bmhicks_homework_5

# bmhicks_homework_5 - Brandy Hicks October 1, 2007 MTHSC...

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Unformatted text preview: Brandy Hicks October 1, 2007 MTHSC 974.001 - Homework # 5 1. Let f : R R satisfy lim | x | f ( x ) = . Assume f is bounded below. a. Prove the minimizing sequence must be bounded. Proof: Let f : R R . Suppose f is bounded below, non-empty and lim | x | f ( x ) = . Then by Theorem 2 . 1, f has a minimal element, inf( f ) R . Now, by the Infimum Tolerance Lemma, we let E = 1 n . Then x n dom( f ) 3 inf ( f ( x )) f ( x n ) &lt; inf ( f ( x )) + 1 n i.e.: We can construct a sequence of points { x n } dom( f ) 3 x 1 dom( f ) 3 inf ( f ( x )) f ( x 1 ) &lt; inf ( f ( x )) + 1 x 2 dom( f ) 3 inf ( f ( x )) f ( x 2 ) &lt; inf ( f ( x )) + 1 2 x 3 dom( f ) 3 inf ( f ( x )) f ( x 3 ) &lt; inf ( f ( x )) + 1 3 . . . x n dom( f ) 3 inf ( f ( x )) f ( x n ) &lt; inf ( f ( x )) + 1 n Hence, as | x | , f ( x n ) inf ( f ( x )). Now, if { x n } was not bounded, we could extract a subsequence n x n o so that | x n | + . Then by assumption, f ( x n ) . But since f ( x n ) inf ( f ( x )), then f ( x n ) inf ( f ( x )), so ( ) we have reached a contradiction. The minimizing sequence { x n } , must be bounded. b. What additional assumptions do you need to put on f to make sure an absolute minimum exists? Answer: We need x n x dom( f ). Then Lower semi-continuity ( lsc ) of f ( x ) f ( x ) = inf ( f ). So if we assume lsc of f , an absolute minimum exists. c. Give an example of this kind of f . Answer: f ( x ) = p | x | . 1 2. Let f : R R satisfy lim | x | f ( x ) | x | = . Assume f is bounded below. a. Prove the minimizing sequence must be bounded. Proof: Let f : R R . Suppose f is bounded below, non-empty and lim | x | f ( x ) | x | = . Now, since our f is bounded below, we know by Theorem 2 . 1 that f has a minimal element, inf( f ) R . Then, by the Infimum Tolerance Lemma, we choose E 3 E = 1 n . Then x n dom( f ) 3 inf ( f ( x )) f ( x n ) &lt; inf ( f ( x )) + 1 n i.e.: We can construct a sequence of points { x n } dom( f ) 3 x 1 dom( f ) 3 inf ( f ( x )) f ( x 1 ) &lt; inf ( f ( x )) + 1 x 2 dom( f ) 3 inf ( f ( x )) f ( x 2 ) &lt; inf ( f ( x )) + 1 2 x 3 dom( f ) 3 inf ( f ( x )) f ( x 3 ) &lt; inf ( f ( x )) + 1 3 . . . x n dom( f ) 3 inf ( f ( x )) f ( x n ) &lt; inf ( f ( x )) + 1 n Hence, as | x | , f ( x n ) inf ( f ( x )). Now, if { x n } was not bounded, we could extract a subsequence n x n o so that | x n | + . Then by assumption, f ( x n ) . But since f ( x n ) inf ( f ( x )), then f ( x n ) inf ( f ( x )), so ( ) we have reached a contradiction....
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## bmhicks_homework_5 - Brandy Hicks October 1, 2007 MTHSC...

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