(U2)
.
Now consider the second claim. The first part shows T_{1}
(U1 )
= T_{2}
(U2)
. Denote by
u_{i}∈ U_{i}, the point T_{i}^{−1}x. Without loss of generality, it can be assumed x≠0 because if
x = 0, then the definition of T_{i}^{−1} gives T_{i}^{−1}
(x )
= 0. Thus for x≠0 neither u_{i} can equal 0.
I need to verify that