Folland 6
Folland 6
Folland 6
Sebastien Picard
Problem 6.8
Suppose (X) = 1R and f Lp for some p > 0, so that f Lq for 0 < q < p.
a. log
in 3.5, with
F (t) = et ).
R ||fq||q log |f |. (Use Exercise
R 42d
R
b. ( |f | 1)/q logR||f ||q , and ( |f |q 1)/q log |f | as q 0.
c. limq0 ||f ||q = exp( log |f |).
Solution:
(a) Define g = q log |f |.
R
Case 1: Suppose g
/ L1 . Then | log |f |q | = . We can split up the integral
Z
Z
Z
Z
Z
q
q
q
q
| log |f | | = log |f | |f |>1 log |f | |f |<1 |f | log |f |q |f |<1 .
R
R
q
q
Since
f
L
,
we
must
have
log
|f
|
=
.
Therefore,
since
log |f |q |f |>1 is finite,
|f
|<1
R
q log |f | = and the inequality holds trivially.
Case 2: Suppose g L1 . Let F (t) = et . We know that F (t) is convex on the real line since
the exponential function is absolutely continuous on every compact interval and F (t) = F (t) > 0.
Thus we can apply Jensens inequality:
Z
Z
1
log |f | =
log |f |q
q
Z
1
=
g
q
R
1
= log e g
q
Z
1
log eg
q
Z
1
= log |f |q = log ||f ||q .
q
(b)
The first inequality follows directly from the fact that log(x) x 1.
Z
Z
1
1
q
log ||f ||q = log |f | ( |f |q 1).
q
q
In order to prove the limit, we will need the fact that the map h : (0, ) R defined by
h(x) = (ax 1)/x is a monotone increasing function for any a 0. If a = 0, this is immediate. If
a > 0, we prove monotonicity by showing that h (x) 0. Indeed,
1
ax log ax ax + 1
h (x) =
,
x2
hence the result will follow if we show that w : (0, ) R, w(x) = x log x x is greater than or
equal to 1. Since w only vanishes at x = 1, and w (1) > 0, we see that the minimum of w occurs
at x = 1 and hence w 1.
We now prove the limit. By lHopitals rule, pointwise we have
log |f | |f |q
|f |q 1
= lim
= log |f |.
q0
q0
q
1
lim
As shown before, (|f |q 1)/q is monotone increasing, hence the limit as q 0 is monotone
decreasing. By the Monotone Convergence Theorem, we have
Z
Z
Z
1
1
q
q
(|f | 1) = lim (|f | 1) = log |f |.
lim
q0 q
q0
q
(c) From
part (a), after composing both sides with the exponential function, we see that ||f ||q
R
exp( log |f |) for all 0 < q < p. From part (b), we see that
Z
Z
q
|f | 1 /q log |f |.
lim sup log ||f ||q lim sup
q0
q0
R
Therefore,
lim
sup
||f
||
exp(
log |f |). Therefore, the limit exists, and limq0 ||f ||p =
q
q0
R
exp( log |f |).
Problem 6.10
Suppose 1 p < . If fn , f Lp and fn f a.e., then ||fn f ||p 0 iff ||fn ||p ||f ||p. (Use
Exercise 20 in 2.3.)
Solution:
Suppose ||fn f ||p 0. Then by the reverse triangle inequality,
| ||fn ||p ||f ||p | ||fn f ||p 0
as n .
On the other hand, suppose ||fn ||p ||f ||p. Notice
|fn f |p 2p (|f |p + |fn |p ).
We can define gn = 2p (|f |p + |fn |p ) and g = 2p+1|f |p L1 . Then gn g a.e, and furthermore,
2
lim
gn = 2
|f | + lim 2
|fn |
p+1
|f | =
g.
By Exercise 20 in 2.3,
lim
|fn f | =
lim |fn f |p = 0.
Problem 6.20
Suppose supn ||fn ||p < and fn f a.e.
a. If 1 < p < , then fn Rf weakly in Lp . (Given g Lq , where q is conjugate to p, and > 0,
there
exists (i) > 0 such that E |g|q < whenever (E) < , (ii) A X such that (A) < and
R
|g|q < , and (iii) B A such that (A\B) < and fn f uniformly on B.)
X\A
b. The result of (a) is false in general for p = 1. (Find counterexamples in L1 (R, m) and l1 .)
It is, however, true for p = if is finite and weak convergence is replaced by weak* convergence.
Solution:
a. Claim (i) follows from Corollary 3.6, and claim (iii) is Egoroffs theorem (2.33). We first prove
claim (ii). By Proposition 2.20, P = {x : |g|q > 0} is finite. Hence we can assume that P =
i Pi
where (Pi ) < and Pi are disjoint. Therefore,
Z
|g| =
|g| =
Z
X
i=1
Pi
|g|q < .
P R
i=N
Pi
We now prove that fn f weakly in Lp . Take g Lq and any > 0. Since supn ||fn ||p C0 < ,
we have that f Lp by Fatous lemma:
Z
Z
p
|f | lim inf |fn |p C0p < .
By (i), choose > 0 such that
Z
|g| <
6C0
q
when (E) < . By (ii), choose A X such that (A) < and
q
Z
q
|g| <
.
6C0
X\A
By (iii), choose B A and N N such that (A\B) < and
3(B)1/p ||g||
,
q
X\A
|fn f | |g| +
||fn f ||p
Z
X\A
|g|
Z
|fn f | |g| +
A\B
1/q
q
q
+ ||fn f ||p
1/q Z
|g|
3(B)1/p ||g||q
B
1/q
Z
Z
q
2C0
|g|
+ 2C0
+
X\A
A\B
|fn f | |g|
Z
A\B
|g|
1/q
1/p
|g|
1/q
< /3 + /3 + /3 = .
0
1
if i 6= n,
if i = n.
We
R see that ||fn ||l1 =
R 1 for all positive integers n, and fn f a.e., where f = 0. But
lim fn 1 d = 1 and f 1 d = 0, where is the counting measure. Therefore fn does not
converge to 0 weakly in l1 .
Now we show that if supn ||fn || = C0 < , fn f a.e., and is finite, then fn f
in the weak* sense on L . The first thing we show is that ||f || < . Indeed, for a.e. x, there exists
a positive integer n such that |fn (x) f (x)| < 1. Then
|f (x)| |f (x) fn (x)| + |fn (x)| 1 + C0 .
Hence ||f || 1 + C0 .
4
|g| <
3(2C0 + 1)
when (E) < . By fact (ii), choose A X such that (A) < and
Z
.
|g| <
3(2C0 + 1)
X\A
By fact (iii), choose B A and N N such that (A\B) < and
|fn (x) f (x)| <
,
3||g||1
|fn f | |g| +
|fn f | |g| +
|fn f | |g|
A\B
B
Z
Z
Z
|g|
||fn f ||
|g| + ||fn f ||
|g| +
3||g||1 B
X\A
A\B
< (2C0 + 1)
||g||1 = .
+ (2C0 + 1)
+
3(2C0 + 1)
3(2C0 + 1) 3||g||1
X\A
Problem 6.22
Let X = [0, 1], with Lebesgue measure.
a. Let fn (x) = cos 2nx. Then fn 0 weakly in L2 (see Exercise 63 in 5.5), but fn 9 0 a.e. or
in measure.
b. Let fn (x) = n(0,1/n) . Then fn 0 a.e. and in measure, but fn 9 0 weakly in Lp for any p.
Solution:
a. Let un (x) = 2 cos 2nx. We note the following two identities, where n 6= m.
Z 1
Z 1
Z 1
1 1
2
2
un = 2
(cos 2nx) dx = 2
( + cos 4nx)dx = 1.
2
0
0
0 2
Z
un um = 2
1
0
fn2
R1
0
cos(2nx) = 1/2.
Finally, we show fn 9 0 in measure. We explicitely calculate the measure of the set of all
x [0, 1] such that | cos 2nx| 1/2. Denote
Ek = 2k,
5
3
7
+ 2k
+ 2k, + 2k + 2k,
+ 2k
+ 2k, 2(k + 1) .
4
4
4
4
Then | cos 2nx| 1/2 iff 2nx Ek for some k N. Therefore, using the 2 periodicity of
cosine, we see that | cos 2nx| 1/2 when x [0, 1] lies in one of the 4n disjoint intervals of length
1/8n. Hence
1
{x [0, 1] : | cos 2nx| 1/2} = .
2
Since the measure is constant at 1/2 for all n, fn 9 0 in measure.
(b) First we show that fn (x) = n(0,1/n) 0 a.e. Fix x (0, 1]. Then fn (x) = 0 for all n > 1/x.
Next, we show that fn (x) = n(0,1/n) 0 in measure. For all > 0, we have
{x [0, 1] : |fn (x)| } (0, 1/n) = 1/n 0
as n . Lastly, we show fn 9 0 weakly in Lp for any p. Since 1 Lq ([0, 1]), let be the bounded
R1
linear functional on Lp defined by (f ) = 0 f . Then
Z 1
1
lim (fn ) = lim
fn (x)dx = lim n = 1.
n
0
Therefore, fn does not converge weakly to zero in Lp .
1/p
|K(x, 1)|x
dx =
1
0
1
+
x1/p (x + 1)
x1/p (x + 1)
1
x1/p
1
x1/p x
< .
Hence the conditions of Theorem 6.20 are satisfied, and the desired inequality follows directly from
the theorem.
Problem 6.29
Suppose that 1 p < , r > 0, and h is a nonnegative measurable function on (0, ). Then:
Z x
p
p Z
Z
p
r1
x
h(y)dy dx
xpr1 h(x)p dx,
r
0
0
0
Z
p
p Z
Z
p
xp+r1 h(x)p dx.
xr1
h(y)dy dx
r
0
0
x
(Apply Theorem 6.20 with K(x, y) = x1 y (0,) (y x), f (x) = x h(x), and g(x) = x h(x) for
suitable , , .)
Solution:
We start by showing the first inequality. Consider K(x, y) = x1 y (0,) (y x) and f (x) =
x h(x), where = (r + 1)/p and = 1 (r + 1)/p. If f
/ Lp (0, ), the right-hand side of the
inequality is infinite and the inequality holds trivially. Hence we assume that f Lp (0, ). It is clear
that K is a Lebesgue measurable function on (0, ) (0, ) such that K(x, y) = 1 K(x, y) for
all > 0. Using the notation from Theorem 6.20, we evaluate C:
Z
Z
Z 1
p
1/p
1
1/p
C=
|K(x, 1)|x
dx =
x (0,) (1 x)x
dx =
x1+(r/p) dx = .
r
0
0
0
By Theorem 6.20, ||T f ||p C||f ||p, where
Z
Z
1
(r+1)/p
Tf =
x y (0,) (y x)x h(x)dx = y
0
h(x)dx.
(r+1)
Z
0
y
p
p Z
p
xp(r+1) h(x)p dx.
h(x)dx dy
r
0
and f. Consider
To show the second inequality, we reapply the theorem with different functions K
K(x,
y) = x1 y (0,) (x y) and f(x) = x h(x), where = (1 r)/p and = 1 + (r 1)/p. If
f
/ Lp (0, ), the right-hand side of the inequality is infinite and the inequality holds trivially. Hence
we assume that f Lp (0, ).
We evaluate C:
Z
Z
1/p
C=
|K(x, 1)|x
dx =
0
1/p
(0,) (x 1)x
dx =
(1+r)/p
T f =
x y (0,) (x y)x h(x)dx = y
0
p
x1(r/p) dx = .
r
h(x)dx.
y
r1
Z
p
p Z
p
xp+r1 h(x)p dx.
h(x)dx dy
r
0
and
p
=
r
=
1.
If
f
L
for
j
=
1,
.
.
.
,
n,
then
j
j
j
1
1 fj L and
Qn
Qn
|| 1 fj ||r 1 ||fj ||pj . (First do the case n=2.)
Solution:
We do the case n = 2 and then proceed by induction. When n = 2, we apply Holders inequality
with conjugate exponents p1 /r and p2 /r.
Z
r
||f1 f2 ||r = |f1 f2 |r || |f1 |r ||p1 /r || |f2 |r ||p2 /r = ||f1 ||rp1 ||f2||rp2 .
The result follows by taking rth roots of both sides.
Suppose theP
result is true for (n 1). First we Holders inequality with conjugate exponents
1
pn /r and (1/r)( 1n1 p1
j ) .
n
r
Y
fj r =
1
|fn |
n1
Y
j=1
|fj |r
Y
n1
r
|fj |r (1/r)(Pn1 p1 )1
|fn | pn /r
1
j=1
Y r
r n1
= fn
fj Pn1
pn
j=1
1
p1
j )
We can
now take rth roots of both sides, and apply the induction hypothesis with r such that
P
n1
(r )1 = 1 p1
j .
n
n
Y
Y
n1
Y
||fj ||pj .
fj r
fj r fn pn
1
j=1
j=1
Problem P
6.38
kp
k
p
f L iff
2 f (2 ) < .
Solution:
Suppose f Lp . Using that f is decreasing, and Theorem 2.15 to exchange an infinite sum and
an integral, we compute the following:
Z 2k
X
X
p
k
k kp
f (2 )
f (2 )2 =
p1 d
p
12
2k1
Z 2k
X
p
f ()p1d
p
12
2k1
Z
p
f ()p1d
=
p
12
Z0
1
=
|f |p <
p
12
where the last step follows from Proposition 6.24. On the other hand, if
have the following:
Z
Z
p
|f | = p
f ()p1d
=p
Z
X
2k+1
f ()p1d
2k
k
f (2 )
= (2 1)
2k+1
p1 d
2k
Problem 6.39
If f Lp , then lim0 p f () = lim p f () = 0. (First suppose f is simple.)
Solution:
P
First, suppose f = n1 aj Ej , where the Ej are disjoint P
and (Ej ) < for all j. Let amax =
max{aj } and amin = min{aj }. Then when < amin , f () = (Ej ) < . Hence
9
lim p f () = lim p
(Ej ) = 0.
< /2 + /2 = .
Therefore, lim0 p f () = 0. Similarly, for all large enough, we have p () < /2p+1 . The
same calculation shows:
1
1
lim sup p f () lim sup { p f ( ) + p ( ) }
2
2
p
p
p p
lim sup { 2 ||f ||p + 2 () }
< /2 + /2 = .
iii) Let ||f ||r < for some r < . Prove that limr ||f ||r = ||f || .
Solution:
(i) Let t (0, 1), and p, q E. By Holders Inequality, we know that
Z
|f |pt|f |q(1t) || |f |pt ||1/t || |f |q(1t) ||1/(1t) .
Therefore, we can take the logarithm and obtain
Z
|f |tp+(1t)q
(tp + (1 t)q) = log
Z
Z
p t
q 1t
|f |
|f |
log
Z
Z
p
|f |q
|f | + (1 t) log
= t log
= t(p) + (1 t)(q).
(ii) We construct a function f : (0, ) C such that E = {1}. We first use the fact that for > 0,
we have
11
1
(0,1) <
x
X
1
an
f=
1 (0,1) ,
n
(1+1/n)
2
x
n=1
where
an =
1
x(1+1/n)1
(0,1) 1 .
X
bn
1
(1,) ,
n x(1+1/n)
2
n=1
where
bn =
1
x(1+1/n)
(1,) 1 .
12
Z
Z
|f |qr |f |r
1/q
1/q
qr
||f ||
|f |r
q
Z
1/q
(qr)/q
|f |r
= lim ||f ||
lim
= ||f ||.
Therefore, if the limit exists we have limq ||f ||q ||f ||. Now, for all > 0, there is a set
E X of positive measure such that |f (x)| ||f || for all x E. Then when q > r we have
Z
1/q
|f |q
(||f || )((E))1/q .
||f ||q
E
Now (E) is finite, so taking the limit as q we get limq ||f ||q ||f || . Since this
holds for all > 0, we conclude that limq ||f ||q = ||f ||.
Next, suppose ||f || = . I have a marvelous proof of this case, which this margin is too
narrow to contain.
13