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18.100pset7

# 18.100pset7 - 18.100B/C Fall 2008 Solutions to Homework 7...

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18.100B/C: Fall 2008 Solutions to Homework 7 1.(a) For a given ε and x , δ f ( x, ε ) is the supremum of a set of positive values. Thus, δ f ( x, ε ) > 0 if and only if this set is nonempty. So δ f ( x, ε ) > 0 if and only if there exists some δ > 0 such that for all t X , d X ( x, t ) < δ implies d Y ( f ( x ) , f ( t )) < ε . That this statement is true for every ε > 0 is exactly the definition of continuity of f at x , so we’re done. (b) First suppose that f is unformly continuous and choose ε > 0. Then there exists some δ > 0 such that for any x, y X we have d Y ( f ( x ) , f ( y )) < ε whenever d X ( x, y ) < δ . It follows that δ f ( x, ε ) δ for every x X , so inf x X δ f ( x, ε ) δ > 0. Conversely, suppose inf x X δ f ( x, ε ) > 0, and define δ = 1 2 · inf x X δ f ( x, ε ). It follows that 0 < δ < δ f ( x, ε ) for every x X . By the properties of the supremum, this means that for each x X there exists some c > δ such that c belongs to the set of which δ f ( x, ε ) is the supremum, i.e. d X ( x, t ) < c implies d Y ( f ( x ) , f ( t )) < ε for every t X . Since δ < c we have that { t X | d X ( x, t ) < δ } ⊂ { t X | d X ( x, t ) < c } and thus d X ( x, t ) < δ implies d Y ( f ( x ) , f ( t )) < ε for all x, t X . But this is exactly the definition of uniform continuity for f , so we’re done. (We introduced a factor of 1 2 in this last paragraph not suggested by the hint. It may be possible to set δ = inf x X δ f ( x, ε ) directly, without the 1 2 factor, and still produce a valid proof. The one problem to watch out for is the following situation: for fixed ε > 0, what if δ f ( x, ε ) = d is constant but d negationslash∈ { δ > 0 | ∀ t Xd X ( x, t ) < δ = d Y ( f ( x ) , f ( t )) < ε } for some x (i.e., the supremum is not achieved)?) (c) For any x, t X we have | f ( x ) f ( t ) | = | x 2 t 2 | = | x t | · | x + t | < δ · (2 x + δ ) , so 2 + δ 2 is an upper bound for {| f ( x ) f ( t ) | | t X, d X ( x, t ) < δ } . It addition, the limit as t approaches x + δ from below of | f ( x ) f ( t ) | is equal to δ · (2 x + δ ), so it is in fact the least upper bound, as desired. (d) It follows from part (c) that δ f ( x, ε ) is the supremum of the set of δ such that 2 + δ 2 < ε . This supremum will simply be the larger of the two roots of the equation 2 + δ 2 = ε , and one can quickly solve the quadratic to find that this root is δ f ( x, ε ) = x 2 + ε x . Since we have for every x X that δ f ( x, ε ) 2 + 2 f (

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18.100pset7 - 18.100B/C Fall 2008 Solutions to Homework 7...

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