Exercise 7.18

Exercise 18: Let { f n } be a uniformly bounded sequence of functions which are Riemann-integrable on [ a , b ] , and put

F n ( x ) = a x f n ( t ) dt ( a x b ) .

Prove that there exists a subsequence { F n k } which converges uniformly on [ a , b ] .

Answers

This follows from Theorem 7.25 if we show that { F n } is pointwise-bounded and equicontinuous on [ a , b ] . Let | f n | K on [ a , b ] . Then for all n

| F n ( x ) | a x | f n ( t ) | dt K ( x a ) ,

so { F n } is pointwise-bounded on [ a , b ] . And if 𝜀 > 0 and x < y are points of [ a , b ] such that y x 𝜀 K , then

| F n ( x ) F n ( y ) | x y | f n ( t ) | dt K ( y x ) 𝜀 ,

showing that { F n } is equicontinuous on [ a , b ] .

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2023-08-07 00:00
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