14-15-ban-cont

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

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

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