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泛函分析讲义（下册）第七章习题解答.pdf
1
Infinitesimal Generators
1.1 Let fT (t) : t 0g is a semi-group of bounded operators in Banach space X , i.e., it satisfies that T (t)T (s) =
T (t + s) for all s; t > 0 and T (0) = I. Let f (t) = ln∥T (t)∥. Suppose that f (t) is bounded on [0; a], show that
(1) f (t) is sub-additive, i.e., f (t + s) f (t) + f (s) for all t; s > 0.
(2) limt!1 1
Proof.
(1) f (t + s) = ln∥T (t + s)∥ = ln∥T (t)T (s)∥ ln(∥T (t)∥∥T (s)∥) = ln∥T (t)∥ + ln∥T (s)∥ = f (t) +
t f (t) = inft>0
f (s);
(2) It is not difficult to see that f (t) is bounded on any finite interval [0; s]. Suppose that f (t) is bounded by Ms
on [0; s]. Fix s. Any t can be written as t = ns + r, where n is an integer and 0 r < a. en we have from
subadditivity of f that
t f (t).
1
f (t) nf (s) + f (r) nf (s) + Ms:
Divide it by t,
Let t ! 1,
f (t)
t
n
ns + r
f (s) +
Ms
t
f (s)
s
+
Ms
t
Take infimum on the right-hand side,
Notice that it holds trivially
e proof is now complete.
limsup
t!1
f (t)
t
f (s)
s
:
limsup
t!1
f (t)
t
inf
s>0
f (s)
s
:
liminf
t!1
f (t)
t
inf
s>0
f (s)
s
:
1.2 Let fT (t) : t 0g is a semigroup of bounded operators, such that T (0) = I and strong continuity at t = 0, i.e.,
s-limt!0+ T (t) = I. Show that the semi-group is strongly continuous.
Proof. We shall show that t 7! T (t)x is continuous for all x 2 X . It is easy to show right strong continuity.
∥T (t t0)x x∥ = 0:
∥T (t0)T (t t0)x x∥ ∥T (t0)∥ lim
t!t+
∥T (t)x T (t0)x∥ = lim
t!t+
lim
t!t+
0
0
0
To prove the left strong continuity, it suffices to show that ∥T (t)∥ to be uniformly bounded when t is near t0. In
fa, it holds that ∥T (t)∥ M e!t for some M and !. Refer to the text before Lemma 7.1.6.
1.3 Let fT (t) : t 0g is a semigroup of bounded operators on Hilbert space H and satisfies T (0) = I and weak
continuity at t = 0. Show that the semigroup is strongly continuous.
Proof. Since T (t)x ⇀ x, the uniform boundedness theorem tells us that ∥T (t)x∥ is uniformly bounded in a neigh-
bourhood of t = 0. Again by the uniform boundedness principle, it holds that ∥T (t)∥ is uniformly bounded near
t = 0. It is then easy to see that for a fixed x0 2 X , x(t) = T (t)x0 is bounded on any compa interval of t.
Suppose that 0 a < t < b < ϵ < , where ϵ > 0. Since x() = T ()x0 = T (t)T ( t)x0 = T (t)x( t),
we have that
∫
∫
(b a)x() =
b
b
T (t)x( t)d
x()dt =
a
a
1
and so, by supatb
∥T (t)∥ < 1, we obtain
(b a)∥x( ϵ) x()∥ =
∫
b
a
T (t)(x( ϵ t) x( t))dt
sup
atb
∥T (t)∥
∥x(t ϵ) x(t)∥dt
∫
a
b
e right hand side tends to zero as ϵ ! 0+, as may be seen by approximating x(t) by finite-valued funions.
So far we have proved that x(t) is strongly continuous at t > 0. Now we prove the strong continuity at t = 0. For
any positive rational number r we have T (t)x(r) = x(t + r), and thus s-limt!0+ T (t)x(r) = x(r). Let M denote
the set consisting of all finite linear combinations with rational coeffcients of x(r)'s, then s-limt!0+ T (t)x = x for
all x 2 M. On the other hand, for any t 2 [0; 1]
∥x(t) x0∥ ∥T (t)x x∥ + ∥x x0∥ + ∥T (t)(x0 x)∥ ∥T (t)x x∥ +
∥x0 x∥
(
sup
t2[0;1]
)
∥T (t)∥ + 1
(
)
and thus
∥x(t) x0∥
limsup
t!0+
(1)
for any x 2 M. It is clear that fx(t) : t 0g M from the weak closedness of M and weak right-continuity of
fx(t)g. erefore the right side of (1) can be made arbitrarily close to 0, concluding that x(t) ! x0 strongly as
t ! 0+.
sup
t2[0;1]
∥T (t)∥ + 1
∥x0 x∥
1.4 Let fT (t) : t 0g be a strongly continuous semi-group of operators on X and A its infinitesimal generator. Show
that the following three conditions are equivalent:
(1) D(A) = X ;
(2) limt!0+ ∥T (t) I∥ = 0;
(3) A 2 L(X ) and T (t) = exp(tA).
Proof. (3))(1): It follows easily from series manipulation that Atx ! Ax for all x as t ! 0.
(1))(2): Since Atx ! x as t ! 0, by uniform boundedness principle, At is uniformly bounded, say by M, in a
small neighbourhood of t = 0. en ∥T (t) I∥ M t ! 0 as t ! 0.
(2))(3): It is easy to verify that
∫
r
t
lim
s!0
1
s
1
t
∫
∫
s+r
T (t)dt = T (r)
< 1
In particular, there exists such that
for all 0 < t < , then 1
t
t
∫
r+s
r
∫
s
0 =
(because
∫
0
T (s)ds I
∫
∫
0 T (s)ds is invertible for t 2 (0; ). Now,
∫
∫
0 ). It follows that for t 2 (0; ),
T (t)dt 1
s
(
∫
T (t)dt =
1
s
1
s
r+s
0
s
r
r
r+s
s
(T (s) I) =
1
s
1
s
r
r+s
T (t)dt 1
s
∫
2
(T (s) I)
)(∫
0
s
(2)
r
T (t)dt
)1
r
T (t)dt
T (t)dt
0
0
∫
∫
r
0
e right-hand side tends to (T (r) I)(
1, so the left-hand side As converges to some bounded linear
operator when s ! 0. It is obvious that this limit operator must be A. Taking limit s ! 0 in (2), we obtain that
r
0 T (t)dt)
Iterated substitution gives
T (r) I = A
+ +
Let n ! 1, we see that T (r) = exp(rA) for all r 0.
T (r) = I + A +
A2
2
An
n!
+
T (s)ds
∫
An+1
n!
0
r
(r t)nT (t)dt
1.5 Let X = C0[0;1) = ff 2 C[0;1) : limx!+1 f (x) = 0g, ∥f∥ = supjf (s)j. Define on X a linear operator
T (t) : a() 7! a(t + ):
Show that fT (t) : t 0g is a strongly contraion semigroup on X .
Proof. It is obvious that T (t + s) = T (t) + T (s) and T (0) = I. Now we show that ∥T (t)a T (t0)a∥ ! 0. is
is because ∥T (t)a T (t0)a∥ = sups
ja(t + s) a(t0 + s)j and a is uniformly continuous. Finally, it is obvious that
∥T (t)∥ 1 for all t 0.
∫ 1
1.6 Let X = L2(R), for x 2 R and y 2 R+, define
((T (y)f )(x) =
1
y
1
(x )2 + y2 f ()d;
T (0)f = f:
y > 0;
Show that fT (y) : y 0g is a strongly continuous semigroup on X and ∥T (y)∥ = 1. (Remark. e integral gives
a harmonic funion on the upper plane with boundary value f)
Proof. First we show that T (y)f 2 L2(R). Indeed, by Cauchy-Schwarz inequality,
2
)(∫ 1
dx
1
∫ 1
∫ 1
(∫ 1
∫ 1
∫ 1
∫ 1
1
1
y
(x )2 + y2 f ()d
(x )2 + y2 d
y
1
1
1
jf ()j2dxd
(x )2 + y2
(
(
)
Hence ∥T (y)∥ 1. On the other hand,
1
1
y
1
2
1
=
=∥f∥2
(
2
T (y)[R;R]
(x) =
1
and thus
∥T (y)[R;R]∥2
2
[R;R]∥2
2
∫
∫
1
R
1
R
1
2
(
R/2
R/2
R/2
R/2
y
R x
(
arctan
(
arctan
R x
)2
y
dx
2
arctan R
2y
3
)
dx
y
(x )2 + y2
jf ()j2d
)
+ arctan
(
R + x
y
)
(
+ arctan
))
))2
R + x
y
dx
(
=
2
arctan R
2y
)2 ! 1
as R ! 1, which implies that ∥T (y)∥ = 1 for y > 0. It is trivial that T (0) = I and ∥T (0)∥ = 1.
When f 2 S (R), Notice that T (y)f is exaly u(; y) that satisfies
∆u = 0;
y > 0
u(x; 0) = f (x; 0)
It is then obvious that T (t + s) = T (t) + T (s) for f 2 S (R), which can be extended to the entire L2(R) easily
because S (R) is dense in L2(R) and ∥T (y)∥ 1.
Now we show ∥T (y)f f∥ ! 0 as y ! 0+. It suffices to show this for f 2 C
1
0 (R) and density of test funions
allows us to extend this result to L2(R). First we show that T (y)f ! f uniformly pointwise as y ! 0+. Let ϵ > 0
be given. Since f is uniformly continuous, there exists such that jf (x) f (y)j < ϵ whenever jx yj < . en
j(T (y)f )(x) f (x)j =
∫
(x )2 + y2 (f () f (x))d
y
yjf () f (x)j
(x )2 + y2 d +
jxj>
yjf () f (x)j
(x )2 + y2 d
where
and
I ϵ
∫
∫
R
jxj<
=: I + J;
∫
J 2∥f∥1
= 2∥f∥1
∫
jxjK
y
R
(x )2 + y2 d = ϵ
∫
)
(
jxj>
2arctan
y
(x )2 + y2 d
y
! 0
∫
∫
jxjK
j(T (y)f )(x)j2dx
uniformly w.r.t. x as y ! 0. Hence T (y)f ! f uniformly w.r.t. x. Suppose that supp f 2 [K; K]. Now,
∥T (y)f f∥2
j(T (y)f )(x) f (x)j2dx +
j(T (y)f )(x)j2dx
∥(T (y))f f∥1 2K +
∫
e first term goes to 0 as y ! 0+. For the second term, we have
∫
∫
jxjK
∫
j(T (y)f )(x)j2dx 1
(
jf ()j2
jjK
jxjK
y
(x )2 + y2 dxd
jxjK
arctan K
jf ()j2
arctan K +
! 0 as y ! 0+ by Dominated Convergence eorem:
jjK
y
y
)
d
erefore we conclude that (T (y) I)f ! 0 as y ! 0+, whence the strong continuity condition is satisfied by
Problem 1.2.
1.7 Let fT (t) : t 0g be a strongly continuous semi-group on X . Suppose that x 2 X, w-limt!0+
show that x 2 D(A) and y = Ax.
t (T (t)I)x = y,
1
4
Proof. Let f 2 X
to e
and > 0 large enough. It is easy to see that e
tf (T (t)(y x)). e derivative is continuous in t, we have on integration
tf (T (t)x) has right derivative, which equals
∫ 1
0
f (x) =
tf (T (t)(y x))dt;
e
for all f 2 X
. erefore,
∫ 1
tT (t)(y x) = (I A)
e conclusion follows immediately by multiplying (I A) on both sides.
x =
e
0
1(y x)
1.8 Let fT (t) : t 0g be a strongly continuous semi-group on a Hilbert space H . Suppose that A is its generator and
T (t) is a normal operator for all t 0. Show that A is normal using Gelfand transform.
1.9 Prove Hille-Yosida-Phillips eorem (eorem 7.1.7): A densely-defined closed linear operator A is an infinitesimal
generator of some strongly continuous semigroup fT (t) : t 0g if and only if
(1) 9!0 > 0 such that (!0;1) (A);
(2) 9M > 0 such that
∥( A)
n∥ M
( !)n ; n = 1; 2; : : :
whenever > ! > !0.
Proof. e necessity has been proved in Lemma 7.1.6. e proof of sufficiency follows the same outline as in
eorem 7.1.5 by defining
B = 2(!0 + A)
1 I
for > 0.
1.10 Let fT (t) : t 0g be a strongly continuous semi-group and A its infinitesimal generator. Suppose that !0 2 R
satisfies f : ℜ > !0g (A)g. Show that
(1) e set fR(A)x : x 2 D(A)g is dense in D(A), where ℜ > !0;
(2) e range of R(A)n is dense for all n 1, where ℜ > !0;
(3) D(An) is dense for all n 1.
(1) Let x 2 D(A2). It follows from R(A)( A)x = x that x 2 R(R(A)jD(A)). Hence D(A2)
Proof.
R(R(A)jD(A)). e conclusion follows immediately from the density of D(A2).
(2) It follows from R(A)( A)x = x (x 2 D(A)) that x 2 R(R(A)), that is, D(A) R(R(A)) and
R(R(A)) is therefore dense. Now, it R(A)n( A)nx = x for all x 2 D(An), whence it follows that
D(An) R(R(A)n). Part (3) shows that D(An) is dense, and hence R(R(A)n) is dense.
(3) is statement aually hold for any strongly continuous semi-group with no further assumptions. To see this,
let ϕ be any funion in C
1
0 [0; 1] and define for any x
∫
1
xϕ =
ϕ(t)T (t)xdt:
en
T (h) 1
h
xϕ =
1
h
∫
(∫
1
0
0
t
′
ϕ
0
)
(s)ds
5
(T (t + h) T (t))xdt
)
ds
1
(T (t + h)x T (t)x)dt
∫
1+h
T (t)xdt
s
T (t)xdt 1
h
s+h
∫
T (t)xdt
1
′
ϕ
(s)
0
s
)
ds
s+h
T (t)xdtds
∫
1+h
1
s+h
(s)ds
′
ϕ
(s)
T (t)xdtds
0
s
∫
∫
∫
=
=
1
h
1
h
=
1
h
= 1
h
(s)
(s)
′
′
′
1
0
1
0
1
∫
0
ϕ
ϕ
ϕ
1
(∫
(∫
∫
∫
s
1
∫
(s)T (s)xds:
1
0
′
ϕ
∫
∫
Since 1
h ! 0+ and equals to
s+h
s
h
T (t)xds ! T (s)x as h ! 0+ and ϕ
′ is bounded, the limit of the right-hand side exists as
Hence xϕ 2 D(A) and
∫
Now it is clear that xϕ 2 D(An) for all n. We can choose a sequence fϕjg C
0 ϕj = 1 and supp ϕj tends to 0. It is not difficult to see that xϕj
Axϕ =
(s)T (s)xds:
ϕ
′
1
1
0
0 [0; 1] such that ϕj 0,
1
! x. Hence D(An) is dense.
1.11 Let fT (t) : t 0g be a strongly continuous semi-group and A its infinitesimal generator. Suppose that f 2
C 1([0;1); X ). Show that the differential equation of operators
Proof. e first term T (t)x0 on the right of (5) satisfies the homogeneous differential equation and the initial
condition, it suffices to show that the second term satisfies 3 with initial value 0.
∫
t
T (t s)f (s)ds =
(3)
(4)
(5)
dx(t)
dt
= Ax(t) + f (t);
x(0) = x0 2 D(A)
+; X ), which is given by
∫
t
T (t s)f (s)ds:
)
∫
x(t) = T (t)x0 +
0
∫
(∫
0
t
=
0
f (0) +
t
T (t s)
)
T (t s)ds
∫
T (t) T (r) =
∫
t
r
A
r
T (t s)ds = T (t r) I;
∫
′
(r)(T (t r) I)dr
t
f
0
(
6
s
0
′
∫
f
(∫
ds
)
dr
t
T (t s)ds
(r)dr
t
′
f
(r)
f (0) +
0
r
t
AT (s)ds;
has a unique solution in C(R1
+; D(A)) \ C 1(R1
0
Note that
it follows that
and then
∫
A
0
t
T (t s)f (s)ds = (T (t) I)f (0) +
∫
0
t
′
(r)T (t r)dr
f
(6)
(7)
= T (t)f (0) f (t) +
∫
′
(t s)ds:
T (s)f
0
On the other hand,
d
dt
Comparing (6) and (7), we see that
∫
t
0
∫
d
dt
T (t s)f (s)ds = T (t)f (0) +
t
∫
t
T (t s)f (s)ds = A
t
T (t s)f (s)ds + f (t):
0
0
It is clear that the initial value of the second term of (5) is 0. Hence x(t) given in (5) is a solution to the differential
′. It is clear that
equation indeed. e continuity of x
x(t) 2 D(A).
In fa, if x(t) 2 C(R1
(t) can be easily concluded from (7) using the continuity of f
+; X ) is a solution to (3) and (4), then
+; D(A)) \ C 1(R1
′
T (t s)x(s) = T
′
(t s)x(s) + T (t s)x
′
d
ds
Integrate on both sides, we obtain that
(s) = T (t s)Ax(s) + T (t s)x
∫
x(t) T (t)x(0) =
t
T (t s)f (s)ds;
′
(s) = T (t s)f (s):
which is exaly (5). e uniqueness of the solution is proved.
0
∑jcnj2 < 1g, where D is the open disc in the complex
∑1
2 Examples of Infinitesimal Generators
n=0 cnzn;∥f∥2 =
2.1 Let X = ff : D ! C : f (z) =
1∑
plane. Define on X
(T (t)f )(z) =
(n + 1)
tcnzn:
n=0
n+1 (n 1) are eigenvalues of A.
Show that fT (t) : t 0g is a strongly continuous semi-group of positive self-adjoint operators. Find its infinitesimal
generator A and show that ln 1
Proof. It is obvious that ∥T (t)∥ 1, T (t + s) = T (t)T (s) and T (0) = I. Now we show that T (t)f ! f strongly
for all f 2 X . Given ϵ > 0 and f =
2 . Choose t
small enough such that 1 ( 1
N∑
∑1
n=0 cnzn, choose N big enough such that
(
2∥f∥. en
1
)2 jcnj2
∥f T (t)f∥2 =
(
1
N +1 )t < ϵp
jcnj2 < ϵ2
1∑
∑
n>N
1
1
n=N +1
(n + 1)t
∑
that is, ∥f T (t)f∥ ϵ when t is sufficiently small. erefore fT (t) : t 0g is a strongly continuous semi-group.
It is straightforward to verify that T (t) is positive and self-adjoint (note that the inner produ (
dnzn) =
cnzn;
∑
∑
cndn).
)2 jcnj2 +
1∑
jcnj2
(n + 1)t
∥f∥2 +
n=N +1
= ϵ2;
n=0
ϵ2
2∥f∥2
ϵ2
2
ϵ2
2
+
7
Define a linear operator A as
1∑
n=0
{
Af =
∑
cn ln
n + 1
1
∑
zn
}
jcnj2 ln2(n + 1) < 1
:
∑
D(A) =
f =
∑
It is clear that D(A) is dense because xn 2 D(A) for all n. We claim that A is closed. Suppose that fn ! f,
Afn ! g, fn =
dnzn. Since Afn ! g,
cnzn and g =
cnkzk, f =
on
or,
1
k=0
k + 1
cnzn 2 X :
∑
cnk ln
1∑
2
cnk dk
1∑
cnk dk
1∑
1∑
ln 1
k+1
k=1
k=1
ln 1
k+1
2 ! 0;
dk
ln2
1
! 0;
k + 1
2 ! 0:
jcnk ckj2 ! 0:
(because d0 must be 0) which implies that
Comparing with fn ! f, or, equivalently,
we obtain that
k=0
dn = cn ln
1
n + 1
for all n 0, that is, f 2 D(A) and g = Af. erefore A is a densely-defined closed operator.
Next we show that A generates a contraion semigroup. For > 0, it is easy to verify that I A is injeive, and
∥f Af∥ ∥f∥, thus I A is invertible and ∥R(A)∥
1. By Hille-Yosida eorem we know that A
generates a contraion semigroup.
Now, to show that A is the infinitesimal generator of fT (t)g, it suffices to show that Atf ! Af on D(A). Let
f 2 D(A), f =
cnzn, then
∑
∥Atf Af∥2 =
=
(n + 1)
et ln 1
∑
jcnj2
∑
jcnj2
∑
jcnj2 ln2
t
t
t 1
ln
n+1 1
t
1
n + 1
ln
! 0
2
1
n + 1
2
1
n + 1
(T (t)f )() =
G( ; t)f ()d;
t > 0
T (0)f = f;
∫
1
2
∑1
n=1 e
as t ! 0+, where we used ex 1 x + x2 for all x 1.
2.2 Let X = L2(; ). Define
where the integral kernel G(; t) = 1 + 2
semi-group. Is it a contraion semi-group?
n2t cos n. Show that fT (t) : t 0g is a strongly continuous
8
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