# HW5-7 - ANSWERS TO SELECTED EXERCISES HW5-7 Exercise 8.1.5...

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Unformatted text preview: ANSWERS TO SELECTED EXERCISES: HW5-7 Exercise 8.1.5 Let f n ( x ) = ( x/n ) e- x/n , x ∈ [0 , ∞ ). a. Show that lim n →∞ f n ( x ) = 0 for all x ∈ [0 , ∞ ). b. Given ε > 0, does there exist an integer n ∈ N such that | f n ( x ) | < ε for all x ∈ [0 , ∞ ) and all n > n . c. Answer the same question as in part (b) for x ∈ [0 ,a ], a > 0. Proof : a. We have lim n →∞ x n e- x/n = lim n →∞ x n lim n →∞ e- x/n = 0 · 1 = 0 . b. We will determine the maximum of the positive function f n on the interval [0 , ∞ ); to this end, we will compute the derivative f n : f n ( x ) = 1 n e- x/n- x n 2 e- x/n = 1 n e- x/n 1- x n . We observe that the sign of f n is determined by the expression 1- ( x/n ), therefore f n ( x ) > 0 if x < n , f n ( x ) < 0 if x > n and f n ( x ) = 0 if x = n . By the first derivative test for monotonicity, this shows that f n attains its global maximum at x = n , which is f n ( n ) = e- 1 . Therefore, the ε given in the exercise, cannot be arbitrarily small, i.e. less than e- 1 (which shows that f n does not converge to zero uniformly on [0 , ∞ ), only pointwise). c. We have seen above that f n ( x ) > 0 when x < n , hence f n is increasing on the interval [0 ,n ]. If we restrict our attention to the interval [0 ,a ], we deduce that f n is increasing on [0 ,a ], for all n > a . Hence, the maximum of f n on [0 ,a ] is f n ( a ) = a n e- a/n . By (a), we know that lim n →∞ f n ( a ) = 0, therefore, given ε > 0, there exists some n such that for all n > n we have | f n ( a ) | < ε. 1 So, if we consider all n > max( a,n ), the following holds for all x ∈ [0 ,a ]: | f n ( x ) | < ε, thus we can answer the above question in the affirmative. Exercise 8.2.2 a. If { f n } and { g n } converge uniformly on E , prove that { f n + g n } converges uniformly on E . b. If { f n } and { g n } converge uniformly on E , and there exist constants M and N such that | f n ( x ) | 6 M and | g n ( x ) | 6 N for all n ∈ N and all x ∈ E , prove that { f n g n } converges uniformly on E . Proof : a. Let ε > 0 be arbitrary. Since { f n } and { g n } converge uniformly on E , there exist n and n 1 such that | f n ( x )- f ( x ) | < ε 2 for all n > n and all x ∈ E , and | g n ( x )- g ( x ) | < ε 2 for all n > n 1 and all x ∈ E , where f and g are the (uniform) limits of { f n } and { g n } on E . Thus, for all n > max( n ,n 1 ) and all x ∈ E , we have | ( f n + g n )( x )- ( f + g )( x ) | 6 | f n ( x )- f ( x ) | + | g n ( x )- g ( x ) | < ε, thus proving that { f n + g n } converges uniformly to f + g on the set E . b. Again, let ε > 0 be arbitrary. Since { f n } and { g n } converge uniformly on E , there exist n and n 1 such that | f n ( x )- f ( x ) | < ε 2 N for all n > n and all x ∈ E , and | g n ( x )- g ( x ) | < ε 2 M for all n > n 1 and all x ∈ E . Then, for all n > max( n ,n 1 ) and all x ∈ E , we have | ( f n g n )( x )- ( fg )( x ) | = | ( f n g n )( x )- ( fg n )( x ) + ( fg n...
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## This homework help was uploaded on 04/16/2008 for the course MATH 131B taught by Professor Hitrik during the Winter '08 term at UCLA.

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HW5-7 - ANSWERS TO SELECTED EXERCISES HW5-7 Exercise 8.1.5...

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