1873_solutions

# N f n 0 there exists an integer n such that the

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Unformatted text preview: on f n is continuous on S then the family f n n  1, 2, 3,  is equicontinuous on S. Suppose that x S and that  0. Choose an integer N such that the inequality sup|f n f |  5 holds whenever n N. Using the fact that the function f N is continuous at the number x, choose a number  1  0 such that the inequality 368 fN x |  |f N t 5 holds whenever t S and |t x |   1 . Now we use the fact that there are only finitely many positive integers n  N to choose a number  2  0 such that the inequality |f n t f n x |  5 holds whenever t S and |t x |   2 and 1 n  N. Finally, we define  to be the smaller of the two numbers  1 and  2 . Now given any member t of S satisfying the condition |t x |  , the condition  |f n t f n x |  5 is already known for 1 n  N and if n N then we have |f n t f n x |  |f n t f t  f t f N t  f N t f N x  f N x f x  f x f n x | |f n t f t |  |f t f N t |  |f N t f N x |  |f N x f x |  |f x fn x |  . 4. Invent a meaning for equi-uniform continuity of a family on a set S and decide whether or not your definition provides an analogue of the preceding exercise. A family of functions on a set S is said to be equi-uniformly continuous on S if for every  0 there exists a number   0 such that whenever t and x belong to S and |t x |   and f we have |f t f x |  . The point of this exercise is to invite the student to show that if f n is a uniformly convergent sequence of uniformly continuous functions on a set S then the family f n n  1, 2, 3,  is equi-uniformly continuous on S. The proof is similar to the one used in Exercise 3. Some Exercises on Bounded Convergence 1. Prove that the sequence f n in the bounded convergence theorem for derivatives actually converges boundedly to the function f. Using the fact that the sequence f n converges boundedly, choose a number p such that the inequality |f n x |  p holds for every positive integer n and every x a, b . Using the fact that the sequence f n c , being convergent, is bounded, choose a number q such that |f n c |  q for every positive integer n. Now given any positive integer n and any member x of a, b we have |f n x |  f n c  Þ f n t dt x c |f n c |  Þ c f n t dt x qp b a. 2. Suppose that a n is a strictly increasing sequence of positive integers and that f n x  sin a n x sin a n1 x 2 for every positive integer n and every number x. a. Work out the integral 2 Þ0 f n x dx and deduce that there must be at least one number x converge to zero. For each n and x we have 369 0, 2 such that the sequence f n x does not |f n x |  sin a n x sin a n1 x 2 4. If the sequence f n converged pointwise (and therefore boundedly) on the interval 0, 2 to the constant function 0 then the bounded convergence theorem would give us lim nÝ 2 Þ0 f n x dx  2 Þ0 0  0. However, 2 Þ0 2 Þ0 f n x dx  sin a n1 x 2 dx  2 sin a n x for each n and we conclude that the sequence f n cannot converge pointwise to 0. b. Prove that there must be at least one number x...
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## This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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