14-15-ban-cont

# 14-15-ban-cont - Lectures 14-15 5 Banach spaces of...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Lectures 14-15 5. Banach spaces of continuous functions In this section we discuss a examples of Banach spaces coming from topology. Notation. Let K be one of the fields R or C , and let Ω be a topological space. We define C K b (Ω) = { f : Ω → K : f bounded and continuous } . In the case when K = C we use the notation C b (Ω). Proposition 5.1. With the notations above, if we define k f k = sup p ∈ Ω | f ( p ) | , ∀ f ∈ C K b (Ω) , then C K b (Ω) is a Banach space. Proof. It is obvious that C K b (Ω) is a linear subspace of ` ∞ K (Ω), and the norm is precisely the one coming from ` ∞ K (Ω). Therefore, it suffices to prove that C K b (Ω) is closed in ` ∞ K (Ω). Start with some sequence ( f n ) n ≥ 1 ⊂ C K b (Ω), which convergens in norm to some f ∈ ` ∞ K (Ω), and let us prove that f : Ω → K is continuous (the fact that f is bounded is automatic). Fix some point p ∈ Ω, and some ε > 0. We need to find some neighborhood V of p , such that | f ( p )- f ( p ) | < ε, ∀ p ∈ V. Start by choosing n such that k f n- f k < ε 3 . Use the fact that f n is continuous, to find a neighborhood V of p , such that | f n ( p )- f n ( p ) | < ε 3 , ∀ Ω ∈ V. Suppose now Ω ∈ V . We have | f ( p )- f ( p ) | ≤ | f n ( p )- f ( p ) | + | f n ( p )- f n ( p ) | + | f n ( p )- f ( p ) | ≤ | f n ( p )- f n ( p ) | + 2 sup q ∈ Ω | f n ( q )- f ( q ) | < 2 ε 3 + ε 3 = ε. A first application of Banach space techniques is the following: Lemma 5.1 (Urysohn type density) . Let Ω be a topological space, let C ⊂ C R b (Ω) be a linear subspace, which contains the constant function 1 . Assume (u) for any two closed sets A,B ⊂ Ω , with A ∩ B = ∅ , there exists a function h ∈ C , such that h A = 0 , h B = 1 , and h (Ω) ∈ [0 , 1] , for all Ω ∈ Ω . Then C is dense in C R b (Ω) , in the norm topology. 101 102 LECTURES 14-15 Proof. The key step in the proof will be the following: Claim : For any f ∈ C R b (Ω) , there exists g ∈ C , such that k g- f k ≤ 2 3 k f k . To prove this claim we define α = inf p ∈ Ω f ( p ) and β = sup p ∈ Ω f ( x ) , so that f ( p ) ⊂ [ α,β ], and k f k = max {| α | , | β |} . Define the sets A = f- 1 α, 2 α + β 3 and B = f- 1 α + 2 β 3 ,β . so that both A and B are closed, and A ∩ B = ∅ . Use the hypothesis, to find a function h ∈ C , such that h A = 0, h B = 1, and h ( p ) ∈ [0 , 1], for all p ∈ Ω. Define the function g ∈ C by g = 1 3 α 1 + ( β- α ) k . Let us examine the difference g- f . Start with some arbitrary point p ∈ Ω. There are three cases to examine: Case I: p ∈ A . In this case we have h ( p ) = 0, so we get g ( p ) = α 3 . By the construction of A we also have α ≤ f ( p ) ≤ 2 α + β 3 , so we get 2 α 3 ≤ f ( p )- g ( p ) ≤ α + β 3 ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 18

14-15-ban-cont - Lectures 14-15 5 Banach spaces of...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online