Uniform Convergence of Sequences of Functions.pdf

# That is k 0 t k 1 k 1 log 1 t on 1 1 404 9 uniform

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That is, k =0 t k +1 k + 1 = log (1 t ) on ( 1 , 1) ,
404 9 Uniform Convergence of Sequences of Functions and the convergence is uniform on any closed subinterval of ( 1 , 1). It turns out that the new series also converges (by the alternating series test) at one of the endpoints, t = 1. The following result is almost obvious and is in fact a straightforward application of Corollary 9.30 . Theorem 9.35 (Term-by-term integration in power series). Suppose that f ( x ) = k =0 a k x k converges for | x | < R . Then for any closed interval [ c, x ] contained in ( R, R ) , the integral x c f ( t ) d t exists and can be obtained by integrating the power series term by term. In particular, we have x 0 f ( t ) d t = k =0 a k k + 1 x k +1 + constant for | x | < R, and the convergence is uniform on any closed subinterval of ( R, R ) . Theorem 9.35 tells us that inside the interval of convergence (not neces- sarily at the endpoints), a power series can be integrated term by term. Next, we see that term-by-term differentiation is also possible. In Theorem 9.15 , we needed for each f n to be continuous on [ a, b ]. Is it possible to weaken the hypothesis to permit integrable functions rather than just continuous functions? Recall that if f and g are two Riemann integrable functions on [ a, b ], then b a f ( x ) d x b a g ( x ) d x, whenever f ( x ) g ( x ) on [ a, b ]. Theorem 9.36 (Integration of sequences of integrable functions). Suppose that { f n } is a sequence of Riemann integrable functions defined on an interval [ a, b ] . If f n f uniformly on [ a, b ] , then f is Riemann integrable on [ a, b ] , and lim n →∞ b a f n ( x ) d x = b a f ( x ) d x. Also, for each t [ a, b ] , t a f n ( x ) d x t a f ( x ) d x uniformly on [ a, b ] . Proof. Because of the argument in Theorem 9.15 , we need only show that the limit function f is integrable on [ a, b ]. We see that the following statements hold: f n is bounded, because each f n is integrable on [ a, b ].
9.2 Uniform Convergence of Series 405 f is bounded, because | f ( x ) | ≤ | f n ( x ) f ( x ) | + | f n ( x ) | ≤ δ n + | f n ( x ) | , where δ n = sup x [ a,b ] | f n ( x ) f ( x ) | → 0 (by Theorem 9.6 ). Because f n f uniformly on [ a, b ], given > 0, there exists an N such that | f n ( x ) f ( x ) | < 3( b a ) for all x [ a, b ] and all n N . (9.4) Since f N is integrable, there exists a partition P of [ a, b ] such that U ( P, f N ) L ( P, f N ) < 3 . For each x [ a, b ], ( 9.4 ) with n = N implies that f N ( x ) 3( b a ) < f ( x ) < f N ( x ) + 3( b a ) , and therefore L ( P, f N ) 3 < L ( P, f ) U ( P, f ) < U ( P, f N ) + 3 . Consequently, U ( P, f ) L ( P, f ) < U ( P, f N ) L ( P, f N ) + 2 3 < 3 + 2 3 = , showing that f is integrable on [ a, b ]. Finally, for n N and for each t [ a, b ], ( 9.4 ) implies that t a f n ( x ) d x t a f ( x ) d x t a | f n ( x ) f ( x ) | d x ( t a ) 3( b a ) ( b a ) 3( b a ) = 3 , and the proof is complete. The limit of a uniformly convergent series of integrable functions is inte- grable, and so term-by-term integration is permissible for such a series. More precisely, we have the following result concerning integration of a series of integrable functions.

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• Winter '17
• Weirui Zhao
• Calculus, lim, Uniform convergence, Pointwise convergence, Modes of convergence

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