ANALYSIS QUALIFYING EXAM PROBLEMS BRIAN LEARY ...

ANALYSIS QUALS 13
= 1
√ |f(w)|
πr D
2
1/2 < 1
√ ,
πr
since f ∈ F. Thus, let A = 1 √ , which depends only on K and does not
πr
depend on x or f.
Spring 2010
Problem 1: (a) Let 1 ≤ p < ∞. Show that if a sequence of real-valued
functions {fn}n≥1 converges in L p (R), then it contains a subsequence
that converges almost everywhere.
(b) Give an example of a sequence of functions converging to zero in L 2 (R)
that does not converge almost everywhere.
Solution. (a) By Chebyshev’s inequality,
µ ({x : |f(x)| > α}) ≤ fp p
.
αp Then we have that
µ ({x : |fn(x) − f(x)| > ε}) ≤ fn − f p
p
εp → 0 as n → ∞,
so fn → f in measure.
Then we may choose a subsequence {fnj } such that
µ x : fnj (x) − fnj+1 (x) > 2 −j ≤ 2 −j .
Let Ej = {x : |fnj (x) − fnj+1 (x)| > 2−j }. Let Fk = ∞
j=k Ej. Then
µ(Fk) ≤
∞
µ(Ej) ≤ 2 1−k .
j=k
For x /∈ Fk and k ≤ j ≤ i, we have that
fnj (x) − fni (x) ≤
i−1
fnl (x) − fnl+1 (x) i−1
≤
l=j
l=j
2 −l ≤ 2 1−j ,
so {fnj } is pointwise Cauchy on F C k . Let F = ∞
k=1 Fk. Then µ(F ) =
0 and fnj converges on F C , so fnj
converges almost everywhere.
(b) We use the example of the moving dyadic intervals. Let f1 = χ [0,1],
f2 = χ [0,1/2], f3 = χ [1/2,1], and in general, for n = 2 k + j < 2 k+1 ,
let fn = χ [j/2 k ,(j+1)/2 k ]. Then fn → 0 in L 2 , but for every x ∈ [0, 1],
fn(x) → 0 as n → ∞.
Problem 3: For an f : R → R belonging to L1 (R), we define the Hardy-
Littlewood maximal function as follows:
1
(Mf)(x) := sup
h>0 2h
x+h
x−h
|f(y)|dy.
Prove that it has the following property: There is a constant A such that
for any λ > 0,
|{x ∈ R (Mf)(x) > λ}| ≤ A
λ f L 1