solutionshw4-555-2011

solutionshw4-555-2011 - Solutions homework 4. Page 108...

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Solutions homework 4. Page 108 Problem 4: Let ( x n ) be a Cauchy sequence and let ± > 0. Then there exists δ > 0 such that | x - y | < δ implies | f ( x ) - f ( y ) | < ± . For this δ there exist N such that | x n - x m | < δ for all n,m N . Hence | f ( x n ) - f ( x m ) | < ± for all n,m N , i.e., ( f ( x n )) is a Cauchy sequence. Page 108 Problem 5. a. If f and g are uniformly continuous, then so is f + g . Proof: Let ± > 0 then there exists δ 1 > 0 such that | x - y | < δ 1 implies | f ( x ) - f ( y ) | < ± 2 . Similarly there exists δ 2 > 0 such that | x - y | < δ 2 implies | g ( x ) - g ( y ) | < ± 2 . let δ = min { δ 1 2 } . Then δ > 0 and | x - y | < δ implies | f ( x ) + g ( x ) - f ( y ) - g ( y ) | < ± . For the proof that | f | is uniformly continuous, let δ > 0 such that | x - y | < δ implies | f ( x ) - f ( y ) | < ± . Then the inequality || f ( x ) | - f ( y ) || ≤ | f ( x ) - f ( y ) | implies that | x - y | < δ implies || f ( x | ) -| f ( y ) || < ± , i.e. | f | is uniformly continuous. Now sup(
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This note was uploaded on 02/05/2012 for the course MATH 555 taught by Professor Staff during the Spring '07 term at South Carolina.

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