s2 - Math 8601 REAL ANALYSIS Fall 2010 Homework#2 Problems...

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Unformatted text preview: Math 8601: REAL ANALYSIS. Fall 2010 Homework #2. Problems and Solutions. #1. Let f ( x ) be a continuous function on [- 1 , 1], such that f (- 1) < < f (1) . Show that f ( c ) = 0 for some c ∈ (- 1 , 1). Proof. Take c := sup { x ∈ [- 1 , 1] : f ( x ) ≤ } ∈ (0 , 1). We claim that f ( c ) = 0. Indeed, if f ( c ) 6 = 0, then we have one of two possible cases: (i) f ( c ) > 0, or (ii) f ( c ) < 0. Since f is continuous, in the case (i) we have f > 0 on [ a,b ] := [ c- δ,c + δ ] ⊂ (0 , 1) for some δ > 0. By definition of c , this yields a contradiction c ≤ a = c- δ . In the case ii) f ( c ) < 0, we have f < 0 on [ a,b ] := [ c- δ,c + δ ] ⊂ (0 , 1), and we again get a contradiction c ≥ b = c + δ . Therefore, we must have f ( c ) = 0. 2. Let ( X 1 ,ρ 1 ) be a compact metric space, i.e. any sequence { x j } ⊂ X 1 contains a convergent subsequence { x j k } → x ∈ X 1 , which means ρ 1 ( x j k ,x ) → 0 as k → ∞ . Let f be a continuous mapping from ( X 1 ,ρ 1 ) to another metric space ( X 2 ,ρ 2 ). Show that f is uniformly continuous on X 1 , i.e. ∀ ε > , ∃ δ > , such that from x,y ∈ X 1 ; ρ 1 ( x,y ) < δ it follows ρ 2 ( f ( x ) ,f ( y )) < ε. Proof. Suppose that the above property is not true, i.e. ∃ ε > 0 such that ∀ δ > 0, there are x,y ∈ X 1 with ρ 1 ( x,y ) < δ and ρ 2 ( f ( x...
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s2 - Math 8601 REAL ANALYSIS Fall 2010 Homework#2 Problems...

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