Linear Operators
For reference purposes, we will collect a number of useful results regarding bounded and
unbounded linear operators.
Bounded Linear Operators
Suppose T is a bounded linear operator on a Hilbert space H. In this case we may suppose
that the domain of T, DÝTÞ, is all of H. For suppose it is not. Then let DÝTÞ CL denote the
closure of DÝTÞ, and extend T to the closure by continuity. That is, if x 5 DÝTÞ CL then there
exists a sequence áx n â Ð DÝTÞ such that x n ¸ x in H. Since T is continuous, Tx n converges
to some limit f 5 H and we can define Tx = f. It is not difficult to show that the value f
depends on x but does not depend on the sequence áx n â. If DÝTÞ CL is not all of H, then it is
a closed subspace of H and, as such, has an orthogonal complement in H. Then extend T
by zero to this orthogonal complement to obtain a linear operator which is now defined on
all of H. It is easy to show that this extended operator is still continuous and has the same
norm as T.
Recall that the linear operator T is said to be well defined on H if
Tv = f, and Tv = g iff f = g. This is equivalent to the statement that there is no nonzero
element f such that T0 = f. The linear operator T is said to be one to one on H if
Tv = f, and Tu = f iff u = v. This is equivalent to the statement that Tu = 0 iff u = 0, Ýonly
the zero element is mapped to zero).
Adjoint of a Bounded Linear Operator
For T a bounded linear operator on Hilbert space H and a fixed v in H, let
FÝuÞ = ÝTu, vÞ H -u 5 H.
Clearly this defines a bounded linear functional F on H and it follows then from the Riesz
theorem that there exists a unique element z F 5 H such that
ÝTu, vÞ H = Ýu, z F Þ H -u 5 H.
If the correspondence relating z F to v is denoted by z F = T D v, then
ÝTu, vÞ H = Ýu, T D vÞ H -u 5 H.
Then it is not hard to show T D is well defined on H with values in H and, moreover, T D is
linear and bounded with ||T D || = ||T||. For example, if T D v = f, and T D v = g then
ÝTu, vÞ H = Ýu, T D vÞ H = Ýu, fÞ H and
ÝTu, vÞ H = Ýu, T D vÞ H = Ýu, gÞ H
hence
Ýu, f ? gÞ H = 0 -u 5 DÝTÞ = H.
Then f = g and T D is well defined. The operator T D is called the adjoint of T and we have
seen it is a well defined and bounded linear operator given only that T is bounded. If, in
addition, T is onto, then the adjoint is one to one. To see this suppose T is onto and
T D v = 0 for some v in H. Then for arbitrary f in H, there is some u in H such that Tu = f and
Ýf, vÞ H = ÝTu, vÞ H = Ýu, T D vÞ H = 0;
i.e. veH which implies v is zero and so T D is one to one. Conversely, suppose T D is one to
1
one and f e Rng T. That is,
-u 5 H
0 = ÝTu, fÞ H = Ýu, T D fÞ H
D
D
But this implies T f e H, or T f = 0. But we assumed T D is one to one and it follows that
f = 0. Then f e Rng T implies f = 0 which is to say, T is onto.
A bounded linear operator T such that ÝTu, vÞ H = Ýu, TvÞ H for all u,v in H is said to be
self adjoint. We are going to be interested in differential operators which are self adjoint but
differential operators are typically not bounded from H into itself.
Unbounded Linear Operators
Consider the operator
TuÝxÞ = u v ÝxÞ,
for u 5 DÝTÞ = C 1 ß0, 1à Ð L 2 ß0, 1à = H.
Then DÝTÞ is dense in H but it is not all of H. Note that ád n ÝxÞâ = ásinÝn^xÞâ is a sequence
in DÝTÞ that is bounded in H but the sequence áTd n ÝxÞâ = áÝn^Þ cosÝn^xÞâ is an unbounded
sequence in H. Then T is densely defined and unbounded on H. Note that if we took H to be
the Hilbert space H 1 ß0, 1à then T is defined on all of H but T does not map H into H. Note
further that if we define
SuÝxÞ = u v ÝxÞ,
for u 5 DÝSÞ = H 1 ß0, 1à Ð L 2 ß0, 1à = H,
then DÝTÞ Ð DÝSÞ Ð H and Tu = Su for all u in DÝTÞ. Then T is a restriction of S and S is an
extension of T and both of these operators are densely defined in H. When dealing with
differential operators on a Hilbert space of functions, we will generally try to choose the
largest possible domain for the operator. However, as we shall see, this is not always easy
to do.
Adjoint of an Unbounded Linear Operator
Suppose T is an unbounded but densely defined linear operator on H. Then we define
y 5 DÝT D Þ Ð H with T D y = v to mean
ÝTu, yÞ H = Ýu, vÞ H
-u 5 DÝTÞ.
T D is well defined because T is densely defined. To see this, suppose T D y = v and T D y = w.
Then
ÝTu, yÞ H = Ýu, vÞ H = Ýu, wÞ H
-u 5 DÝTÞ.
-u 5 DÝTÞ, and since DÝTÞ is dense in H, this implies
But in this case
Ýu, v ? wÞ H = 0
D
is
a nonempty subspace of H (i.e. DÝT D Þ contains at
v = w. It is easy to show that DÝT Þ
D
least the zero element) and T is a linear operator from DÝT D Þ into H.
For example, consider
TuÝxÞ = u v ÝxÞ,
for
u 5 DÝTÞ = áu 5 C 1 ß0, 1à : uÝ0Þ = uÝ1Þ = 0â Ð L 2 ß0, 1à = H.
2
Then
1
1
v
v
ÝTu, vÞ H = X 0 u v ÝxÞ vÝxÞdx = uÝxÞ vÝxÞ| x=1
x=0 ? X 0 uÝxÞ v ÝxÞ dx = Ýu, ? v Þ H
Since this string of equalities is valid for any v in DÝTÞ = C 1 ß0, 1à, it follows that DÝT D Þ
contains DÝTÞ, and moreover, T D v = ?v v ÝxÞ. In fact, we can show that
with no boundary conditions .
DÝT D Þ = C 1 ß0, 1à
On the other hand, consider
TuÝxÞ = u v ÝxÞ,
u 5 DÝTÞ = áu, u v 5 L 2 ß0, 1à : uÝ0Þ = 0â.
for
Then y 5 DÝT D Þ Ð H with T D y = v means
ÝTu, yÞ H = Ýu, vÞ H
-u 5 DÝTÞ.
x
FÝxÞ = X v, so that F v = v and FÝ1Þ = 0.
Let
1
Then
1
1
v
Ýu, vÞ H = X 0 uÝxÞ F v ÝxÞdx = F u| x=1
x=0 ? X 0 u ÝxÞ FÝxÞdx = ?ÝTu, FÞ H
and
ÝTu, y + FÞ H = 0
Now choose
-u 5 DÝTÞ.
x
u 0 ÝxÞ = X ß yÝsÞ + FÝsÞà ds,
0
which implies
u 0 5 L 2 ß0, 1à,
u v0 = y + F 5 L 2 ß0, 1à, and u 0 Ý0Þ = 0;
i.e., u 0 5 DÝTÞ. Moreover, then
which leads to
ÝTu 0 , y + FÞ H = ||y + F|| 2H = 0,
x
yÝxÞ = ?FÝxÞ = ? X v,
1
or
? y v ÝxÞ = vÝxÞ = T D y,
y, y v 5 L 2 ß0, 1à,
yÝ1Þ = 0.
We can deduce from this that T D y = ?y v ÝxÞ, DÝT D Þ = áy, y v 5 L 2 ß0, 1à,
yÝ1Þ = 0â
A densely defined linear operator T on H is said to be symmetric if
ÝTu, vÞ H = Ýu, TvÞ H
-u, v 5 DÝTÞ.
In this case DÝTÞ Ð DÝT D Þ and T D u = Tu -u 5 DÝTÞ. Then T D is an extension of T. If it is
the case that T is symmetric and, in addition, DÝTÞ Ð DÝT D Þ, then we say T is self-adjoint.
3
For any bounded, symmetric operator T, DÝTÞ = DÝT D Þ = H so the operator is then self
adjoint; i.e., for bounded operators symmetry and self adjointness are equivalent. For
unbounded operators, they are in general different. For example
T 1 uÝxÞ = u”ÝxÞ with DÝT 1 Þ = áu 5 C 2 ß0, 1à : uÝ0Þ = uÝ1Þ = 0â
can be easily shown to be self adjoint, in contrast to
T 2 uÝxÞ = u”ÝxÞ with DÝT 2 Þ = áu 5 C 2 ß0, 1à : uÝ0Þ = uÝ1Þ = 0, u v Ý0Þ = u v Ý1Þ = 0â
which is symmetric but not selfadjoint. Similarly,
T 3 uÝxÞ = i u v ÝxÞ with DÝT 3 Þ = áu 5 C 1 ß0, 1à : uÝ0Þ = uÝ1Þ = 0â
is symmetric but not selfadjoint while
T 4 uÝxÞ = i u v ÝxÞ with DÝT 4 Þ = áu 5 C 1 ß0, 1à : uÝ0Þ = 0â
is neither symmetric nor selfadjoint
Closed Linear Operators
Recall that a linear operator T on H is said to be bounded if there exists a constant C > 0
such that ||Tx|| H ² C ||x|| H for all x in H. T is said to be continuous if
x n ¸ x in H implies Tx n ¸ Tx in H. For linear operators, these two notions are equivalent. A
densely defined linear operator T is said to be closed if
x n 5 DÝTÞ
x n ¸ x in H
Tx n ¸ f in H
implies
x 5 DÝTÞ
Tx = f
Note that T continuous implies T closed but not conversely. Differential operators are not
continuous but are closed, (if the domain is properly chosen). We have seen that if T is
linear and densely defined, then the adjoint T D is well defined. In addition, T D is closed,
whether or not T itself is closed. To see this, suppose
x n 5 DÝT D Þ, x n ¸ x in H and T D x n ¸ f in H. Then for any z 5 DÝTÞ,
ÝTz, x n Þ H = Ýz, T D x n Þ H ,
ÝTz, x n Þ H ¸ ÝTz, xÞ H , and
Ýz, T D x n Þ H ¸ Ýz, fÞ H ,
which is to say, ÝTz, xÞ H = Ýz, fÞ H -z 5 DÝTÞ. But this implies x 5 DÝT D Þ and T D x = f, so T D is
closed. Note that it was not necessary to suppose that T was closed.
We have seen in previous examples that it may be possible to specify the domain of a
differential operator in several different ways. For example, TuÝxÞ = u v ÝxÞ could have any of
the following subspaces of L 2 ß0, 1à as its domain,
4
D 1 = C K ß0, 1à Ð D 2 = C 1 ß0, 1à Ð D 3 = H 1 ß0, 1à.
We would like to choose the domain as large as possible, and moreover to have T on this
domain be closed. To see how this might be done, recall that if T is linear and densely
defined then the adjoint T D is linear and well defined. Now if T D is also densely defined, then
the adjoint of T D , denoted T DD , is also well defined and it is closed since it is an adjoint.
More precisely, z 5 DÝT DD Þ with T DD z = w means
Ýz, T D yÞ H = Ýw, yÞ H = ÝT DD z, yÞ H
-y 5 DÝT D Þ.
But, for any y 5 DÝT D Þ, Ýz, T D yÞ H = ÝTz, yÞ H -z 5 DÝTÞ. This implies that
DÝTÞ Ð DÝT DD Þ and T DD z = Tz -z 5 DÝTÞ; i.e., T DD is an extension of T, and it is closed.
Thus if the densely defined operator T is not closed but its adjoint T D is densely defined,
then T has a closed extension which is just the adjoint of T D . The domain of this extension
must contain the limit points of all sequences áu n â in DÝTÞ for which the sequence áTu n â
also converges. This is not the same thing as the closure of DÝTÞ as the following theorem
shows.
Theorem (Closed Graph Theorem) A closed operator on a closed domain is necessarily
bounded.
The theorem shows that if DÝT DD Þ were equal to DÝTÞ CL then the closed operator T DD on the
closed domain DÝTÞ CL would be necessarily bounded. But this would contradict the fact that
T DD z = Tz -z 5 DÝTÞ and T is not bounded. Then, what we have instead is the following,
DÝTÞ Ð DÝT DD Þ Ð DÝTÞ CL = H,
where T is extended to the larger domain DÝT DD Þ which is also dense in H but is not closed.
In summary, we can say that if T is a densely defined linear operator on H then
a) if T D is also densely defined then T DD is well defined
and T DD is a closed extension of T
b) if T is closed then T D is densely defined and T = T DD
Solvability of Linear Equations
Let A denote a densely defined linear operator on H, and let
RÝAÞ = áy 5 H : y = Ax for some x 5 DÝAÞâ
NÝAÞ = áx 5 DÝAÞ : Ax = 0â.
Since A is densely defined, A D is well defined and we will let
RÝA D Þ = áy 5 H : y = A D x for some x 5 DÝA D Þâ
5
NÝA D Þ = áx 5 DÝA D Þ : A D x = 0â
Then we have the following facts relating these subspaces:
1. RÝAÞ e = NÝA D Þ
2. ÝRÝAÞ e Þ e = RÝAÞ CL = NÝA D Þ e
To see that 1 is true, suppose first that z 5 RÝAÞ e . Then ÝAu, zÞ H = 0 -u 5 DÝAÞ. Then
z 5 DÝA D Þ, with A D z = 0 since
0 = ÝAu, zÞ H = Ýu, A D zÞ H -u 5 DÝAÞ.
This proves that RÝAÞ e Ð NÝA D Þ. now suppose z 5 NÝA D Þ, i.e., Ýx, A D zÞ H = 0 -x 5 H. In
particular, if x 5 DÝAÞ then
0 = Ýx, A D zÞ H = ÝAx, zÞ H
-x 5 DÝAÞ.
But this implies z e RÝAÞ and we have proved that NÝA D Þ Ð RÝAÞ e . The two inclusions
together imply 1. The proof of 2 is left as an exercise. Note that 2 implies that if A has
closed range, then the equation Au = f is solvable if and only if f e NÝA D Þ. In fact, we can
show that
RÝAÞ = RÝAÞ CL
if and only if
RÝA D Þ = RÝA D Þ CL
and if this is the case, then
RÝAÞ = NÝA D Þ e
and
RÝA D Þ = NÝAÞ e
Lemma 1 Suppose A is closed and densely defined, and for some C > 0,
|| Au|| H ³ C || u|| H
-u 5 DÝAÞ.
Then A has closed range.
Proof- It is an exercise to show that the null space of any closed linear operator is a closed
subspace of H. Then for A a closed linear operator, we have H = NÝAÞ ã NÝAÞ e . Now let
áv n â Ð DÝAÞ be such that Av n = f n ¸ f in H; i.e., f is a limit point of RÝAÞ.
Now v n can be written uniquely in the form
vn = un + wn
Then for each n,
u n 5 NÝAÞ e
and
w n 5 NÝAÞ.
Av n = Au n
6
and
|| f n ? f m || H = || Au n ? Au m || H ³ C|| u n ? u m || H .
But then áu n â is a Cauchy sequence in H with limit point u 5 H, and since A is closed we
have
Au n ¸ f
un ¸ u
implies
u 5 DÝAÞ
Au = f
.
Then f 5 RÝAÞ and it follows that A has closed range.n
Corollary- Suppose A is closed and densely defined, and for some C > 0,
Ý Au, uÞ H ³ C || u|| 2H
-u 5 DÝAÞ.
Then A has closed range.
An operator A with the property of the corollary is said to be accretive (and -A is said to be
dissipative).
Lemma 2- Suppose A is densely defined and self adjoint, and for some C > 0,
Ý Au, uÞ H ³ C || u|| 2H
-u 5 DÝAÞ V NÝAÞ e .
Then precisely one of the following alternatives must hold:
a) Au = f has a unique solution for every f 5 H
b) Au = f has no solution unless f e NÝA D Þ.
If f e NÝA D Þ then solutions exist but are not unique.
It is evident why this is true. The hypotheses imply that RÝAÞ is closed, hence
RÝAÞ = NÝA D Þ e . Then RÝAÞ is either all of H or else the orthogonal complement of RÝAÞ is
just NÝA D Þ. In addition, since A is self adjoint, NÝAÞ = NÝA D Þ so the following statements are
all equivalent, NÝA D Þ = á0â, RÝAÞ = H, NÝAÞ = á0â.
Invertibility of A
For an unbounded operator A, we define the inverse operator, A ?1 , by
f 5 DÝA ?1 Þ with A ?1 f = u
if and only if
u 5 DÝAÞ with Au = f.
7
Then for u, v 5 DÝAÞ,
A ?1 f = u
A ?1 f = v
é
Au = f
Av = f
ì AÝu ? vÞ = 0
from which it is clear that A ?1 is well defined if and only if A is one to one.
In particular, if A is densely defined and if there exists C > 0 such that
-u 5 DÝAÞ.
Ý Au, uÞ H ³ C || u|| 2H
then A ?1 exists. In addition, in this case A ?1 is necessarily bounded. To see this note that if
A ?1 were not bounded then we could find a sequence áu n â Ð DÝAÞ with
|| u n || H = 1 and || Au n || H ¸ 0. This would contradict the accretiveness assumption.
If A is one to one and closed, then A ?1 is well defined and is also closed. To see this,
suppose áf n â Ð RÝAÞ is such that f n ¸ f, and A ?1 f n ¸ u. To show A ?1 is closed we have to
show f 5 RÝAÞ and A ?1 f = u. Then let u n = A ?1 f n and note that since A is closed, we have
that
u n ¸ u,
Au n ¸ f
implies
u 5 DÝAÞ
Au = f
But this means f 5 RÝAÞ and A ?1 f = u and we are done.
If A is one to one and closed, then RÝAÞ is closed if and only if A ?1 is bounded. To see
why this is true suppose first that A ?1 is bounded and let áf n â a sequence in RÝAÞ with f n ¸ f.
Since A ?1 is bounded, A ?1 f n = u n is a Cauchy sequence in H with limit u. Then, since A is
closed,
u n ¸ u,
Au n = f n ¸ f
implies
u 5 DÝAÞ
Au = f
Then f 5 RÝAÞ and it follows that RÝAÞ is closed. Conversely, if RÝAÞ is closed then it follows
from the closed graph theorem that A ?1 is bounded (since A ?1 is known to be closed).
If A is closed and densely defined on H then
and
RÝAÞ = H iff ÝA D Þ ?1 is bounded
RÝA D Þ = H iff ÝAÞ ?1 is bounded.
To see this, suppose first RÝAÞ = H. Since ÝAx, yÞ H = Ýx, A D yÞ H -x 5 DÝAÞ, we conclude
that A D y = 0 iff y = 0. Then ÝA D Þ ?1 is well defined, and since A D is closed, ÝA D Þ ?1 is
closed. But RÝAÞ = H is closed so RÝA D Þ is also closed, and then, by the closed graph
theorem, ÝA D Þ ?1 is bounded. Conversely, suppose ÝA D Þ ?1 is bounded.Then NÝA D Þ = á0â
and it follows that RÝA D Þ is closed. But then RÝAÞ is closed and equal to NÝA D Þ e = H.
Spectral Theory for Closed Operators
Let A be closed and densely defined on H. Then for every complex number V, VI ? A is
also closed and has the same domain as A. We define the following terms
8
V belongs to the resolvent set for A if ÝVI ? AÞ ?1 exists and is bounded with
RÝVI ? AÞ = H
V belongs to the point spectrum for A if ÝVI ? AÞ ?1 fails to exist;
i.e., 0u ® 0 such that ÝVI ? AÞu = H = 0. Then V is an eigenvalue
for A and x 5 NÝVI ? AÞ is an eigenvector for A
with
V belongs to the continuous spectrum for A if ÝVI ? AÞ ?1 exists but is not bounded
RÝVI ? AÞ ® H but RÝVI ? AÞ CL = H
V belongs to the residual or compression spectrum for A if RÝVI ? AÞ CL ® H
We collect here several facts about the resolvent set and spectrum of A.
If A is closed, densely defined and symmetric then essentially the same proofs that
work in linear algebra for symmetric matrices can be used to show that
ÝAu, uÞ H is real for all u in DÝAÞ
all eigenvalues of A are real
eigenvectors corresponding to distinct eigenvalues are orthogonal;
i.e., V ® W implies NÝVI ? AÞ e NÝWI ? AÞ.
If, in addition, A is self adjoint on H, then the resolvent set of A contains the complement of
the real axis. Moreover, for ImÝVÞ ® 0, ÝVI ? AÞ ?1 is bounded and satisfies
ÝVI ? AÞ ?1
²
LÝHÞ
1
|ImÝVÞ|
and
ImÝÝA ? VIÞx, xÞ H = ImÝVÞ || x|| 2H
-x 5 DÝAÞ.
This can be seen by writing
||ÝA ? VIÞx|| H ||x|| H ³ |ÝÝA ? VIÞx, xÞ H |
³ ImÝÝA ? VIÞx, xÞ H ³ | ImÝVÞ| || x|| 2H
-x 5 DÝAÞ.
Here we used that ÝAx, xÞ H is real for all x in DÝAÞ. Then ÝVI ? AÞ ?1 exists for ImÝVÞ ® 0. In
addition, RÝVI ? AÞ must be dense in H if ImÝVÞ ® 0. For suppose not. Then there exists a
y ® 0 which is orthogonal to RÝVI ? AÞ. But in that case we would have
ÝÝA ? VIÞx, yÞ H = 0
-x 5 DÝAÞ
and since A is self adjoint, this is the same as
9
Ýx, ÝA ? V# IÞyÞ H = 0
-x 5 DÝAÞ.
But DÝAÞ is dense in H so this implies ÝA ? V# IÞy = 0, which is to say Ay = V# y. Since ÝAy, yÞ H
is real for all y there can be no such y.
If A is a closed, densely defined linear operator on H, then the resolvent set is an open
subset of the complex plane. In each component of the resolvent set, ÝVI ? AÞ ?1 is an
analytic function of V with values in LÝHÞ. This can be seen by noting that for V 0 in the
resolvent set of A, ÝV 0 I ? AÞ ?1 is a bounded linear operator on H with domain equal to
RÝVI ? AÞ = H. Fix V 0 in the resolvent set and define
K
SÝVÞ = ÝV 0 I ? AÞ ?1 I + >ÝV 0 ? VÞ n ÝV 0 I ? AÞ ?1
n
.
n=1
This series converges in the norm of LÝHÞ provided V lies in the disc |ÝV 0 ? VÞ|
ÝV 0 I ? AÞ ?1 < 1. Then the series defines an analytic function of V in this disc. Now it is an
exercise to show that
ÝVI ? AÞ SÝVÞ = ßÝV ? V 0 ÞI + ÝV 0 I ? AÞà SÝVÞ = I
SÝVÞ ÝVI ? AÞ = I.
Then SÝVÞ = ÝVI ? AÞ ?1 and SÝVÞ is analytic in a disc.
Note that if both V, W belong to the resolvent set for A then the following analogue of a
partial fractions decomposition is valid.
ÝVI ? AÞ ?1 = ÝVI ? AÞ ?1 ÝWI ? AÞÝWI ? AÞ ?1
= ÝVI ? AÞ ?1 ßÝW ? VÞI + ÝVI ? AÞàÝWI ? AÞ ?1
and
i.e.,
= ÝW ? VÞÝVI ? AÞ ?1 ÝWI ? AÞ ?1 + ÝWI ? AÞ ?1 ,
ÝVI ? AÞ ?1 ? ÝWI ? AÞ ?1 = ÝW ? VÞÝVI ? AÞ ?1 ÝWI ? AÞ ?1 ;
ÝVI ? AÞ ?1 ÝWI ? AÞ ?1 =
1
ÝVI ? AÞ ?1 ? ÝWI ? AÞ ?1 .
W?V
.
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