hw1-sol

# hw1-sol - Modern Analysis Homework 1 Possible solutions 1...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Modern Analysis, Homework 1 Possible solutions 1. Consider ε = 1, and let f = lim n →∞ f n . By uniform convergence, there exists N ∈ N such that n > N implies k f n- f k u ≤ 1. Since each f n is bounded, this shows that f is bounded and that k f n k u ≤ k f k u + 1 if n > N . Set M = max max n =1 ,...,N k f n k u , k f k u + 1 . Then k f n k u ≤ M for all n ∈ N . 2. (a) Let f = lim n →∞ f n , g = lim n →∞ g n , and ε > 0. By uniform convergence, there exists N f ,N g ∈ N such that if n > N f and x ∈ E , then | f n ( x )- f ( x ) | < ε/ 2, and if n > N g and x ∈ E , then | g n ( x )- g ( x ) | < ε/ 2. It follows that if n > max( N f ,N g ) and if x ∈ E , then | ( f n ( x )+ g n ( x ))- ( f ( x )- g ( x )) | ≤ | f n ( x )- f ( x ) | + | g n ( x )- g ( x ) | < ε . (b) As in (a), let f = lim n →∞ f n , g = lim n →∞ g n , and ε > 0. By problem 1, we know that { g n } is uniformly bounded, ie there exists G > 0 such that for all n ∈ N and all x ∈ E , we have | g n ( x ) | < G . By the uniform convergence of { f n } , there exists N f ∈ N such that n > N f and x ∈ E implies that | f n ( x )- f ( x ) | < ε 2 G . By the uniform convergence of { g n } , we may find N g ∈ N such that n > N g and x ∈ E implies that | g n ( x )- g ( x ) | < ε 2(sup x ∈ E | f ( x ) | +1) . It follows that if n > max( N f ,N g ) and x ∈ E , then we have | f n ( x ) g n ( x )- f ( x ) g ( x ) | < | f n ( x )- f ( x ) | · | g n ( x ) | + | f ( x ) | · | g n ( x )- g ( x ) | < ε. 3. Many examples are possible; here’s one: let’s define two functions f,g : (0 , ∞ ) → R by f ( x ) = x if x ≤ 1 /n 1 /n if x > 1 /n and simply g n ( x ) = 1 /x for all n ∈ N . Since k f n k u ≤ 1 /n for all n ∈ N , it follows that f n converges uniformly to the zero function (and of course g n converges uniformly to 1 /x ). Then f n ( x ) g n ( x ) = 1 if x ≤ 1 /n 1 nx if x > 1 /n This function converges pointwise to the zero function, but does not do so uniformly: for a given n ∈ N , we have f n (1 /n ) g n (1 /n ) = 1. 4. (a) First note (say, by looking at the graph) that { x } is discontinuous precisely on Z . Similarly, for a given n ∈ N , { nx } is discontinuous precisely on 1 n Z (integral multiples of 1 n )). In fact, if p ∈ Z , then lim x → p/n- { nx } = 1 while lim x → p/n + { nx } = 0. For n ∈ N , let g n ( x ) = { nx } n 2 . Then | g n ( x ) | ≤ 1 /n 2 for all x ∈ R , so by the Weierstrass M-test, since ∑ 1 n 2 converges, it follows that the sequence f n ( x ) def = 1 ∑ n k =1 g k ( x ) converges uniformly to the function f . Also each f n ( x ) is clearly Riemann-integrable on any bounded domain [ a,b ], so by uniform convergence, we get that f too is Riemann-integrable and Z b a f ( x ) dx = lim n →∞ Z b a f n ( x ) dx = lim n →∞ Z b a n X k =1...
View Full Document

{[ snackBarMessage ]}

### Page1 / 5

hw1-sol - Modern Analysis Homework 1 Possible solutions 1...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online