Complex and Functional analysis Aidan Sims

51 Pages·2016·0.58 MB·English

Complex and Functional analysis
Draft Notes, March 2, 2016
Aidan Sims
Abstract. These course notes were developed by Aidan Sims in Spring of 2013 as
the basis for an honours-level course covering the fundamentals of complex analysis and
functional analysis. They are based heavily on parts of Serge Lang’s text Complex
Analysis [1] and on parts of Gert Pedersen’s excellent text Analysis Now [2].
Thesenotes doubtlesscontainplentyoftypo’sand possibly moreseriouserrors, and
I would appreciate hearing about them as you find them.
Thanks to Hui Li for his ongoing reading and correction of the notes; the remaining
mistakes are of course my own fault.
Contents
Chapter 1. Functional Analysis 5
1. Normed spaces and Banach spaces 5
2. Baire category and consequences 9
3. Linear functionals and dual spaces 12
4. Weak topologies 16
5. The weak* topology for normed spaces 19
Chapter 2. Complex Analysis 23
1. Background 23
2. Cauchy’s Theorem 25
3. Analytic vs holomorphic functions 34
4. Consequences of Holomorphicity 38
5. Winding Numbers and the Global Cauchy Formula 41
6. Invertibility, Openness and the Maximum Modulus Principle 47
Bibliography 51
3

CHAPTER 1
Functional Analysis
In this chapter, we change tack completely and consider functional analysis. The
broad idea of functional analysis is that one can profitably study the properties of vector
spaces V by concentrating on the space of linear maps from V into the base field (for us
the base field will always be C).
1. Normed spaces and Banach spaces
In this section we will briefly discuss some basic facts about normed vector spaces and
Banach spaces.
A complex normed linear space is a complex vector space V equipped with a norm
(cid:107)·(cid:107), which is to say a map from V to [0,∞) satisfying
(N1) (cid:107)λx(cid:107) = |λ|(cid:107)x(cid:107) for all x ∈ V and λ ∈ C;
(N2) (cid:107)x+y(cid:107) ≤ (cid:107)x(cid:107)+(cid:107)y(cid:107) for all x,y ∈ V; and
(N3) (cid:107)x(cid:107) = 0 implies x = 0.
If (N1) and (N2) hold, but not (N3), then we call (cid:107)·(cid:107) a seminorm. The normed vector
space V is a Banach space if it is complete in the norm (cid:107)·(cid:107) in the sense that every Cauchy
sequence converges. The triangle inequality gives (cid:107)x(cid:107) = (cid:107)x+y−y(cid:107) ≤ (cid:107)x−y(cid:107)+(cid:107)y(cid:107), and
rearrangingweobtain(cid:107)x(cid:107)−(cid:107)y(cid:107) ≤ (cid:107)x−y(cid:107). Symmetrygives(cid:107)y(cid:107)−(cid:107)x(cid:107) ≤ (cid:107)y−x(cid:107) = (cid:107)x−y(cid:107)
(cid:12) (cid:12)
as well, and so we obtain the so-called “reverse triangle inequality” (cid:12)(cid:107)x(cid:107)−(cid:107)y(cid:107)(cid:12) ≤ (cid:107)x−y(cid:107).
Example 1.1. The modulus function z (cid:55)→ |z| on C is a norm. More generally, for
p ≥ 1, the map (cid:107)x(cid:107) = (cid:0)(cid:80)n |x |p(cid:1)1/p is a norm on Cn, called the p-norm.
p i=1 i
Example 1.2. Let X be a locally compact Hausdorff space. Then the set
C (X) = {f : X → C | f is continuous and there exists
b
B > 0 such that |f(x)| < B for all x ∈ X}
of bounded continuous functions on X equipped with (cid:107)·(cid:107) given by (cid:107)f(cid:107) = sup{|f(x)| :
∞ ∞
x ∈ X} is a normed vector space under pointwise addition and scalar multiplication. We
claim that it is a Banach space.
To see this, suppose that f is a Cauchy sequence. For x ∈ X, we have |f (x) −
n n
f (x)| ≤ (cid:107)f − f (cid:107) → 0 as m,n → ∞, and since C is complete, the sequence f (x)
m n m ∞ n
converges. Define f : X → C by f(x) = limf (x) for all x. We show that f is bounded
n
and continuous. Indeed, choose N so that (cid:107)f −f (cid:107) < 1 for m,n ≥ N. For x ∈ X there
n m ∞
exists N ≥ N such that |f (x)−f(x)| < 1 for n ≥ N , and then we have |f(x)−f (x)| ≤
x n x N
|f(x)−f (x)|+|f (x)−f (x)| < 2, and in particular f(x) < |f (x)|+2 ≤ (cid:107)f (cid:107) +2.
Nx Nx N N N ∞
So (cid:107)f (cid:107) +2 is a bound for f.
N ∞
To see that f → f in (cid:107) · (cid:107) , fix ε > 0, since (f ) is Cauchy, there exists N ∈ N
n ∞ n
such that m,n ≥ N implies (cid:107)f −f (cid:107) < ε/2. Since each f(x) = lim f (x), for each
m n ∞ n→∞ n
5
6 1. FUNCTIONAL ANALYSIS
x ∈ X, there exists n ≥ N such that |f (x)−f(x)| < ε/2. Now for x ∈ X and n ≥ N,
x nx
we have
|f (x)−f(x)| ≤ |f (x)−f (x)|+|f (x)−f(x)| < ε,
n n nx nx
and hence (cid:107)f −f(cid:107) = sup |f (x)−f(x)| ≤ ε.
n ∞ x n
For continuity of f, we show that the preimage under f of an open ball is open.
Suppose that f(x) ∈ B(z,ε); we must find a neighbourhood of x whose image is contained
in B(z,ε). Let δ = 1 inf{|f(x)−w| : w (cid:54)∈ B(z,ε)}. Choose N such that (cid:107)f −f(cid:107) < δ/3.
2 N ∞
Since f is continuous, U := f−1(B(f (x),δ/3)) is open, and it contains x by definition.
N N N
For y ∈ U, we have |f(y)−f(x)| ≤ |f(y)−f (y)|+|f (y)−f (x)|+|f (x)−f(x)| ≤ δ.
N N N N
So f(U) ⊆ B(f(x),δ) ⊆ B(z,ε).
Example 1.3. Again let X be a locally compact Hausdorff space. Let
C (X) = {f : X → C | f is continuous, and for each ε > 0 there exists
0
K ⊆ X compact such that |f(x)| < ε whenever x (cid:54)∈ K}.
We call this the space of functions which vanish at infinity on X. We claim that C (X)
0
is a closed subspace of C (X), and hence a Banach space.
b
To see that C (X) ⊆ C (X), fix f ∈ C (X), choose a compact set K such that
0 c 0
|f(x)| < 1 for x (cid:54)∈ K, and observe that since K is compact and f is continuous, f(K) is
compact and hence bounded above, say by B. Now max{B,1} is a bound for f. To see
that C (X) is closed, suppose that f is a sequence in C (X) converging to f. Fix ε > 0.
0 n 0
Choose n such that (cid:107)f −f(cid:107) < ε/2 and a compact set K such that |f (x)| < ε/2 off K.
n ∞ n
Then for x (cid:54)∈ K, we have |f(x)| ≤ |f(x)−f (x)|+|f (x)| < ε.
n n
A linear operator T : V → W between normed vector spaces is a map such that
T(λx+y) = λTx+Ty for all λ,x,y.
Proposition 1.4. Let T : X → Y be a linear operator between normed vector spaces.
The following are equivalent:
(1) T is bounded in the sense that there exists K ≥ 0 such that (cid:107)Tx(cid:107) ≤ K(cid:107)x(cid:107) for all
x ∈ X;
(2) T is continuous; and
(3) T is continuous at some point x ∈ X.
Proof. For (1) =⇒ (2), first observe that if K = 0 then T is the constant map
x (cid:55)→ 0, and hence continuous. So suppose that K > 0, and fix x ∈ X and ε > 0. If
(cid:107)y −x(cid:107) ≤ ε/K, then (cid:107)Ty −Tx(cid:107) = (cid:107)T(y −x)(cid:107) ≤ K(cid:107)y −x(cid:107) ≤ ε.
The implication (2) =⇒ (3) is trivial.
Finally, supposethatT iscontinuousatx. Thenthereexistsδ > 0suchthat(cid:107)y−x(cid:107) ≤
δ implies (cid:107)T(y−x)(cid:107) = (cid:107)Ty−Tx(cid:107) ≤ 1. Now for z ∈ X\{0}, we have (cid:107)Tz(cid:107) = (cid:107)z(cid:107)T( δz ) =
δ (cid:107)z(cid:107)
(cid:107)z(cid:107)(cid:107)T((x+ δz )−x)(cid:107) ≤ (cid:107)z(cid:107), and so K = 1 is a bound for T. (cid:3)
δ (cid:107)z(cid:107) δ δ
We write B(X,Y) for the set of all bounded linear operators from a normed vec-
tor space X to a normed vector space Y. This set is a vector space under point-
wise addition and scalar multiplication (exercise!). We define a norm on B(X,Y) by
(cid:107)T(cid:107) := sup{(cid:107)Tx(cid:107) : (cid:107)x(cid:107) ≤ 1}, and call this the operator norm. We will usually just
op
write (cid:107)T(cid:107) for (cid:107)T(cid:107) .
op
If T ∈ B(X,Y) and S ∈ B(Y,Z), then S ◦ T ∈ B(X,Z), and (cid:107)S ◦ T(cid:107) ≤ (cid:107)S(cid:107)(cid:107)T(cid:107)
(exercise!).
1. NORMED SPACES AND BANACH SPACES 7
Lemma 1.5. If X is a normed vector space and Y is a Banach space, then B(X,Y)
is a Banach space.
Proof. Suppose that T is a Cauchy sequence in B(X,Y). Fix x ∈ X and ε > 0.
n
There exists N such that (cid:107)T −T (cid:107) ≤ ε/(cid:107)x(cid:107), and hence (cid:107)T x−T x(cid:107) ≤ ε, for m,n ≥ N.
m n m n
So (T x)∞ is a Cauchy sequence in Y, and therefore converges to some vector Tx. The
n n=1
algebra of limits ensures that T(λx + y) = λTx + Ty, and so T : X → Y is linear. To
see that it is bounded, we show that it is continuous at 0. For this, fix ε > 0. The
sequence T is Cauchy, so we can find N such that m,n ≥ N forces (cid:107)T −T (cid:107) ≤ 1, and
n m n
therefore (cid:107)T (cid:107) ≤ (cid:107)T (cid:107)+1 for all n ≥ N. Hence K := max{(cid:107)T (cid:107),...,(cid:107)T (cid:107),(cid:107)T (cid:107)+1}
n N 1 N−1 N
satisfies (cid:107)T (cid:107) < K for all n ∈ N. Fix x with (cid:107)x(cid:107) ≤ min{1, ε }. Choose N such that
n 2K
(cid:107)T x−Tx(cid:107) < ε/2. Then
N
(cid:107)T (cid:107)ε
N
(cid:107)T(x)−0(cid:107) ≤ (cid:107)Tx−T x(cid:107)+(cid:107)T x(cid:107) ≤ (cid:107)Tx−T x(cid:107)+ < ε.
N N N
2K
So T is continuous at 0, and therefore bounded by Proposition 1.4.
It remains to show that T → T in the operator norm. For this, fix ε > 0, fix
n
N so that (cid:107)T − T (cid:107) < ε/2 for every n ≥ m. For each x ∈ X with (cid:107)x(cid:107) ≤ 1, since
m n
T x → Tx, we can choose n ≥ N such that (cid:107)T x − T (cid:107) < ε/2. Fix n ≥ N. For
n x nx x
each x ∈ X with (cid:107)x(cid:107) ≤ 1, we have (cid:107)T x − Tx(cid:107) ≤ (cid:107)T x − T x(cid:107) + (cid:107)T x − Tx(cid:107) ≤ ε.
n n nx nx
Hence (cid:107)T −T(cid:107) = sup (cid:107)T x−Tx(cid:107) ≤ ε for all n ≥ N. That is, T → T in operator
n (cid:107)x(cid:107)≤1 n n
norm. (cid:3)
Using this, we can define quotient spaces.
Proposition 1.6. Let X be a normed vector space, and let V be a subspace of X. Let
Q : x (cid:55)→ x+V be the quotient map from X to X/V. Then
(cid:107)x+V(cid:107) := inf (cid:107)x+v(cid:107)
v∈V
defines a seminorm on X/V, with (cid:107)Q(x)(cid:107) ≤ (cid:107)x(cid:107) for all x ∈ X. If V is closed, then (cid:107)·(cid:107)
is a norm on X/V, and it is a Banach-space norm if X is a Banach space.
Proof. We have (cid:107)λx + V(cid:107) = inf (cid:107)λx + v(cid:107) = inf λ(cid:107)x + λ−1v(cid:107) = λ(cid:107)x + V(cid:107),
v∈V v∈V
giving (N1). For x,y ∈ X and v,w ∈ V we have (cid:107)x+v(cid:107)+(cid:107)y+w(cid:107) ≥ (cid:107)x+y+(v+w)(cid:107) ≥
(cid:107)x+y+V(cid:107), and so inf (cid:107)x+v(cid:107)+(cid:107)y+w(cid:107) ≥ (cid:107)x+y+V(cid:107). This gives (N2), so (cid:107)·(cid:107) is a
v,w
seminorm. We have (cid:107)Q(x)(cid:107) = inf (cid:107)x+v(cid:107) ≤ (cid:107)x+0(cid:107) = (cid:107)x(cid:107).
v∈V
Suppose that V is closed and (cid:107)x + V(cid:107) = 0. Then there exist v in V such that
n
(cid:107)x+v (cid:107) → 0. Thus v → −x, and since V is closed, we deduce that x ∈ V and hence
n n
x+V = 0; so X/V is a normed space. Suppose that X is complete, and z is a Cauchy
n
sequence in X/V. Choose a subsequence (w ) such that (cid:107)w − w (cid:107) < 2−n for all n.
n n n+1
Suppose that Q(x ) = w . Choose y with Q(y ) = w . Then (cid:107)Q(x −y)(cid:107) = (cid:107)Q(x )−
n n n n+1 n n
Q(y)(cid:107) < 2−n, and so there exists v ∈ V such that (cid:107)x −y+v(cid:107) < 2−n. Then x = y−v
n n+1
satisfies Q(x ) = w , and (cid:107)x −x (cid:107) < 2−n. By induction, there is a sequence x
n+1 n+1 n n+1 n
in X such that Q(x ) = w and (cid:107)x −x (cid:107) < 2−n for all n. This sequence is Cauchy,
n n n n+1
and therefore converges, say to x. We have (cid:107)w −Q(x)(cid:107) = (cid:107)Q(x −x)(cid:107) ≤ (cid:107)x −x(cid:107) → 0,
n n n
and so w converges to Q(x) in X/V. (cid:3)
n
Lemma 1.7. Suppose that X,Y are normed vector spaces and T ∈ B(X,Y). Suppose
that V is a closed subspace of X and that V ⊆ ker(T). Then there is a bounded linear
˜ ˜ ˜
map T : X/V → Y such that T(x+V) = Tx for all x, and we have (cid:107)T(cid:107) = (cid:107)T(cid:107).
8 1. FUNCTIONAL ANALYSIS
Proof. We have already seen that X/V is a normed linear space. If x+V = y+V,
˜
then x−y ∈ V, and so Tx = T(x+y −y) = Ty +T(x−y) = Ty. So T is well-defined,
and it is linear because the operations in X/V are inherited from X. For x ∈ X and
v ∈ V, we have Tv = 0 and so
˜
(cid:107)TQx(cid:107) = (cid:107)Tx(cid:107) = (cid:107)T(x+v)(cid:107) ≤ (cid:107)T(cid:107)(cid:107)x+v(cid:107),
˜
and since v was arbitrary, we deduce that (cid:107)TQx(cid:107) ≤ inf (cid:107)T(cid:107)(cid:107)x+v(cid:107) = (cid:107)T(cid:107)(cid:107)Qx(cid:107), so
v∈V
˜
that (cid:107)T(cid:107) ≤ (cid:107)T(cid:107). For the reverse inequality, observe that if (cid:107)x(cid:107) ≤ 1, then (cid:107)Qx(cid:107) ≤ 1, and
so (cid:107)T(cid:107) = sup (cid:107)Tx(cid:107) ≤ sup (cid:107)Tx(cid:107) = sup (cid:107)T˜Qx(cid:107) = (cid:107)T˜(cid:107). (cid:3)
(cid:107)x(cid:107)≤1 (cid:107)Qx(cid:107)≤1 (cid:107)Qx(cid:107)≤1
Lemma 1.8. Suppose that X is a normed vector space and V is a closed subspace.
Suppose that both V and X/V are Banach spaces. Then X is a Banach space.
Proof. Fix a Cauchy sequence x in X. Let Q : X → X/V be the quotient map.
n
Then Qx is a Cauchy sequence, so converges to some Qx. Since Q(x −x) → 0 we may
n n
choose v ∈ V such that (cid:107)x +v −x(cid:107) < 1/n+(cid:107)Q(x −x)(cid:107) for all n. Since Q(x −x) → 0,
n n n n n
we deduce that x + v → x. Given ε > 0, choose N such that (cid:107)x + v − x(cid:107) < ε/3
n n n n
for n ≥ N and such that (cid:107)x − x (cid:107) < ε/3 for m,n ≥ N. Then for m,n ≥ N, we
m n
have (cid:107)v − v (cid:107) ≤ (cid:107)(x − v − x )(cid:107) + (cid:107)(x − v − x )(cid:107) + (cid:107)x − x (cid:107) ≤ ε, and so the
n m m m n n m n
sequence v is Cauchy. Since V is a Banach space, we have v → v ∈ V, and then
n n
x = x +v −v → x−v in X. (cid:3)
n n n n
Proposition 1.9. Let X and Y be Banach spaces, and suppose that X is a dense
0
subspace of X. Then every T ∈ B(X ,Y) extends uniquely to an operator T ∈ B(X,Y),
0 0
and (cid:107)T(cid:107) = (cid:107)T (cid:107).
0
Proof. For x ∈ X, choose x in X such that x → X. Then x is Cauchy, and it
n 0 n n
follows that T x is also Cauchy because T is bounded. We define Tx = limT x . If
0 n 0 0 n
y converges to x also, then (cid:107)limT x − limT y (cid:107) = lim(cid:107)T (x − y )(cid:107) = 0 since T is
n 0 n 0 n 0 n n 0
bounded, and so T is independent of the choice of sequence x converging to x. This T
x n
is linear by the algebra of limits, and extends T because for x ∈ X we can choose x to
0 0 n
be the constant sequence x = x. If (cid:107)x(cid:107) = 1, and X (cid:51) x → x, then xn → x also, and
n 0 n (cid:107)xn(cid:107)
so (cid:107)T x(cid:107) = lim(cid:107)T xn (cid:107) ≤ (cid:107)T (cid:107). On the other hand there exist x ∈ X with (cid:107)x (cid:107) ≤ 1
0 (cid:107)xn(cid:107) 0 n 0 n
and (cid:107)Tx (cid:107) = (cid:107)T x (cid:107) → (cid:107)T (cid:107), giving (cid:107)T(cid:107) ≥ (cid:107)T (cid:107) as well. (cid:3)
n 0 n 0 0
Corollary 1.10. Let X be a normed vector space, and suppose that X and Y are
0
Banach spaces, and that T : X → X and S : X → Y are isometric embeddings of X
0 0 0
into dense subspaces of X and Y. Then there is an isometric isomorphism ψ : X → Y
such that ψ ◦T = S.
Proof. Since both S and T are isometric, the formula ψ(Sx) = Tx is well-defined
and isometric from a dense subspace of X to a dense subspace of Y. So the result follows
from Proposition 1.9. (cid:3)
We will see later that every normed vector space X can be embedded isometrically
0
into a Banach space X, called the completion of X . In the words of Pedersen [2, Proof
0
of Proposition 2.1.12], the “most pedestrian” way to prove this is to construct X as the
space of equivalence classes of Cauchy sequences in X (Exercise!); but in this course, we
0
will follow Pedersen’s approach, and use this as an opportunity to emphasise the utility
of dual spaces later in the notes.
2. BAIRE CATEGORY AND CONSEQUENCES 9
2. Baire category and consequences
Recall that a metric space is a set X endowed with a metric d : X ×X → [0,∞) such
that d(x,x) = 0 for all x, d(x,y) = d(y,x), and d(x,z) ≤ d(x,y)+d(y,z) for all x,y,z.
A sequence (x ) converges to x in X if d(x ,x) → 0, and a sequence (x ) is Cauchy if
n n n
d(x ,x ) → 0 as m,n → ∞. The metric space X is complete if every Cauchy sequence
m n
in X converges.
If you are not used to metric spaces, just think of X as a normed vector space with
d(x,y) = (cid:107)x−y(cid:107).
If X is a metric space, then for x ∈ X and r > 0, the open ball B(x,r) is the set
{y ∈ X : d(x,y) < r}. We say that a set U ⊆ X is open if for every x ∈ U there exists
r > 0 such that B(x,r) ⊆ U. The interior of a set Y ⊆ X is the set
Y◦ = {x ∈ Y : there exists r > 0 such that B(x,r) ⊆ Y}.
A set Y ⊆ X is dense in X if every open ball in X has nonempty intersection with Y;
equivalently if every point in X is the limit of a sequence in Y.
Definition 2.1. Let X be a complete metric space. A subset Y of X is said to be
of first category or a meagre set, if there is a sequence (Y ) of closed subsets of X, each
n
(cid:83)
with empty interior, such that Y = Y . Any set which is not meagre is said to be of
n n
second category or nonmeagre.
The so-called Baire category theorem says that meagre sets have empty interior. The
name of the theorem comes from Baire’s (somewhat bland) nomenclature “first category”
and “second category.”
Theorem 2.2 (Baire Category theorem). Let (X,d) be a complete metric space. Sup-
(cid:84)
pose that (U ) is a sequence of open dense subsets of X. Then U is also dense in X.
n n n
Equivalently, any countable union of meagre sets has empty interior.
Proof. Fix x ∈ X and r > 0. We must show that (cid:84) (B(x ,r )∩U ) is nonempty.
0 0 n 0 0 n
To see this, first observe that if U is a dense open subset of X and B = B[x,R] is a
closed ball of radius R in X, then that U is dense implies that B ∩U is nonempty; that
U is open implies that B(x,R)∩U is open; and it then follows that there is a closed ball
B(cid:48) = B[x(cid:48),R(cid:48)] with R(cid:48) < R/2 such that B[x(cid:48),R(cid:48)] ⊆ B(x,r)∩U.
Applying the argument of the preceding paragraph inductively, we obtain a decreas-
ing chain B[x ,r ] ⊃ B[x ,r ] ⊂ B[x ,r ]··· of open balls such that B[x ,r ] ⊆ U ∩
1 1 2 2 3 3 i i i
B(x ,r ) and r < 2−ir . For i ≤ j, we have
i−1 i−1 i 0
j−1 j−1 j−1
(cid:88) (cid:88) (cid:88)
d(x ,x ) ≤ d(x ,x ) ≤ r ≤ 2−ir ≤ 2−i−1r .
i j l l+1 i 0 0
l=i l=i l=i
This implies that the sequence x is Cauchy, and so converges to x ∈ X because X is
i
complete. Since the B[x ,r ] are closed and nested, we have x ∈ B[x ,r ] ⊆ U ∩B(x ,r )
i i i i i 0 0
for all i.
For the final assertion, observe that a set U is dense and open if and only if X \U is
meagre, and that a set is dense if and only if its complement has empty interior. (cid:3)
The Bairecategory theorem, while relativelysimple to prove, is remarkably useful. We
discuss a number of famous consequences. For the first one, recall that a map f : X → Y
between topological spaces is an open map if f(U) is open in Y whenever U is open in X.
10 1. FUNCTIONAL ANALYSIS
Theorem 2.3 (Open Mapping Theorem). Suppose that X,Y are Banach spaces, and
that T ∈ B(X,Y) is surjective. Then T is an open map.
To prove the theorem, we will need a technical lemma. When we say that a set U is
dense in a set V, we mean that every open subset of V intersects U nontrivially, but not
necessarily that U ⊆ V.
Lemma 2.4. Let X, Y be Banach spaces, and consider T ∈ B(X,Y). Suppose that
T(B (0,1)) is dense in B (0,r) with r > 0. Then B (0,r(cid:48)) ⊆ T(B (0,1)) for all r(cid:48) < r.
X Y Y X
Proof. Fix r(cid:48) < r and let ε = 1−r(cid:48)/r. The statement is obvious if r(cid:48) = 0, so we may
assume that ε < 1. Let U := T(B (0,1)). Fix y ∈ B(0,r). Since U is dense in B(0,r),
X
there exists y ∈ U such that (cid:107)y−y (cid:107) < εr. That U is dense in B (0,r) implies that εU
1 1 Y
is dense in εB (0,r) = B (0,εr). Since (cid:107)y−y (cid:107) < εr, it follows that there exists y ∈ εU
Y Y 1 2
such that (cid:107)(y −y )−y (cid:107) < ε2r. Inductively, we obtain a sequence y ∈ εn−1U such that
1 2 n
(cid:107)y −(cid:80)n y (cid:107) < εnr; that is, (cid:80)∞ y converges in Y to y.
i=1 i i=1 i
Each εn−1U = εn−1T(B (0,1)) = T(B (0,εn−1)), so there exists x ∈ B (0,εn−1)
X X n X
such that Tx = y . Since ε < 1, the sequence of partial sums (cid:80)n x is Cauchy, so
n n i=1 i
(cid:80)∞ x converges to some x ∈ X. Continuity gives
i=1 i
(cid:88) (cid:88)
T = Tx = y = y.
x i i
i i
We have (cid:107)x(cid:107) ≤ (cid:80) (cid:107)x (cid:107) ≤ (cid:80) εi−1 = (1 − ε)−1. That is, y ∈ T(B (0,(1 − ε)−1)) =
i i i X
(1−ε)−1T(B (0,1)), andmultiplyingthroughby(1−ε), weobtain(1−ε)y ∈ T(B (0,1)).
X X
Since y ∈ B (0,r) was arbitrary, we deduce that (1−ε)B(0,r) ⊆ T(B (0,1)) as claimed.
Y X
(cid:3)
Proof of Theorem 2.3. Since T is surjective and X = (cid:83)∞ B (0,n), we have
n=1 X
∞
(cid:91)
Y = T(B (0,n)).
X
n=1
Since the T(B (0,n)) are all closed, the Baire Category theorem implies that they cannot
X
all have empty interior, so there exists n > 0 such that T(B (0,n)) contains an open ball
X
B (y,ε). Hence T(B (0,1)) is dense in B (1y, ε).
Y X Y n n
We have
ε 1 ε 1 ε 1 1
B (0, ) ⊆ B ( y, )−B ( y, ) ⊆ T(B (0, )−B (0, )) ⊆ T(B (0,1)).
Y Y Y X X X
n n 2n n 2n 2 2
So T(B (0,1)) is dense in B (0, ε), and then Lemma 2.4 implies that T(B (0,1)) con-
X Y n X
tains B (0, ε ).
Y 2n
Now fix y ∈ TX, say y = Tx. Then
ε ε
B (y, ) = Tx+B (0, ) ⊆ Tx+T(B (0,1)) = T(B (x,1)) ⊆ TX.
Y Y X X
2n 2n
So TX is open. (cid:3)
Corollary 2.5. If X and Y are Banach spaces, and T ∈ B(X,Y) is bijective, then
T−1 ∈ B(Y,X).