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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 ,c + ] (0 , 1) for some > 0. By definition of c , this yields a contradiction c a = c . In the case ii) f ( c ) < 0, we have f < 0 on [ a,b ] := [ c ,c + ] (0 , 1), and we again get a contradiction c b = c + . Therefore, we must have f ( c ) = 0. 2. Let ( X 1 , 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 1 ( x j k ,x ) 0 as k . Let f be a continuous mapping from ( X 1 , 1 ) to another metric space ( X 2 , 2 ). Show that f is uniformly continuous on X 1 , i.e. > , > , such that from x,y X 1 ; 1 ( x,y ) < it follows 2 ( f ( x ) ,f ( y )) < . Proof. Suppose that the above property is not true, i.e. > 0 such that > 0, there are x,y X 1 with 1 ( x,y ) < and 2 ( f ( x...
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 Fall '08
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 Math

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