2Baire category theorem
II Linear Analysis
2.2 Some applications
We are going to have a few applications of the Baire category theorem.
Proposition. R \ Q 6= ∅, i.e. there is an irrational number.
Of course, we can also prove this directly by, say, showing that
√
2
is irrational,
or that noting that
R
is uncountable, but
Q
is not. However, we can also use
the Baire category theorem.
Proof.
Recall that we defined
R
to be the completion of
Q
. So we just have to
show that Q is not complete.
First, note that
Q
is countable. Also, for all
q ∈ Q
,
{q}
is closed and has
empty interior. Hence
Q =
[
q∈Q
{q}
is the countable union of nowhere dense sets. So it is not complete by the Baire
category theorem.
We will show that there are normed vector spaces which are not Banach
spaces.
Proposition. Let
ˆ
`
1
be a normed vector space defined by the vector space
V = {(x
1
, x
2
, ···) : x
i
∈ R, ∃I ∈ N such that i > I ⇒ x
i
= 0},
with componentwise addition and scalar multiplication. This is the space of all
sequences that are eventually zero.
We define the norm by
kxk
ˆ
`
1
=
∞
X
i=1
|x
i
|.
Then
ˆ
`
1
is not a Banach space.
Note that
ˆ
`
1
is not standard notation.
Proof. Let
E
n
= {x ∈
ˆ
`
1
: x
i
= 0, ∀i ≥ n}.
By definition,
ˆ
`
1
=
∞
[
n=1
E
n
.
We now show that
E
n
is nowhere dense. We first show that
E
n
is closed. If
x
j
→ x
in
ˆ
`
1
with
x
j
∈ E
n
, then since
x
j
is 0 from the
n
th component onwards,
x
is also 0 from the
n
th component onwards. So we must have
x ∈ E
n
. So
E
n
is
closed.
We now show that
E
n
has empty interior. We need to show that for all
x ∈ E
n
and
ε >
0, there is some
y ∈
ˆ
`
1
such that
ky − xk < ε
but
y 6∈ E
n
. This
is also easy. Given x = (x
1
, ··· , x
n−1
, 0, 0, ···), we consider
y = (x
1
, ··· , x
n−1
, ε/2, 0, 0, ···).
Then
ky − xk
ˆ
`
1
< ε
but
y 6∈ E
n
. Hence by the Baire category theorem,
ˆ
`
1
is not
complete.
Proposition. There exists an f ∈ C([0, 1]) which is nowhere differentiable.
Proof.
(sketch) We want to show that the set of all continuous functions which
are differentiable at at least one point is contained in a meagre subset of
C
([0
,
1]).
Then this set cannot be all of C([0, 1]) since C([0, 1]) is complete.
Let E
m,n
be the set of all f ∈ C([0, 1]) such that
(∃x)(∀y) 0 < |y − x| <
1
m
⇒ |f(y) − f (x)| < n|y − x|.
(where the quantifiers range over [0, 1]).
We now show that
{f ∈ C([0, 1]) : f is differentiable somewhere} ⊆
∞
[
n,m=1
E
m,n
.
This is easy from definition. Suppose
f
is differentiable at
x
0
. Then by definition,
lim
y→x
0
f(y) − f (x
0
)
y − x
0
= f
0
(x
0
).
Let
n ∈ N
be such that
|f
0
(
x
0
)
| < n
. Then by definition of the limit, there
is some
m
such that whenever 0
< |y − x| <
1
m
, we have
|f(y)−f (x)|
|y−x
0
|
< n
. So
f ∈ E
m,n
.
Finally, we need to show that each
E
m,n
is closed and has empty interior.
This is left as an exercise for the reader.
Theorem
(Banach-Steinhaus theorem/uniform boundedness principle)
.
Let
V
be a Banach space and
W
be a normed vector space. Suppose
T
α
is a collection
of bounded linear maps T
α
: V → W such that for each fixed v ∈ V ,
sup
α
kT
α
(v)k
W
< ∞.
Then
sup
α
kT
α
k
B(V,W )
< ∞.
This says that if the set of linear maps is pointwise bounded, then they are
uniformly bounded.
Proof. Let
E
n
= {v ∈ V : sup
α
kT
α
(v)k
W
≤ n}.
Then by our conditions,
V =
∞
[
n=1
E
n
.
We can write each E
n
as
E
n
=
\
α
{v ∈ V : kT
α
(v)k
W
≤ n}.
Since
T
α
is bounded and hence continuous, so
{v ∈ V
:
kT
α
(
v
)
k
W
≤ n}
is
the continuous preimage of a closed set, and is hence closed. So
E
n
, being the
intersection of closed sets, is closed.
By the Baire category theorem, there is some
n
such that
E
n
has non-empty
interior. In particular, (
∃n
)(
∃ε >
0)(
∃v
0
∈ V
) such that for all
v ∈ B
(
v
0
, ε
), we
have
sup
α
kT
α
(v)k
W
≤ n.
Now consider arbitrary kv
0
k
V
≤ 1. Then
v
0
+
ε
2
v
0
∈ B(v
0
, ε).
So
sup
α
T
α
v
0
+
εv
0
2
W
≤ n.
Therefore
sup
α
kT
α
v
0
k
W
≤
2
ε
n + sup
α
kT
α
v
0
k
.
Note that the right hand side is independent of v
0
. So
sup
kv
0
k≤1
sup
α
kT
α
v
0
k
W
≤ ∞.
Note that this result is not true for general functions. For example, consider
f : [0, 1] → R defined by
x
y
1
2
n−1
1
2
n
1
2
n+1
1
2
n+2
n
Then for all x ∈ [0, 1], we have
sup
n
|f
n
(x)| < ∞,
but
sup
n
sup
x
|f
n
(x)| = ∞.
However, by a proof very similar to what we had above, we have
Theorem
(Osgood)
.
Let
f
n
: [0
,
1]
→ R
be a sequence of continuous functions
such that for all x ∈ [0, 1]
sup
n
|f
n
(x)| < ∞
Then there are some a, b with 0 ≤ a < b ≤ 1 such that
sup
n
sup
x∈[a,b]
|f
n
(x)| < ∞.
Proof. See example sheet.
Theorem
(Open mapping theorem)
.
Let
V
and
W
be Banach spaces and
T
:
V → W
be a bounded surjective linear map. Then
T
is an open map, i.e.
T (U) is an open subset of W whenever U is an open subset of V .
Note that surjectivity is necessary. If
q 6∈ T
(
V
), then we can scale
q
down
arbitrarily and still not be in the image of
T
. So
T
(
V
) does not contain an open
neighbourhood of 0, and hence cannot be open.
Proof. We can break our proof into three parts:
(i)
We first want an easy way to check if a map is an open map. We want
to show that
T
is open if and only if
T
(
B
V
(1))
⊇ B
W
(
ε
) for some
ε >
0.
Note that one direction is easy — if
T
is open, then by definition
T
(
B
V
(1))
is open, and hence we can find the epsilon required. So we are going to
prove the other direction.
(ii) We show that T (B
V
(1)) ⊇ B
W
(ε) for some ε > 0
(iii)
By rescaling the norm in
W
, we may wlog the
ε
obtained above is in fact
1. We then show that if T (B
V
(1)) ⊇ B
W
(1), then T (B
V
(1)) ⊇ B
W
(
1
2
).
We now prove them one by one.
(i)
Suppose
T
(
B
V
(1))
⊇ B
W
(
ε
) for some
ε >
0. Let
U ⊆ V
be an open set.
We want to show that T (U) is open. So let p ∈ U, q = T p.
Since
U
is open, there is some
δ >
0 such that
B
V
(
p, δ
)
⊆ U
. We can also
write the ball as B
V
(p, δ) = p + B
V
(δ). Then we have
T (U) ⊇ T (p + B
V
(δ))
= T p + T (B
V
(δ))
= T p + δT (B
V
(1))
⊇ q + δB
W
(ε)
= q + B
W
(δε)
= B
W
(q, δε).
So done.
(ii) This is the step where we use the Baire category theorem.
Since T is surjective, we can write W as
W =
∞
[
n=1
T (B
V
(n)) =
∞
[
n=1
T (nB
V
(1)) =
∞
[
n=1
T (nB
V
(1)).
We have written
W
as a countable union of closed sets. Since
W
is a
Banach space, by the Baire category theorem, there is some
n ≥
1 such that
T (nB
V
(1))
has non-empty interior. But since
T (nB
V
(1))
=
nT (B
V
(1))
,
and multiplication by
n
is a homeomorphism, it follows that
T (B
V
(1))
has
non-empty interior. So there is some ε > 0 and w
0
∈ W such that
T (B
V
(1)) ⊇ B
W
(w
0
, ε).
We have now found an open ball in the neighbourhood, but we want a ball
centered at the origin. We will use linearity in two ways. Firstly, since if
v ∈ B
V
(1), then −v ∈ B
V
(1). By linearly of T , we know that
T (B
V
(1)) ⊇ B
W
(−w
0
, ε).
Then by linearity, intuitively, since the image contains the balls
B
W
(
w
0
, ε
)
and
B
W
(
−w
0
, ε
), it must contain everything in between. In particular, it
must contain B
W
(ε).
To prove this properly, we need some additional work. This is easy if we had
T
(
B
V
(1))
⊇ B
W
(
w
0
, ε
) instead of the closure of it — for any
w ∈ B
W
(
ε
),
we let
v
1
, v
2
∈ B
V
(1) be such that
T
(
v
1
) =
w
0
+
w
,
T
(
v
2
) =
−w
0
+
w
.
Then v =
v
1
+v
2
2
satisfies kvk
V
< 1 and T (v) = w.
Since we now have the closure instead, we need to mess with sequences.
Since
T (B
V
(1)) ⊇ ±w
0
+
B
W
(
ε
), for any
w ∈ B
W
(
ε
), we can find sequences
(
v
i
) and (
u
i
) such that
kv
i
k
V
, ku
i
k
V
<
1 for all
i
and
T
(
v
i
)
→ w
0
+
w
,
T (u
i
) → −w
0
+ w.
Now by the triangle inequality, we get
v
i
+ u
i
2
< 1,
and we also have
v
i
+ u
i
2
→
w
0
+ w
2
+
−w
0
+ w
2
= w.
So w ∈ T (B
V
(1)). So T (B
V
(1)) ⊇ B
W
(ε).
(iii) Let w ∈ B
W
(
1
2
). For any δ, we know
T (B
V
(δ)) ⊇ B
W
(δ).
Thus, picking δ =
1
2
, we can find some v
1
∈ V such that
kv
1
k
V
<
1
2
, kT v
1
− wk <
1
4
.
Suppose we have recursively found v
n
such that
kv
n
k
V
<
1
2
n
, kT (v
1
+ ··· + v
n
) − wk <
1
2
n+1
.
Then picking
δ
=
1
2
n+1
, we can find
v
n+1
satsifying the properties listed
above. Then
P
∞
n=1
v
n
is Cauchy, hence convergent by completeness. Let
v be the limit. Then
kvk
V
≤
∞
X
i=1
kv
i
k
V
< 1.
Moreover, by continuity of T , we know T v = w. So we are done.
Note that in this proof, we required both
V
and
W
to be Banach spaces.
However, we used the completeness in different ways. We used the completeness
of
V
to extract a limit, but we just used the completeness of
W
to say it is of
second category. In particular, it suffices to assume the image of
T
is of second
category, instead of assuming surjectivity. Hence if we know that
T
:
V → W
is a bounded linear map such that
V
is Banach and
im T
is of second category,
then
T
is open. As a consequence
T
has to be surjective (its image contains a
small open ball which we can scale up arbitrary).
We are now going to look at certain applications of the open mapping theorem
Theorem
(Inverse mapping theorem)
.
Let
V, W
be Banach spaces, and
T
:
V → W
be a bounded linear map which is both injective and surjective. Then
T
−1
exists and is a bounded linear map.
Proof.
We know that
T
−1
as a function of sets exists. It is also easy to show
that it is linear since
T
is linear. By the open mapping theorem, since
T
(
U
) is
open for all
U ⊆ V
open. So (
T
−1
)
−1
(
U
) is open for all
U ⊆ V
. By definition,
T
−1
is continuous. Hence
T
−1
is bounded since boundedness and continuity are
equivalent.
Theorem
(Closed graph theorem)
.
Let
V, W
be Banach spaces, and
T
:
V → W
a linear map. If the graph of T is closed, i.e.
Γ(T ) = {(v, T (v)) : v ∈ V } ⊆ V × W
is a closed subset of the product space (using the norm
k
(
v, w
)
k
V ×W
=
max{kvk
V
, kwk
W
}), then T is bounded.
What does this mean? Closedness of the graph means that if
v
n
→ v
in
V
and
T
(
v
n
)
→ w
, then
w
=
T
(
v
). What we want to show is continuity, which
is a stronger statement that if
v
n
→ v
, then
T
(
v
n
) converges and converges to
T (v).
Proof.
Consider
φ
: Γ(
T
)
→ V
defined by
φ
(
v, T
(
v
)) =
v
. We want to apply
the inverse mapping theorem to this. To do so, we need to show a few things.
First we need to show that the spaces are Banach spaces. This is easy — Γ(
T
)
is a Banach space since it is a closed subset of a complete space, and we are
already given that V is Banach.
Now we need to show surjectivity and injectivity. This is surjective since for
any
v ∈ V
, we have
φ
(
v, T
(
v
)) =
v
. It is also injective since the function
T
is
single-valued.
Finally, we want to show φ is bounded. This is since
kvk
V
≤ max{kvk, kT (v)k} = k(v, T (v))k
Γ(T )
.
By the inverse mapping theorem,
φ
−1
is bounded, i.e. there is some
C >
0 such
that
max{kvk
V
, kT (v)k} ≤ Ckvk
V
In particular, kT (v)k ≤ Ckvk
V
. So T is bounded.
Example.
We define
D
(
T
) to be equal to
C
1
([0
,
1]) as a vector space, but
equipped with the
C
([0
,
1]) norm instead. We seek to show that
D
(
T
) is not
complete.
To do so, consider the differentiation map
T : D(T ) → C([0, 1]),
First of all, this is unbounded. Indeed, consider the sequence of functions
f
n
(x) = x
n
. Then kf
n
k
C([0,1])
= 1 for all n. However,
kT f
n
k
C([0,1])
= sup
x∈[0,1]
nx
n−1
= n.
So T is unbounded.
We claim the graph of
T
is closed. If so, then since
C
([0
,
1]) is complete, the
closed graph theorem implies D(T ) is not complete.
To check this, suppose we have
f
n
→ f
in the
C
([0
,
1]) norm, and
f
0
n
→ g
,
again in the C([0, 1]) norm. We want to show that f
0
= g.
By the fundamental theorem of calculus, we have
f
n
(t) = f
n
(0) +
Z
t
0
f
0
n
(x) dx.
Hence by uniform convergence of f
0
n
→ g and f
n
→ f, we have
f(t) = f (0) +
Z
t
0
g(x) dx.
So by the fundamental theorem of calculus, we know that
f
0
(
t
) =
g
(
t
). So the
graph of T is closed.
Since we know that
C
([0
,
1]) is complete, this shows that
D
(
T
) is not complete.
Example.
We can also use the Baire category theorem to understand Fourier
series. Let
f
:
S
1
→ R
be continuous, i.e.
f
: [
−π, π
]
→ R
which is continu-
ous with periodic boundary condition
f
(
−π
) =
f
(
π
). We define the Fourier
coefficients
ˆ
f : Z → C by
ˆ
f(k) =
1
2π
Z
π
−π
e
−ikx
f(x) dx.
We define the Fourier series by
X
k∈Z
e
ikz
ˆ
f(k).
In particular, we define the partial sums as
S
n
(f)(x) =
n
X
k=−n
e
ikx
ˆ
f(k).
The question we want to ask is: does the Fourier series converge? We are not
even asking if it converges back to
f
. Just if it converges at all. More concretely,
we want to know if S
n
(f)(x) has a limit as n → ∞ for f continuous.
Unfortunately, no. We can show that there exists a continuous function
f
such that
S
n
(
f
)(
x
) diverges. To show this, we can consider
φ
n
:
C
(
S
1
)
→ R
defined by φ
n
(f) = S
n
(f)(0). Assume that
sup
n
|φ
n
(f)|
is finite for all continuous f. By Banach-Steinhaus theorem, we have
sup
n
kφ
n
k
B(C(S
1
),R)
< ∞,
On the other hand, we can show that
φ
n
(f) =
1
2π
Z
π
−π
f(x)
sin
n +
1
2
x
sin
x
2
dx.
It thus suffices to find a sequence f
n
such that kf
n
k
C(S
1
)
≤ 1 but
Z
π
−π
f(x)
sin
n +
1
2
x
sin
x
2
dx
→ ∞,
which is a contradiction. Details are left for the example sheet.
What’s the role of the Banach-Steinhaus theorem here? If we wanted to prove
the result directly, then we need to find a single function
f ∈ C
(
S
1
) such that
φ
n
(
f
) is unbounded. However, now we just have to find a sequence
f
n
∈ C
(
S
1
)
such that φ
n
(f
n
) → ∞. This is much easier.
There is another thing we can ask. Note that if
f
is continuous, then
|
ˆ
f
(
k
)
| →
0 as
k → ±∞
. In fact, this is even true if
f ∈ L
1
(
S
1
), i.e.
f
is Lebesgue
integrable and
Z
π
−π
|f(x)| dx < ∞.
A classical question is: do all sequences
{a
n
} ∈ C
with
|a
n
| →
0 as
n → ±∞
arise as the Fourier series of some
f ∈ L
1
? The answer is no. We let
˜c
0
be
the set of all such sequences. By the inverse mapping theorem, if the map
φ
:
L
1
(
S
1
)
→ ˜c
0
is surjective, then the inverse is bounded. But this is not true,
since we can find a sequence
f
n
such that
kf
n
k
L
1
(S
1
)
→ ∞
but
sup
k
|
ˆ
f
k
(
`
)
| ≤
1
for all `. Details are again left for the reader.