Night at the Museum - Alan SilvestriDescripción completa
Selected Problems from Chapter 7.3 Vinh-Kha Le
Problem 23. Provide an example of a Cauchy sequence of real numbers that
is not rapidly Cauchy. Example. The real-valued sequence defined explicitly by the equation x equation x n = 1/n
is Cauchy but not rapidly Cauchy. Proof. Elementary calculus shows us that { xn } converges to 0. Because { xn } is a conver convergen gentt sequence, sequence, it is also Cauchy Cauchy. Now suppose that there is a sequence of positive numbers { k } such that
|xk+1 − xk | ≤ k2 for all k all k . Explicit Explicit substituti substitution on gives gives us the following relationship relationship for all k all k..
1 1 1 1 ≤ − = = | xk+1 − xk | ≤ k2 2 (k + 1) k(k + 1) k + 1 k Taking the square root of the left-most and right-most expressions shows us that 1 all k.. ≤ k for all k k + 1 Taking the sum of the above for k goes from 1 to ∞ shows us that the series ∞
∞
1 = ∞. k + 1 n=1
≥ k
n=1
In other other words, words, this series diverges. diverges. Therefore, Therefore, { xn } is not rapidly Cauchy. Problem 25. Assume that E has E has finite measure and 1 ≤ p 1 < p2 ≤ ∞. Show
{ f n } → f in L that if {f n } → f in L L p2 (E ), then { L p1 (E ). Proof. Corollary Corollary 3 guarante guarantees es the existence existence of a number number c for which all n. f n − f p1 ≤ c f n − f p2 for all n. Taking the limit of both sides as n → ∞ and applying the convergence condition for { f n } → f in L in L p2 (E ) gives us lim f n − f p1 ≤ lim c f n − f p2 = 0
n→∞
n→∞
by monotonicity of the limit. It follows that { f n } → f in L in L p1 (E ). ). 1
Problem 26. (The L p Dominated Convergence Theorem) Let {f n } be a se-
quence of measurable functions that converges pointwise a.e. on E to f . For 1 ≤ p < ∞, suppose there is a function g in L p (E ) such that for all n, | f n | ≤ g a.e. on E . Prove that {f n } → f in L p (E ). Proof. From the given information, we know that
|f n − f | ≤ |f n | + |f | ≤ 2g a.e. on E by monotonicity of the limit. Taking the pth power of the left-most and rightmost sides yields |f n − f | p ≤ 2 p · g p a.e. on E . Note that the right-hand expression is integrable because
p
p
p
2 · g ≤ 2
E
|g| p is finite as g ∈ L p (E ).
E
We also know from the given information that { f n − f } → 0 a.e. on E , so lim |f n − f | p = 0 a.e. on E
n→∞
by the continuity of x → | x| p . Applying the Lebesgue Dominated Convergence Theorem to the above expression yields lim
n→∞
|f n − f | p = 0.
E
This was what was to be proven. Problem 27. For E a measurable set and 1 ≤ p < ∞, assume {f n } → f in
L p (E ). Show that there is a subsequence {f nk } and a function g ∈ L p (E ) for which |f nk | ≤ g a.e. on E for all k . Proof. Because {f n } converges, it is Cauchy. Therefore, it has a subsequence {f nk } that is rapidly Cauchy. This is to say that ∞
≤ where < ∞.
f
nk+1 − f nk p
2
k
k
k=1
Let the sequence { gk } be defined by the following explicit definition. k
f ≥ g , we have Consider the difference | g − g |. For f = g , so |g − g | ≤ f − f . |f | ≤ g ≤ f ≤ g , we have a similar result: For f − f . g = g , so |g − g | = 0 ≤ f gk = max
k+1
nk
nk+1
k
nk+1
nj
k
j =1
nk+1
k+1
k+1
k
k
nk+1
k
k
k+1
k+1
k
2
nk+1
nk
nk
We can conclude that
|gk+1 − gk | ≤ f nk+1 − f nk everywhere on E . By monotonicity of the Lebesgue integral,
gk+1 − gk p ≤ f nk+1 − f nk
2
≤ . p
k
In other words, { gk } is rapidly Cauchy with respect to L p (E ). Theorem 6 tells us that { gk } converges pointwise to a function g ∈ L p (E ) a.e. on E . Trivially, |f nk | ≤ g a.e. on E for all k .
