Cantor’s Nested Set Theorem
Definition
Functional Analysis
Suppose that (X , d) is a metric space and A is a subset of X . Then
the diameter diam A of A is defined by
Lecture 14: The Theorems by Cantor, Baire, and Banach and
Steinhaus
diam A = sup d(x, y ).
x,y ∈A
Bengt Ove Turesson
Theorem (Cantor’s Nested Set Theorem)
Department of Mathematics
Linköping University
Suppose that (X , d) is a complete metric space and that F1 , F2 , . . .
is sequence of nonempty subsets of X such that
November 5, 2015
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Cantor’s Nested Set Theorem
(i) F1 ⊃ F2 ⊃ . . .;
(ii) Fn is closed for every n ≥ 1;
(iii) diam Fn → 0 as n → ∞.
T
Then the intersection ∞
n=1 Fn contains exactly one element.
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Baire’s Theorem
Theorem
Proof.
If (X , d) is a complete metric space, then the intersection of any
countable collection of open dense subsets of X is dense in X .
Existence
For n = 1, 2, . . . , choose an element xn in Fn .
If m > n, then d(xm , xn ) ≤ diam Fn .
Since diam Fn → 0, this shows that
sequence.
Remark
(xn )∞
n=1
For the case X = Rd , the theorem was proved by René-Louis Baire
in 1894.
is a Cauchy
Because X is complete, this means that xn → x for some
x ∈ X.
Example
Using the facts that xm ∈ Fn for m ≥ n and that every set Fn
is closed,
T∞ it follows that x ∈ Fn for every n ≥ 1, i.e.,
x ∈ n=1 Fn .
Let (xk )∞
k=1 be an enumeration of Q.
For n = 1, 2, . . . , let Fn be the union of all sets of the form
(xk − k −n , xk + k −n ), where k = 1, 2, . . . .
Then every set Fn is open and T
dense in R.
According to Baire’s theorem, ∞
n=1 Fn is dense in R.
Uniqueness
T∞
If x, y ∈ n=1 Fn , then d(x, y ) ≤ diam Fn for every n ≥ 1,
which implies that d(x, y ) = 0, that is, x = y .
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Baire’s Theorem
The Uniform Boundedness Principle
Proof.
Definition
Let {Vn : n = 1, 2, . . .} be any countable collection of open,
dense subsets of X .
Suppose that X and Y are two normed spaces. A sequence
(Tn )∞
n=1 of linear operators from X to Y is said to be pointwise
bounded if
sup kTn xk < ∞ for every x ∈ X .
Let x1 be T
an arbitrary element of X . It suffices to show that
Br1 (x1 ) ∩ ∞
n=1 Vn 6= ∅ for any radius r1 ≤ 1.
n≥1
Since Br1 (x1 ) ∩ V1 is open and nonempty, there exists a ball
Br2 (x2 ) such that Br2 (x2 ) ⊂ Br1 (x1 ) ∩ V1 and r2 ≤ 1/2.
Theorem (The Uniform Boundedness Principle)
By induction, there exists a ball Brn+1 (xn+1 ) such that
Brn+1 (xn+1 ) ⊂ Brn (xn ) ∩ Vn and rn+1 ≤ 1/(n + 1) for any
number n ≥ 1.
T
Cantor’s theorem shows that ∞
n=1 Brn (xn ) contains an
element x.
T
It follows by the construction that x ∈ Br1 (x1 ) ∩ ∞
n=1 Vn .
Suppose that X is a Banach space and Y is a normed space. Then
any pointwise bounded sequence (Tn )∞
n=1 ⊂ B(X , Y ) is bounded,
that is,
sup kTn k < ∞.
n≥1
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The Uniform Boundedness Principle
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The Banach–Steinhaus Theorem
Proof.
S
By the assumption, X = ∞
m=1 Xm , where
Xm = {x ∈ X : kTn xk ≤ m for every n}.
Definition
Since X is complete, it follows from Baire’s theorem that
some set Xm has nonempty interior.
Theorem (The Banach–Steinhaus Theorem)
Suppose that X and Y are two normed spaces. A sequence
(Tn )∞
n=1 of linear operators from X to Y is said to be converge
pointwise to T ∈ L(X , Y ) if Tn x → Tx for every x ∈ X .
Notice that Xm is closed since the norm in X is continuous
and every operator Tn is continuous.
Suppose that X is a Banach space, Y is a normed space, and
(Tn )∞
n=1 is a sequence of bounded linear operators from X to Y . If
Tn converges pointwise to T ∈ L(X , Y ), then
Thus, there exists a vector x0 ∈ Xm and a number r > 0 such
that x ∈ Xm if kx − x0 k ≤ r .
If ky k ≤ 1, then x = x0 + ry ∈ Xm , so
x − x 2m
0 kTn y k = T
≤
r
r
(i) supn≥1 kTn k < ∞;
(ii) T ∈ B(X , Y );
for every n.
(iii) kT k ≤ lim inf n→∞ kTn k.
This shows that kTn k ≤ 2m/r for every n.
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The Banach–Steinhaus Theorem
The Banach–Steinhaus Theorem
Proof.
(i) For any x ∈ X , the sequence (Tn x)∞
n=1 is convergent, so
(Tn x)∞
n=1 is bounded. The conclusion therefore follows from the
Uniform Boundedness Principle.
(ii) Put C = supn≥1 kTn k < ∞. Since kTn xk ≤ C kxk for every
x ∈ X and every n ≥ 1, it follows from the continuity of the
norm in X that kTxk ≤ C kxk for every x ∈ X , and hence that
T ∈ B(X , Y ).
(iii) We have that
Remark
The two examples below show that the inequality in (iii) may
be strict.
Notice that if Tn → T in B(X , Y ), which is a stronger
assumption that in the theorem, then kT k = limn→∞ kTn k.
Indeed,
kT k − kTn k ≤ kT − Tn k → 0 as n → ∞.
kTxk = lim kTn xk = lim inf kTn xk ≤ (lim inf kTn k)kxk
n→∞
n→∞
n→∞
for every x ∈ X , from which it follows that
kT k ≤ lim inf kTn k.
n→∞
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The Banach–Steinhaus Theorem
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The Banach–Steinhaus Theorem
Example
R1
Put Tn x = 0 t n x(t) dt for x ∈ L1 (0, 1), n = 1, 2, . . . .
R1
Then |Tn x| ≤ 0 t n |x(t)| dt ≤ kxk1 for every x ∈ L1 (0, 1), so
kTn k ≤ 1.
If xm (t) = m for 1 − m1 ≤ t ≤ 1 and xm (t) = 0 for 0 ≤ t < 1 −
where m = 1, 2, . . ., then kxm k1 = 1 and
Example
Put Tn x = xn for x ∈ c0 , n = 1, 2, . . . .
Then |Tn x| = |xn | ≤ kxk∞ for every x ∈ c0 , so kTn k ≤ 1.
If xj = 1 for j = n and xj = 0 for j 6= n, then
|Tn x| = 1 = kxk∞ .
Z
1
m,
m 1 n+1 1− 1−
1
n+1
m
1− m
1 1
m n+1
=
1− 1−
+O
=
1
+
O
.
n+1
m
m2
m
kTn k ≥ |Tn xm | = m
Hence, kTn k = 1 for every n.
Since Tn x = xn → 0 as n → ∞ for every x ∈ c0 , Tn
converges pointwise to T = 0.
1
t n dt =
Since this inequality holds for any m ≥ 1, this shows that kTn k ≥ 1.
Hence, kTn k = 1 for every n.
R1
By the Dominated Convergence Theorem, Tn x = 0 t n x(t) dt → 0 for
every x ∈ L1 (0, 1), so Tn converges pointwise to T = 0.
Therefore, kT k < lim inf n→∞ kTn k.
Therefore, kT k < lim inf n→∞ kTn k.
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An Application to Fourier Series
An Application to Fourier Series
Example (end)
Example (beginning)
Define TN : C2π (R) → C by TN f = SN f (0), N = 1, 2, . . . .
Let C2π (R) denote the subspace of C (R) consisting of functions with
period 2π.
We will show that there exists a function f ∈ C2π (R) whose Fourier
series is divergent at t = 0.
To prove that there exists a function f such that the Fourier series
of f is divergent at t = 0, it suffices according to the
Banach–Steinhaus theorem to show that supn≥1 kTN k = ∞.
For f ∈ C2π (R), the Nth partial sum to the Fourier series of f is
Notice that kTN k =
SN f (t) =
N
X
n=−N
1
fb(n)e int =
2π
Z
Now,
π
−π
DN (t − s)f (s) ds,
N = 1, 2, . . . ,
kDN k1 ≥ 4
where DN is the Dirichlet kernel:
DN (s) =
N
X
n=−N
e
ins
1
2π kDN k1 .
1
2 )s
sin(N +
=
,
sin(s/2)
Z π
Z (N+ 1 )π
2
1 ds
du
)s
=
4
| sin u|
−→ ∞
+
sin(N
2
s
u
0
0
as N → ∞ because
s ∈ R.
R∞
0
| sin u|
u
du = ∞.
This shows that supn≥1 kTN k = ∞.
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An Application to Fourier Series
Remark
The result in this example was proved by Paul du Bois-Reymond in
1873. The proof is due to Andrey Kolmogorov.
In 1923, Kolmogorov proved that there exists a function in L1 (−π, π)
whose Fourier series diverges a.e. Kolmogorov showed in 1926 that
”almost everywhere” can be replaced by ”everywhere”.
In 1915, Nikolai Luzin conjectured that the Fourier series of a
function f ∈ L2 (−π, π) converges a.e.
This result was proved by Lennart Carleson as late as 1966.
Carleson’s result was generalized in 1968 by Richard A. Hunt to
Lp (−π, π), where 1 < p < ∞.
Jean-Pierre Kahane and Yitzhak Katznelson showed in 1966 that, for
any subset E of (−π, π) with zero measure, there exists a 2π-periodic,
continuous function on R whose Fourier series diverges on E .
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