Topological Injections and Topological Surjections | 6

ABSTRACT

We begin with the following: 6.1 Definition.

(Pietsch [1980, p.26]). Let E and F be normed spaces over K and T ∊ L(E,F). The injection modulus of T is given by (6.1.1) j ( T ) = sup { τ ≥ 0 : τ ‖ x ‖ ≤ ‖ Tx ‖ ( for all x ∈ E ) } , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq809.tif"/>

and the surjection modulus of T is given by (6.1.2) q ( T ) = sup { τ ≥ 0 : τ U F ⊂ Τ ( U E ) } . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq810.tif"/>

In particular, if T = O (the zero operator) then we put j ( O ) = 0 and q ( O ) = 0. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq811.tif"/>

Remark

It is easily seen that j ( T ) = inf { ‖ Tx ‖ : ‖ x ‖ = 1 } ( for all T ∈ L ( E,F ) ) . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq812.tif"/>

It is also clear that j ( T ) ≤ ‖ T ‖ and q ( T ) ≤ ‖ T ‖ ( for all T ∈ L ( E,F ) ) . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq813.tif"/>

and that q(T) > 0 if and only if T is open.

Most of the results of this section are taken from the excellent book written by Pietsch [1980].

6.a A characterization of injection modulus:

Let E and F be normed spaces and T ∊ L(E,F). Then the following statements are equivalent:

j(T) ≥ τ > 0.

T is one–one and τ T − 1 ( U F ) ⊂ U E https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq814.tif"/> .

T is one–one and ( τ U F ) ∩ ( TE ) ⊂ TU E https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq815.tif"/> (i.e. T is relatively open).

99T has a bounded inverse T^–1:T(E) → E.

As an application of Banach’s open mapping theorem, the surjection modulus can be calculated by the following:

6.2 Lemma

Let E and F be Banach spaces and T ∊ L(E,F). Then q ( T ) = sup { τ ≥ 0 : τ U F ⊂ Τ ( U E ) ¯ } . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq816.tif"/>

Proof.

Suppose that q ( T ) ¯ = sup { τ ≥ 0 : τ U F ⊂ Τ ( U E ) ¯ } . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq817.tif"/>

Then q ( T ) ≤ q ( T ) ¯ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq818.tif"/> . To prove that q ( T ) ¯ ≤ q ( T ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq819.tif"/> , it suffices to show that for any β > 0 with β U F ⊂ T ( U E ) ¯ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq820.tif"/> , we have β ≤ q(T).

In fact, let β be such a number. Then T ( U E ) ¯ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq821.tif"/> is a 0–neighbourhood in F for the norm–topology, hence T is almost open, thus T open (by (4.5) (b)); consequently, δT(U_E) is a 0–neighbourhood in F (for any δ > 0). As β U F ⊂ Τ ( U E ) ¯ = ∩ μ > 0 { Τ ( U E ) + μ U F } = ∩ δ > 0 { T ( U E ) + δ Τ ( U E ) } ⊂ ( 1 + δ ) Τ ( U E ) ( for all δ > 0 ) , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq822.tif"/>

we conclude that β 1 + δ ≤ q ( T ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq823.tif"/> , and hence that β ≤ q(T) (since δ was arbitrary), which obtains our requirement.

Remark.

From the proof of the preceding result, we see that if E and F are only assumed to be normed spaces, then T ∊ L(E,F) is almost open if and only if sup { τ ≥ 0 : τ U F ⊂ Τ ( U E ) ¯ } > 0. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq824.tif"/>

6.3 Defintion

100Let E and F be normed spaces over K and T ∊ L(E,F). We say that T is a :

topological injection or isomorphic embedding), denoted by T : E >→F, if j(T) > 0;

metric injection if j ( T ) = 1 = ‖ T ‖ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq825.tif"/> .

topological surjection, denoted by T : E ↠ F, if q(T) > 0;

metric surjection if q ( T ) = 1 = ‖ T ‖ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq826.tif"/> .

Remark

Suppose that T ∊ L(E,F). Then T is a topological surjection if and only if it is open; T is an isomorphism if and only if it is a topological injection and topological surjection; and T is a metric isomorphism (i.e. isometry) if and only if it is a metric injection and a metric surjection.

6.4 Proposition

Let E and F be Banach spaces and T ∊ L(E,F). Then the following statements are equivalent:

(a) T is a topological injection.

(b) T is one–one and has a closed range T(E).

(c) T admits a factorization https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/page100.tif"/>

where T₀ : E → T(E) is an isomorphism and J_T(E) is the embedding map.

In this case, we have (6.4.1) j ( T ) = ‖ T 0 − 1 ‖ − 1 . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq827.tif"/>

<target id="page_101" target-type="page">101</target>Proof.

(a) ⇒ (b): As j(T) > 0, for any a with 0 < α < j(T), there exists an r > 0 such that (1) r>α and r ‖ x ‖ ≤ ‖ Tx ‖ ( for all x ∈ E ) . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq828.tif"/>

It then follows that T is one–one [since Tx = 0 ⇒ x = 0].

To prove the closedness of T(E), let T_{x_n} → y in F. Then { x_n } is a Cauchy sequence in E(by (1)), hence converges to some x ∊ E (by the completeness of E), thus lim n Tx n = y = Tx ∈ T ( E ) . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq829.tif"/>

(b) ⇒ (c): By Banach’s open mapping theorem, the statement (b) ensures that the map T₀, defined by T 0 x = Tx ( for all x ∈ E ) , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq830.tif"/>

is an isomorphism from E onto the Banach space T(E). Thus the implication follows.

(c) ⇒ (a): Since T₀ has a bounded inverse T 0 − 1 : T ( E ) → E https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq831.tif"/> , it follows that ‖ x ‖ = ‖ T 0 − 1 ( Tx ) ‖ ≤ ‖ T 0 − 1 ‖ ‖ Tx ‖ ( for all x ∈ E ) , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq832.tif"/>

and hence that (2) j ( T ) = ‖ T 0 − 1 ‖ − 1 > 0. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq833.tif"/>

Therefore T is a topological injection.

Finally, we assume that T is a topological injection. Then j ( T ) ≥ ‖ T 0 − 1 ‖ − 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq834.tif"/> (by (2)). In order to verify (6.4.1), it suffices to show that for any r > 0 with r ‖ x ‖ ≤ ‖ Tx ‖ ( x ∈ E ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq835.tif"/> , we have r ‖ T 0 − 1 ‖ − 1 ≤ 1. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq836.tif"/>

102Indeed, let r > 0 be such a number. Then r ‖ T 0 − 1 ‖ − 1 = r sup { ‖ T 0 − 1 ( Tx ) ‖ : ‖ Tx ‖ ≤ 1 } = sup { r ‖ x ‖ : ‖ Tx ‖ ≤ 1 } ≤ sup { ‖ Tx ‖ : ‖ Tx ‖ ≤ 1 } ≤ 1 , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq837.tif"/>

which obtains our assertion.

6.5 Proposition

Let E and F be Banach spaces and T ∊ L(E,F). Then the following statements are equivalent:

(a) T is a topological surjection.

(b) T is onto.

(c) T admits a factorization https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/page102.tif"/>

where T₀ : E/Ker T → F is an isomorphism and Q is the quotient map.

In this case, we have (6.5.1) q ( T ) = ‖ T 0 − 1 ‖ − 1 . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq838.tif"/>

Proof.

(a) ⇒ (b): Trivial (Open mappings must be onto).

(b) ⇒ (c): Follows from Banach’s open mapping theorem and the fact that T₀ is continuous and open (since T is onto and E/Ker T and F are Banach spaces).

Finally, to prove (6.5.1), we first notice that if τ > 0 is such that τ U F ⊂ T ( U E ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq839.tif"/> , then ‖ T 0 − 1 ‖ − 1 = sup { ‖ T 0 − 1 y ‖ : y ∈ U F } ≤ sup { ‖ T 0 − 1 ( τ − 1 Tx ) ‖ : x ∈ U E } = τ − 1 sup { ‖ Qx ‖ : x ∈ U F } ≤ τ − 1 , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq840.tif"/>

103hence τ ≤ ‖ T 0 − 1 ‖ − 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq841.tif"/> , and consequently, q ( T ) ≥ ‖ T 0 − 1 ‖ − 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq842.tif"/> . Conversely, let r = ‖ T 0 − 1 ‖ − 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq843.tif"/> . We claim that rO F ⊂ T ( U E ) , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq844.tif"/> ,

so that rU F = rO F ⊆ T ( U E ) ¯ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq845.tif"/> ; consequently, r ≤ q(T).

In fact, for any y ∊ O_F, there is an x ∊ E with ry = Tx [since T is onto]. As T = T₀Q and T₀ is an isomorphism, it follows from ry = Tx that rT 0 − 1 y = T 0 − 1 x = Qx https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq846.tif"/> , and hence that ‖ Qx ‖ = r ‖ T 0 − 1 y ‖ ≤ ‖ T 0 − 1 ‖ ‖ y ‖ = ‖ y ‖ < 1. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq847.tif"/>

Notice that Q ( O E ) = O E / Ker T https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq848.tif"/> . There exists an u ∊ O_E such that Qx = Qu; it then follows from T = T ∘ ∘ Q https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq849.tif"/> and ry = Tx that ry = Tx = ( T o Q ) x = T o ( Qu ) = Tu ∈ T ( O E ) ⊂ T ( O E ) , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq850.tif"/> which obtains our assertion.

For criteria of topological injections (resp. topological surjections) between normed spaces, we quote the following two results:

6.b Topological injections and subspaces (see <xref ref-type="bibr" rid="ref54">Junek [1983]</xref>):

Let E and F be normed spaces and T ∊ L(E, F). Then the following statements are equivalent

(i) T is a topological injection.

104(ii) T is one–one and relatively open.

(iii) (Universal property of topological injections and subspaces) T is one–one, and for any normed space G and any S ∊ L(G,F) with S ( G ) ⊂ Τ ( E ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq851.tif"/>

there exists exactly one S₀ ∊ L(G,E) such that S = TS₀ (i.e., S can be factorized through T from the left).

6.c Universal property of topological surjections and quotients (see <xref ref-type="bibr" rid="ref54">Junek [1983]</xref>):

Let E and F be normed spaces and T ∊ L(E,F). Then T is a topological surjection (i.e., open) if and only if T is onto, and for any normed space F₀ and any S ∊ L(E,F₀) with Ker T ⊂ Ker S, https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq852.tif"/>

there exists exactly one R₀ ∊ L(F,F0) such that S = R₀T (i.e. S can be factorized through T from the right).

We now describe the duality relations between topological injections and topological surjections as follows:

6.6 Theorem (<xref ref-type="bibr" rid="ref86">Pietsch [1980]</xref>)

Let E and F be Banach spaces and T ∊ L(E,F). Then j ( T ) = q ( T ′ ) and j ( T ′ ) = q ( T ) . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq853.tif"/>

In particular, T is a topological injection (resp. metric injection) if and only if its dual operator T' is a topological surjection (resp. metric surjection); T is a topological surjection (resp. metric surjection) if and only if T' is a topological injection (resp. metric injection). Consequently, T is an isomorphism (resp. metric isomorphism) if and only if T' 105is an isomorphism (resp. metric isomorphism).

Proof.

The equality j(T) = q(T') clearly follows from (1) τ ‖ x ‖ ≤ ‖ Tx ‖ ( x ∈ E ) ⇔ τ U E ′ ⊂ T ′ ( U F ′ ) . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq854.tif"/>

Thus, we are going to show that (1) is true.

In fact, let τ > 0 be such that τ ‖ x ‖ ≤ ‖ Tx ‖ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq855.tif"/> (x ∊ E). For any u ′ ∈ U E https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq856.tif"/> , the functional f, defined by < Tx,f> = <x, u ′ > ( x ∈ E ) , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq857.tif"/>

is a linear functional on T(E) (since T is one–one). As τ ‖ x ‖ ≤ ‖ Tx ‖ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq858.tif"/> (x ∊ E), it follows that | < Tx,f> | = | < x, u ′ > | ≤ ‖ x ‖ ‖ u ′ ‖ ≤ ‖ x ‖ τ − 1 ‖ Tx ‖ ; https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq859.tif"/>

in other words, f is continuous with ‖ f ‖ T ( E ) ≤ τ − 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq860.tif"/> . By Hahn–Banach’s extension theorem, f has an extension y′ ∊ F' with ‖ y ′ ‖ = ‖ f ‖ T ( E ) ≤ τ − 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq861.tif"/> . In particular, < x,T' y ′ > = < Tx, y ′ > = < Tx,f > = < x, u ′ > ( x ∈ E ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq862.tif"/>

thus u′ = T' y′ and τ u ′ = T ' ( τ y ′ ) ∈ T ' ( U F ′ ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq863.tif"/> , which shows that τ U E ′ ⊂ T ′ ( U F ′ ) . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq864.tif"/>

Conversely, let τ > 0 be such that τ U E ′ ⊂ T ′ ( U F ′ ) . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq865.tif"/> For any x ∊ E, we have τ ‖ x ‖ = τ sup { | < x, u ′ | : u ′ ∈ U E ′ } ≤ τ sup { | < x, τ ′ T ′ v ′ > | : v ′ ∈ U F ′ } = sup { | < Tx, v ′ > | : v ′ ∈ U F ′ } = ‖ Tx ‖ . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq866.tif"/>

106Therefore (1) holds.

By (6.2), the equality j(T') = q(T) follows from (2) τ U F ⊂ Τ ( U E ) ¯ ⇔ τ ‖ v ′ ‖ ≤ ‖ T ′ v ′ ‖ ( v ′ ∈ F ′ ) . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq867.tif"/>

We are going to verify (2).

In fact, let τ > 0 be such that τ U F ⊂ Τ ( U E ) ¯ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq868.tif"/> .For any v′ ∊ F', we have τ ‖ v ′ ‖ = τ sup { | < y, v ′ > | : y ∈ U F } ≤ τ sup { | τ − 1 < Tx, v ′ > | : x ∈ U E } = sup { | < x,T' v ′ > | : x ∈ U E } = ‖ T' v ′ ‖ . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq869.tif"/>

Conversely, let τ > 0 be such that (3) τ ‖ v ′ ‖ ≤ ‖ T' v ′ ‖ . ( v ′ ∈ F' ) . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq870.tif"/>

To prove that τ U F ⊂ Τ ( U E ) ¯ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq871.tif"/> , we assume on the contrary that there is an y ∊ U_F such that τ y ∉ T ( U E ) ¯ . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq872.tif"/>

Then the strong separation theorem (see(8.3) in §8) ensures that there is an g ∊ F' such that sup { | < Tx,g | : x ∈ U E } ≤ 1 < | < τ y,g> | . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq873.tif"/>

Therefore ‖ T'g ‖ = sup { | < x,T'g> | : x ∈ U E } ≤ 1 < | < τ y,g> | ≤ τ ‖ y ‖ ‖ g ‖ ≤ τ ‖ g ‖ , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq874.tif"/>

which contradicts (3).

6.d <target id="page_107" target-type="page">107</target>The composition of topological injections:

Let E, F and G be Banach spaces and given the following two operators: E → S F → T G https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq875.tif"/>

Then TS is a topological injection if and only if

S : E → F is a topological injection, and

( TJ S ( E ) F :S ( E ) → G https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq876.tif"/> is a topological injection.

Dually, one can verify the following:

6.e The composition of topological surjections:

Then TS is a topological surjection if and only if

T: F → G is a topological surjection, and

Q Ker T F S:E → F / Ker T https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq878.tif"/> is a topological surjection.

As other important applications of (3.j), we have the following:

6.f Extension property and lifting property:

A Banach space F is said to have the extension property (resp. metric extension property), if for any Banach spaces E and E₀, any topological injection (resp. metric injection), J : E₀ >→ E and any S₀ ∊ L(E₀,F) there exists an extension S ∊ L(E.F) such that S 0 = SJ ( resp .S 0 =SJ and ‖ S ‖ = ‖ S 0 ‖ ) . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq879.tif"/>

Dually, a Banach space E is said to have the lifting property (resp. metric lifting property) 108if for any Banach spaces F and F₀, any topological surjection (resp. metric surjection and ∊ > 0)Q:F↠F₀ and any T₀ ∊ L(E,F₀) there exists a lifting T ∊ L(E,F) such that T 0 = QT ( resp . T 0 = QT and ‖ T ‖ ≤ ( 1 + ϵ ) ‖ T 0 ‖ ) . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq880.tif"/>

For any index set Λ, the space ℓ ∞ ( Λ ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq881.tif"/> has the metric extension property; while the space ℓ 1 ( Λ ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq882.tif"/> has the metric lifting property.

Let E be a Banach space. The map J E : E > → ℓ ∞ ( U E ′ ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq883.tif"/> , defined by J E x = [ < x , f > , f ∈ U E ′ ] ( x ∈ E ) , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq884.tif"/>

is a metric injection from E into ℓ ∞ ( U E ′ ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq885.tif"/> . (Consequently, any Banach space is metrically isomorphic to a subspace of some Banach space with the metric extension property.) The map Q E : ℓ 1 ( U E ′ ) ↠ E https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq886.tif"/> , defined by Q E ( [ ξ x , x ∈ U E ] ) = ∑ U E ξ x ⋅ x ( [ ξ x , x ∈ U E ] ∈ ℓ ′ ( U E ) ) , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq887.tif"/>

is a metric surjection from ℓ 1 ( E ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq888.tif"/> onto E. (Consequently, any Banach space is metrically isomorphic to a quotient space of some Banach space with the metric lifting property.)

E ≺ ℓ ∞ ( U E ′ ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq889.tif"/> if and only if E has the extension property.

E ≺ ℓ 1 ( Λ ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203749807/1e575708-9d79-4f43-a374-4d458ec8fff6/content/eq890.tif"/> (for some Λ) if and only if E has the lifting property.