NOTES ON BANACH SPACES

GESTUR ÓLAFSSON

1. Banach Spaces

1.1. Banach Spaces. Banach spaces are special cases of topological vector spaces which we will encounter later on. In short those are spaces that are endowed with two structure:

1. Algebraic structure: i.e., vector spaces.

2. Topological structure.

And those structures are compatible in the sense that the algebraic operations are continuous.

Definition 1. Let V be a vector space over the field 𝕂 = ℝ, ℂ. A nonnegative real valued function ∥·∥ on V is called a norm if the following holds for all x, y ∈ V and all α ∈ 𝕂:

1. ∥x∥ = 0 if and only if x = 0;

2. ∥x + y∥ ≤ ∥x∥ + ∥y∥;

3. ∥αx∥ = |α| ||x||.

A normed vector space becomes a metric space be defining

d(x, y) := ||x − y||

Definition 2. Let V be a normed vector space of ℝ or ℂ. Then V is called a Banach space if V is complete.

From now on 𝕂 will always stand for ℝ or ℂ. Let V be a vector space over 𝕂. A subset W ⊂ V is a subspace if for all x, y ∈ W and all α ∈ 𝕂 we have

αx + y ∈ S .

If the subspace W is closed in V then W is a closed subspace. If V is a Banach space, then every closed subspace is also a Banach space.

Let A ⊂ V. Let F be the collection of all subspaces containing A. Then F ≠ ∅ because V ∈ F. We let

Span(A) = ⋂_{W∈F} W

We denote Span(A) sometimes by < A >. Notice that < A > is the smallest subspace of V containing A. We can also describe < A > as the space of finite linear combinations α_{1}x_{1} + ⋯ + α_{n}x_{n}. The smallest closed subspace containing A is the space \overline{< A >}.

If A, B ⊂ V and λ ∈ 𝕂. Then we use the following notation:

1. A + B := {a + b | a ∈ A, b ∈ B};

2. A − B := {a − b | a ∈ A, b ∈ B};

3. λA := {λa | a ∈ A}.

2 GESTUR ÓLAFSSON

Here are few examples of Banach spaces: Example 1. Let 1 ≤ p < ∞. The spaces L^{p}(ℝ) are Banach spaces with norm

||f||_{p} := (∫ |f(t)|^{p} dt)^{1/p}

For p = ∞ the norm is the supremum norm. Example 2. Let 1 ≤ p < ∞. Define on the space of real sequences the norm

||{a_{n}}_{n=0}^{∞}||_{p} := (∑_{j=0}^{∞} |a_{j}|^{p})^{1/p}

Then

ℓ^{p} := {{a_{n}} | ||{a_{n}}||_{p} < ∞}

The spaces are Banach spaces. 1.2. The Example C(X) and the Stone-Weierstrass Theorem. Let X be a compact Hausdorff space. In the following we will only deal with 𝕂 = ℝ and leave the simple generalization to ℂ to the reader. Define a norm ||·||_{∞} on C(X) by

||f||_{∞} := sup_{x∈X} |f(x)|

This is finite because X is compact. The norm ||·||_{∞} is called the supremum norm. Lemma 1. The space (C(X), ||·||_{∞}) is a Banach space. Proof. The only nontrivial statement is, that C(X) is complete in the supremum norm. Let {f_{n}} be a Cauchy sequence in C(X). Then for x ∈ X the sequence {f_{n}(x)} is a Cauchy sequence in ℝ. Thus

f(x) := lim_{n→∞} f_{n}(x)

exists. We have to show that f ∈ C(X).

Let x ∈ X and ϵ > 0 be given. Let 0 < ϵ′ < ϵ/3. Chose N such that

||f_{n} − f_{m}||_{∞} < ϵ′ ∀n, m ≥ N

Then in particular |f_{n}(y) − f_{m}(y)| < ϵ′ for all y ∈ X. It follows that

|f_{n}(y) − f(y)| ≤ ϵ′ < ϵ/3 , ∀y ∈ X, ∀n ≥ N

Thus

||f_{n} − f||_{∞} < ϵ/3

Chose U ∋ x open such that

|f_{N}(x) − f_{N}(y)| < ϵ/3 , ∀y ∈ U .

Then we have for y ∈ U:

|f(x) − f(y)| ≤ |f(x) − f_{N}(x)| + |f_{N}(x) − f_{N}(y)| + |f_{N}(y) − f(y)| < ϵ Thus f is continuous which shows that C(X) is a Banach space.

Suppose that B is a Banach space endowed with the structure of an algebra, i.e., a”multiplication” B × B → B satisfying the usual condition on the compatibility between the three operation addition, multiplication by scalars and the new multiplication. We call B for a Banach algebra if

||xy|| ≤ ||x|| ||y|| ∀x, y ∈ B

It follows by the definition that the multiplication is continuous.

Theorem 1. Let X be a compact topological space. Then C(X) is a Banach algebra.

Proof. We know by the above lemma that C(X) is a Banach space. Let x ∈ X. Then

|f(x)g(x)| = |f(x)| |g(x)| ≤ ||f||_{∞} ||g||_{∞}

Taking the supremum it follows that ||fg||_{∞} ≤ ||f||_{∞} ||g||_{∞}

Definition 3. A family F of function on a topological space Y is said to separate points in Y if for every x, y ∈ Y, x ≠ y there is a function f ∈ F such that f(x) ≠ f(y).

Lemma 2. Let X be a compact Hausdorff space. Then the space C(X) separates points.

Proof. This follows form the fact that X is normal and the Urysohn’s Lemma.

The space C(X) has also the structure of a lattice. Let f, g ∈ C(X). Define

f ∧ g(x) := min{f(x), g(x)} and f ∨ g(x) := max{f(x), g(x)} .

Then f ∧g, f ∨g ∈ C(X). A subspace L ⊂ C(X) is called a lattice if for all f, g ∈ L we have f ∧ g, f ∨ g ∈ L. We leave the following to the reader:

Theorem 2 (Stone-Weierstrass). Let X be a compact space and let A ⊂ C(X) be an algebra that separates points and contains the constant functions. Then A is dense in C(X), i.e., given f ∈ C(X) and ϵ > 0, there exists a g ∈ A such that ||g − f||_{∞} < ϵ.

Corollary 1. Let B ⊂ ℝ be closed and bounded and let f : B → ℝ be continuous. Let ϵ > 0. Then there exists a polynomial function p(x) such that sup_{x∈B} |p(x) − f(x)| < ϵ.

We will proof the Stone-Weierstrass Theorem in few Lemmas.

Lemma 3. There exists a sequence of polynomial functions {p_{n}(t)}_{n} such that on the interval [0, 1] the following holds:

1. 0 ≤ p_{n}(t) ≤ p_{n+1}(t) ≤ \sqrt{t}.

2. lim_{n→∞} p_{n}(t) = \sqrt{t} unifomly on [0, 1].

Proof. Define p_{n}(t) inductively by p_{1}(t) = 0 and

p_{n+1}(t) := p_{n}(t) + \frac{1}{2} (t − p_{n}(t)^{2}) .

Then

\sqrt{t} − p_{n}(t) = \sqrt{t} − p_{n}(t) − \frac{1}{2} (\sqrt{t} − p_{n}(t)) (\sqrt{t} + p_{n}(t))

= (\sqrt{t} − p_{n}(t)) (1 − \frac{1}{2}(\sqrt{t} + p_{n}(t)))

By induction hypothesis \sqrt{t} ≥ p_{n}(t). Hence \sqrt{t} − p_{n}(t) ≥ 0 and (\sqrt{t} + p_{n}(t)) ≤ 2\sqrt{t} ≤ 2. It follows that

\sqrt{t} − p_{n}(t) ≥ 0 .

Furthermore

p_{n+1}(t) − p_{n}(t) = \frac{1}{2} (t − p_{n}(t)^{2}) ≥ 0

because t − p_{n}(t)^{2} ≥ 0.

It follows that for each t ∈ [0, 1] the sequence {p_{n}(t)}_{n} is monotone increasing and bounded. Hence convergent. Hence:

0 = lim_{n→∞} p_{n+1}(t) − lim_{n→∞} p_{n}(t) = \frac{1}{2} (t − lim_{n→∞} p_{n}(t)^{2})

It follows that lim p_{n}(t) = \sqrt{t}. But a monotone bounded function sequence is uniformly bounded on compact intervals. Hence the lemma

Lemma 4. There exists a sequence of polynomials {q_{n}(t)}_{n} that converges uniformly on the interval [−1, 1] to the function t ↦ |t|.

Proof. Let q_{n}(t) := p_{n}(t^{2}).

Lemma 5. Let A be as in the Stone-Weierstrass Theorem. Let f ∈ \overline{A} then |f| ∈ \overline{A} and \overline{A} is a lattice.

Proof. Let f ∈ \overline{A}. Assume first that ||f||_{∞} ≤ 1. Let q_{n}(t) = a_{n,m}t^{m}+a_{n.m−1}t^{m−1}+ ⋯ + a_{n,o} be as in the above lemma. Then

q_{n}(f) = a_{n,m}f^{m} + a_{n.m−1}f^{m−1} + ⋯ + a_{n,o} ∈ \overline{A}

Hence lim_{n→∞} q_{n}(f) = |f| ∈ \overline{A} because q_{n}(t) converges uniformly to |t| on [0, 1]. For general f we have |f| = ||f||_{∞} (\frac{|f|}{||f||_{∞}} ). Hence |f| ∈ \overline{A}.

Let now f, g ∈ \overline{A}. Then

f ∧ g = \frac{1}{2}(f + g) − \frac{1}{2} |f − g| ∈ \overline{A}

f ∨ g = \frac{1}{2}(f + g) + \frac{1}{2} |f − g| ∈ \overline{A}

by the first part and the fact that with f, g ∈ \overline{A} we have f ± g ∈ \overline{A}.

Lemma 6. Let A be as in the Stone-Weierstrass Theorem. Let x, y ∈ X, x ≠ y, and let a, b ∈ ℝ with a ≤ b. Then there exists a f ∈ \overline{A} such that: f(x) = a and f(y) = b. Furthermore we can choose f such f(z) ≤ b for all z ∈ X.

Proof. Choose g ∈ A such that g(x) ≠ g(y). This is possible because A separates points. Let

f_{1}(t) := a + (b − a)\frac{g(t) − g(x)}{g(y) − g(x)} .

Then f ∈ A because A contains the constants. Furthermore f_{1}(x) = a and f_{1}(y) = b. We let now

f := (f_{1} ∧ b)

Then f ∈ \overline{A} because of the last lemma and f(x) = a because a ≤ b, f(y) = b and f(z) ≤ b.

Lemma 7. Let A be as before. Let ϵ > 0 and x ∈ X be given. Let g ∈ C(X). Then there exists a f ∈ \overline{A} such that

g(x) = f(x)
f(y) ≤ g(y) + ϵ , ∀y ∈ X

Proof. For z ∈ X let h_{z} ∈ \overline{A} be such that

h_{z}(x) = g(z)
h_{z}(z) = g(z) + ϵ/2

This is possible by the last lemma. Both h_{z} and g are continuous. Hence there exists an open subset U_{z} ∋ z such that

h_{z}(y) < g(y) + ϵ , ∀z ∈ U_{z}

We use now that X is compact. Hence there exists finitely many U_{i} = U_{z_{i}} such that X = ∪_{i=1}^{n}U_{i}. Let h_{i} = h_{x_{i}}, and let

f = h_{1} ∧ ⋯ ∧ h_{n} ∈ \overline{A}

Then f(x) = g(x). Let y ∈ X. Then there exists a i such that y ∈ U_{i}. Hence

f(y) ≤ h_{i}(y) < g(y) + ϵ

We can now proof the Stone-Weierstrass Theorem: Let g ∈ C(X). Let ϵ > 0 be given. For each x ∈ X we choose f_{x} ∈ \overline{A} such that

g(x) = f_{x}(x)
f_{x}(y) < g(y) + ϵ

By the first equation there is an open V_{x} ∋ x such that

f_{x}(y) > g(y) − ϵ , ∀y ∈ V_{x} .

Again we use that X is compact to find finitely many V_{i} = V_{x_{i}} such that X = ∪_{i=1}^{k}V_{i}. Let f_{i} = f_{x_{i}} and

f = f_{1} ∨ ⋯ ∨ f_{k}.

Let z ∈ X. Let i be such that z ∈ V_{i}. Then

g(z) − ϵ < f_{i}(z) < g(z) + ϵ .

Hence

g(z) − ϵ ≤ f(z) ≤ g(z) + ϵ .

It follows that ||f − g||_{∞} ≤ ϵ, which shows that \overline{A} is dense.

1.3. Linear Operators. Let V and W be two vector spaces. A map T : V → W is linear if

T (λx + y) = λT (x) + T (x) , ∀λ ∈ℝ , ∀x, y ∈ V .

If V and W are normed vector spaces then the linear operator T is bounded if there exists a constant M > 0 such that

||T (x)|| ≤ M ||x|| , ∀x ∈ V .

If T is bounded we set

||T || := sup_{x∈X} \begin{matrix}| & |T (x)| & | \\ | & |x| & | \end{matrix} ≤ M < ∞ .

The number ||T || is called the norm of T. A simple calculation shows that

||T || = sup_{||x||=1} ||T (x)|| = sup_{||x||≤1} ||T (x)||

Lemma 8. If T : V → W and S : W → Z are both bounded. Then the composition ST : V → Z. is also bounded.

Proof. Let x ∈ V. Then ||(ST)(x)|| ≤ ||S(T x)|| ≤ ||S|| ||T (x)|| ≤ ||S|| ||T || ||x||.

A bounded operator T : V → W is called invertible if there exists a bounded linear operator S : W → V such that

ST = I_{V} and T S = I_{W} .

Here I_{X} stands for the identity map x ↦ x. The kernel Ker(T) of a linear operator is

Ker(T) = {x ∈ V | T (x) = 0} .

Ker(T) is a subspace of V.

Example 3. In the finite dimensional case a linear map T : V → V then the following three properties are equivalent:

1. T is invertible;

2. T is injective;

3. T is surjective.

This is different in the infinite dimensional case. Let 1 ≤ p < ∞. Let V = ℓ_{p}. Define the shift operators S and T by:

S((r_{n})) := (0, r_{0}, r_{1}, …)

and

T ((r_{n})) := (r_{1}, r_{2}, r_{3}, …)

Then S is injective, T is surjective, and T S = I. But the kernel of ST is exactly

Ker(T S) = {(r, 0, 0, …) | r ∈ ℝ} .

Proposition 1. A bounded linear operator is uniformly continuous. If a linear operator is continuous at one point it is bounded.

Let V, W be normed spaces. Let B(V, W) be the space of bounded operators from V to W. Let S, T ∈ B(V, W) and x ∈ V. Then

||(S + T)(x)|| = ||S(x) + T (x)||
≤ ||S(x)|| + ||T (x)||
≤ (||S|| + ||T ||) ||x||

Hence ||S + T || ≤ ||S|| + ||T ||. If ||T || = 0 then ||T (x)|| = 0 for all x ∈ V. Hence T = 0. We also have ||λT || = |λ| ||T ||. Hence ||·|| is a norm on B(V, W).

Proposition 2. Let V, W be normed spaces. If W is a Banach space, then B(V, W) is also a Banach space.

1.4. Linear Functionals and the Hahn-Banach Theorem. Before we proof the Hahn-Banach Theorem recall the Hausdorff Maximum Principle: A relation R on a set X is antisymmetric and transitive relation on a set X is called a partial ordering (on X). Partial ordering will usually be denoted by ≺. A partial ordering ≺ is called a linear ordering if for any two elements x, y ∈ X we have x ≺ y or y ≺ x. If x ≺ y then we say that x precedes y. Sometimes we say in this case that x is smaller than y. If ≺ is a linear a partial ordering x ⪯ y means that either x ≺ y or x = y. The following maximum principle is equivalent to the axiom of choice:

Axiom 1 (Hausdorff Maximal Principle). Let ≺ be a partial ordering on a set X. Then there is a maximal linearly ordered subset S ⊂ X

Proposition 3 (Hahn-Banach). Let p be a real-valued function defined on the vector space V satisfying

p(x + y) ≤ p(x) + p(y) and p(λx) = λp(x)

for all x, y ∈ V and all λ ≥ 0. Suppose that f is a linear functional on defined on an subspace W ⊂ V, and satisfying

f(x) ≤ p(x) , ∀x ∈ W .

Then there is a linear functional F defined on V such that

F (x) ≤ p(x) , ∀x ∈ V

and such that

F (x) = f(x) , ∀x ∈ W .

Proof. Let X be the set of all pair (S, g) such that S ⊂ V is a subspace containing W and g : S → ℝ is linear such that g(x) ≤ p(x) for all x ∈ S and g|W = f. Then X ≠ ∅ because (W, f) ∈ X. Define a linear ordering ≺ on X by

(S, g) ≺ (R, h) ⇔ S ⊂ R and h|S = g

By the maximum principle there is a maximal linearly ordered family S in X. Let S = {(S_{i}, g_{i})}_{i∈I}. Let

S := ⋃_{i∈I} S_{i} .

We claim that S is a subspace of V. Let x, y ∈ S, λ ∈ ℝ. There exists S_{i} and S_{j} such that x ∈ S_{i} and y ∈ S_{j}. Because S is linearly ordered we can assume that i = j. Hence λx + y ∈ S_{i} ⊂ S.

Define g : S → ℝ in the following way: If x ∈ S then there is a S_{j} such that x ∈ S_{j}. Let

g(x) := g_{j}(x) .

We have to show that g is well defined. Assume that we also have x ∈ S_{k}. Then we can assume that S_{k} ⊂ S_{j}. Hence g_{j}|S_{j} = g_{k}. In particular g_{j}(x) = g_{k}(x). That g is proved in the same way as S is a vector space. By construction we have

∀(R, h) ∈ S : (R, h) ⪯ (S, g)

Because of the maximality of S it follows that (S, g) ∈ S. As in the book it is now shown that actually S = V. Hence the theorem.

Proposition 4. Let the notation be as in the Hahn-Banach Theorem. Let G be an Abelian semigroup of linear operators on V such for every A ∈ G we have

p(Ax) ≤p(x) ∀x ∈ V

and

f(Ax) = f(x) ∀x ∈ W

Then there exists a linear extension of f such that F (Ax) = F (x) for all x ∈ V.

Proposition 5. Let 0 ≠ x ∈ V, then there exists a bounded linear functional f : V →ℝ such that

− ||y|| ≤ f(y) ≤ ||y|| ∀y ∈ V

and f(x) = ||x||. In particular ||f|| = 1.

Proof. As in the book.

Lemma 9. Let V be a normed vector space. Then the set of bounded linear functionals on V separates points in V.

Proof. Let s, t ∈ V, such that s ≠ t. Take x = s − t. Then the above lemma says that there exists a linear functional f such that f(s − t) ≠ 0. But that is the same as f(s) ≠ f(t).

Let X be a metric space and A ⊂ X. For y ∈ X we define

d(y, A) := inf_{a∈A} d(y, a).

Then 0 ≤ d(y, A) < ∞. We call the number d(y, A) the distance of y to A. Assume that d(y, A) = 0. Then we can find a sequence {a_{n}}_{n} in A such that

d(y, a_{n}) → 0 n → ∞

But then a_{n} → y and y ∈ \overline{A}. Hence

\overline{A} = \{y ∈ X | d(y, A) = 0\} .

In particular we have defined the notation of a the distance of a point from a subspace in V.

Proposition 6. Let W ⊂ V be a subspace. Let y ∈ V with d(y, W) ≥ δ >0. Then there is a f : V → ℝ such that f(y) = δ, f|W = 0 and ||f|| ≤ 1.

Let V be a normed vector space. The dual V^{∗} of V is the Banach space of all bounded linear maps V → ℝ. If v ∈ V and f ∈ V^{∗} then we will sometimes write

< f, x >= f(x)

The dual V^{∗∗} of V is defined as V is a Banach space. Define a map φ : V → V^{∗∗} by

< φ(x), f >=< f, x > .

Then φ is linear. Assume that φ(x) =0. Then < f, x >=0 for all f ∈ V^{∗}. By the above lemma it follows that x = 0. Thus φ is injective. We have

|< φ(x),f >| = |< f, x >|
≤ ||f|| ||x||

Hence ||φ(x)|| ≤ ||x||. So in particular ||φ|| ≤ 1. But if x ≠ 0 then there exists a f ∈ V^{∗} of norm ||f|| = 1 such that f(x) = ||x||. Hence

< φ(x), f >= ||x||

which implies that ||φ(x|| = ∥x∥ and ∥φ∥ = 1.

Definition 4. Let V, W be normed spaces. Let T : V → W be linear. Then T is an isometry if ||T (x)|| = ||x|| for all x ∈ V. We say that V and W are isometric if there exists an isometry of V onto W.

Let us remark the following:

1. If T is an isometry, then T is injective.

2. If T is an isometry, then ||T || = 1.

3. If T is an isometry onto W then T is invertible and T^{−1} : W → V is also an isometry. (Define S : W → V by

S(y) = x if T (x) = y

S is well defined because T is bijective. It is also clear that S is linear. Furthermore we have

||S(y)|| = ||x|| = ||T (x)|| = ||y||

Hence S is an isometry. We have S(T (x)) = x be definition. We also have T (S(y)) = T (x) = y. Hence T is invertible with inverse S.

What we have just seen is the following:

Lemma 10. φ is an isometry of V into V^{∗∗}.

We know that V^{∗∗} is a Banach space. Hence \overline{φ(V)} is also a Banach space. Thus:

Lemma 11. Let V be a normed vector space. Then there exists a Banach space B and an isometry T : V → B such that T (V) is dense in B, i.e., each normed vector space can be isometrically embedded into a Banach space.

Definition 5. The normed vector space V is reflexive if φ : V → V^{∗∗} is an isometric isomorphism.

Notice that if V is reflexive, then V is a Banach space.

Example 4. The spaces L^{p}([0, 1]) are reflexive for 1 < p < ∞: Recall that (L^{p})^{∗} ≃ L^{q} if 1 ≤ p < ∞ and 1/p + 1/q = 1. Thus for 1 < p < ∞ we get

L^{p∗∗} ≃ L^{q∗} ≃ L^{p}

Hence L^{p} is reflexive for 1 < p < ∞.

For p = 1 we have L^{1∗∗} = L^{∞∗} ⊃ L^{1}. We claim that L^{∞∗} ≠ L^{1}. Let V = L^{∞}([0, 1]) and let W =C([0, 1]). Then W is a closed subspace of V. Define δ ∈ W^{∗} by

< δ, g >= g(0)

Then δ(g) ≤ ||g||_{∞} =: p(g). By the Hahn-Banach theorem Theorem there exists a F ∈ L^{∞}([0, 1])^{∗} such that ||F || ≤ 1 and F (g) = g(0) for each g ∈ C([0, 1]).

Suppose we have y ∈ L^{1}([0, 1]) such that

< F, g >= ∫_{0}^{1} g(t)y(t) dt ∀g ∈ L^{∞}([0, 1])

Define x_{n}(t) := (1−t)^{n}. Then pointwise (and in L^{1}) x_{n} → 0. We have |x_{n}(t)y(t)| ≤ |y(t)| and y is integrable. We have < F, x_{n} >= x_{n}(0) = 1 for all n. Hence

1 = lim < F, x_{n} >
= lim ∫_{0}^{1} x_{n}(t)y(t) dt
= ∫_{0}^{1} lim x_{n}(t)y(t) dt

= 0

which is impossible. Hence F ∉ φ(L^{1}([0, 1])) ⊂ L^{1}([0, 1])^{∗∗} = L^{∞}([0, 1])^{∗}. Hence L^{1}([0, 1]) is not reflexive.

2. Baire Category Theorem

Definition 6. Let X be a topological space and let E ⊂ X.

1. E is nowhere dense if \overline{E}^{o} = ∅. (This is the same as \overline{X \ \overline{E}} = X \ \overline{E}^{o} = X, or X \ \overline{E} is dense.)

2. E is of the first category if and only if E is a finite or countable union of nowhere dense sets.

3. E is of the second category if E is not of the first category.

Example 5. Let a ∈ ℚ. Then E_{p} := {p} is nowhere dense. It follows that ℚ = ∪_{q∈ℚ}E_{q} is of the first category, even if ℚ is dense in ℝ.

Lemma 12. If E is of the second category, then \overline{E} contains an open set.

Proof. If \overline{E} does not contain an open set, then \overline{E}^{o} = ∅. Thus E would be nowhere dense.

Theorem 3 (Bair). Let X be a complete metric space and {O_{k}} a countable collection of dense open sets. Then ⋂ O_{k} is dense.

Theorem 4 (Bair Category Theorem). Let X be a complete metric space. Then no nonempty open subset of X is of first category.

An application of the theorem is the following uniformly boundness principle:

Theorem 5. Let X be a metric space and let F ⊂ C(X). Suppose that for each x ∈ X there is a number M_{x} such that |f(x)| ≤ M_{x} for all f ∈ F. Then there is a nonempty set O ⊂ X and a constant M > 0 such that

|f(x)| ≤ M ∀f ∈ F , ∀x ∈ O .

Proof. For N ∈ ℕ let

E_{N} = {x ∈ X | ∀f ∈ F : |f(x)| ≤ N}

Then

E_{N} = ⋂_{f∈F} {x ∈ X | |f(x)| ≤ N}

a intersection of closed sets. Hence E_{N} is closed. Let x ∈ X and choose N ≥ M_{x}. Then x ∈ E_{N}. It follows that X = ∪E_{N}. As X is not of the first category it follows that one of the E_{N} contains an open set O. Hence |f(x)| ≤ N for all x ∈ O.

3. The Closed Graph Theorem

Definition 7. Let X and Y be topological spaces. A map f : X → Y is open if f(O) is open in Y for all O ⊂ X open.

The main result in this section is the following open mapping theorem:

Proposition 7. Let V and W be Banach spaces. A continuous surjective linear transformation T ∈ B(V, W) is open. In particular if T is bijective, then T is an isomorphism.

Let us name few applications before we proof this:

Proposition 8. Let V be a vector space which is complete in the norms ||·||_{1} and ||·||_{2}. Suppose there exists a constant C_{2} such that

||x||_{1} ≤ C_{2} ||x||_{2} ∀x ∈ V .

Then the norm are equivalent, i.e., there is a constant C_{1} such that

||x||_{2} ≤ C_{1} ||x||_{1} ∀x ∈ V

Proof. The identity map V ∋ x ↦ x ∈ V is a continuous map (V, ||·||_{2}) → (V, ||·||_{1}). Hence an isomorphs, thus the inverse (V, ||·||_{1}) → (V, ||·||_{2}), x ↦ x is bounded, which is exactly the statement.

Let T : V → W be linear map. Then the graph of T is the subspace G(T) ⊂ V × W defined by:

G(T) := {(v, T (v) | v ∈ V}

Proposition 9 (Closed Graph Theorem). Let V and W be Banach spaces, and let T : V → W be linear. Then T is continuous if and only if G(T) is closed.

Proposition 10 (Uniform boundness Theorem). Let V be a Banach space, and let W be a normed space. Let F ⊂ B(V, W) be such that for each v ∈ V there exists a constant M_{v} such that

T (v) ≤ M_{v} ||v|| ∀T ∈ F

Then the operators in F are uniformly bounded; i.e., there is a constant M such that

||T || ≤ M ∀T ∈ F .

12 GESTUR ÓLAFSSON

Department of Mathematics, LSU, Baton Rouge, LA 7083

E-mail address: olafsson@math.lsu.edu