Brandon Behring
Functional Analysis HW 4
Exercise 12.4 Give an example of a monotonic decreasing sequence of nonnegative functions converging point wise to a function f that the equality
in Theorem 12.33 does not hold.
Solution: Let fn = n1 χ0≤n , then fn → f = 0 but
Z
Z
Z n
1
1
fn dµ =
χ0≤n dµ =
dµ = 1
n
0 n
but
Z
Z
fn dµ =
so
1
χ0≤n dµ =
n
Z
lim
n→∞
Z
0
n
1
dµ = 1
n
Z
fn dµ = 1 6=
f dµ = 0.
Exercise 12.6 Use the dominated convergence theorem to prove Corollary
12.36 for differentiation under an integral sign.
Proof: Let us take all the assumptions of Corollary 12.36. For a fixed
t ∈ I, let n be big enough that t + 1/n ∈ I then define
fn (x, t) ≡
f (x, t + n1 ) − f (x, t)
1/n
By assumption (b), f (x, ·) is differentiable almost everywhere in I so
fn (x, t) → ∂f (x, t)/∂t as n → ∞. We can now also use the mean value
theorem to see that for some c ∈ (t, t + 1/n) such that
|f (x, t + 1/n) − f (x, t)| ≤ |
∂f
1
1
(x, c)| · | | ≤ g(x)
∂t
n
n
where the second inequality comes from assumption (c), that g(x) is an
integrable function that dominates the partial derivatives with respect to
t for every t ∈ I.
Thus, |fn (x, t)| ≤ g(x) for all x ∈ X and we can apply the Lebesgue
Dominated Convergence to bring the limit inside the integral
Z
Z
lim
fn dµ = f dµ.
n→∞
1
(note: assumption (a) tells us that each fn does exist) Noting that
Z
ρ(t) =
f (x, t)dµ(x)
X
than
ρ(t + 1/n) − ρ(t)
=
1/n
Z
X
f (x, t + n1 ) − f (x, t)
dµ(x) =
1/n
Z
fn (x, t)dµ(x).
X
Taking the limits as n → ∞, we have
dρ
(t)
dt
ρ(t + 1/n) − ρ(t)
1/n
Z
fn (x, t)dµ(x)
= lim
n→∞ X
Z
=
lim fn (x, t)dµ(x)
n→∞
ZX
∂f (x, t)
=
dµ(x)
∂t
X
=
lim
n→∞
as claimed.
Exercise 12.8 Let fn : X → C be a sequence of measurable functions converging to f pointwise almost everywhere. Suppose there exists g ∈ Lp (X)
such |fn | ≤ g a.e. Prove that fn → in the Lp -norm.
Proof: Note that for all n, |fn | ≤ g almost everywhere so we must also
have |f | ≤ g almost everywhere. Thus
|f − fn | ≤ |f | + |fn | ≤ g + g = 2g
almost everywhere and thus
|f − fn |p ≤ (2g)p = 2p g p .
Since g ∈ Lp (X), we can use the dominated convergence theorem to bring
the limit inside the integral
Z
lim
n→∞
|f − fn |p dµ =
Z
lim |f − fn |p dµ =
n→∞
2
Z
0dµ = 0.
Exercise 12.10 If x, y ≥ 0 and > 0 is any positive number, then
2
1
x + y2 .
2
2
xy ≤
Proof:
If x, y ≥ 0 and > 0, then
0
≤
=
2
1
√
x −
y
1
x2 − 2xy + y 2 .
√
Bringing 2xy to the left hand side and dividing both sides by two gives us
1
2
x + y2 .
2
2
xy ≤
Exercise 12.12 Prove the following generalization of Hölders inequality: if
1 ≤ pi ≤ ∞, where i = 1, . . . , n, satisfy
n
X
1
= 1,
p
i=1 i
and fi ∈ Lpi (X, µ), then f1 · · · fn ∈ L1 (X, µ) and
Z
|
f1 · · · fn dµ| ≤ ||f1 ||p1 · · · ||fn ||pn .
Proof: Let us proceed by induction. For n = 1 this is basic and for n = 2
it is Holder’s inequality. Let us assume this is true for n functions and see
this implies the inequality holds for n+1 functions. Assume fn+1 ∈ Lpn+1 ,
P
n+1 1
i=1 pi = 1 and let g(x) = f1 f2 · · · fn . Then by Holder,
Z
|
where
(f1 · · · fn )fn+1 dµ| = ||g||Q ||fn+1 ||pn+1
1
1
+
=1
Q pn+1
or
Q=
We also have
3
pn+1
.
pn+1 − 1
n+1
X
i=1
n
X 1
1
1
=
+
=1
pi
p
pn+1
i=1 i
so
n
X
1
pn+1 − 1
1
1
=1−
=
=
p
p
p
Q
i
n+1
n+1
i=1
This tells us that
Pn
Q
i=1 pi
= 1.
1
Q
Noting that if ||g Q ||1 = ||g||Q we can apply our inductive hypothesis on
Pn
fiQ ∈ Lpi /Q noting that i=1 pQi = 1
Z
|
(f1 · · · fn )fn+1 dµ| = ||g||Q ||fn+1 ||pn+1
1
= ||g Q ||1Q ||fn+1 ||pn+1
1
= ||f1Q f2Q · · · fnQ ||1Q ||fn+1 ||pn+1
1
1
1
= ||f1Q ||pQ1 /Q · ||f2Q ||pQ2 /Q · · · ||fnQ ||pQn /Q ||fn+1 ||pn+1
= ||f1 ||p1 ||f2 ||p2 · · · ||fn ||pn ||fn+1 ||pn+1
Exercise 12.15 If f ∈ Lp Rn ∩ Lq (Rn ), where p < q, then f ∈ Lr (Rn ) for any
p < r < q, and show that
1/r−1/q
1/p−1/r
||f ||r ≤ (||f ||p ) 1/p−1/q (||f ||q ) 1/p−1/q
This result is one of the simplest examples of an interpolation inequality.
Proof: Let λ =
1/p−1/r
1/p−1/q ,
then we can rewrite this as
||f ||r ≤ (||f ||p )1−λ (||f ||q )λ .
Noting that if we define
P
=
Q =
then
4
p
r(1 − λ)
q
rλ
1
1
+
P
Q
=
=
=
=
=
=
=
r(1 − λ) rλ
+
p
q
r(1 − λ) rλ
+
p
q
r 1/r − 1/q
r 1/p − 1/r
+
p 1/p − 1/q q 1/p − 1/q
1 − r/q
r/p − 1
+
1 − p/q q/p − 1
q−r
r−p
+
q−p q−p
q−r+r−p
q−p
q−p
=1
q−p
so P are Q are conjugate.
Using Holder’s inequality
Z
Z
|f |r dµ =
|f |(1−λ)r |f |λr dµ
||f (1−λ)r ||P ||f λr ||Q
Q1
Z
P1 Z
λQr
(1−λ)P r
|f |
dµ
=
|f |
dµ
≤
Z
=
=
r(1−λ)
Z
rλ
p
q
q
|f | dµ
|f | dµ
p
(||f ||p )(1−λ)r (||f ||q )λr .
Taking the r-th roots of both sides gives us
Z
r1
r
||f ||r =
|f | dµ
≤ (||f ||p )1−λ (||f ||q )λ .
Exercise 12.16 A function f : Rn → C is said to be Lp continuous if τh f → f
in Lp (Rn ) as h → 0 in Rn , where τh f (x) = f (x − h) is the translation of
f by h. Prove that, if 1 ≤ p < ∞, every f ∈ Lp (Rn ) is Lp - continuous.
Give a counter-example to show this result is not true when p = ∞.
Proof:
We wish to prove τh f → f in Lp , that is as h → 0 we have
||τh f − f ||p → 0.
5
Because Cc (Rn ) are dense in Lp (R)n , for any > 0 we can approximate
f by a continuous function of compact support g such ||f − g||p < /3.
Additionally, doing a simple substitution u = x − h
||τh f −
τh g||pp
Z
|τh f (x) − τh g(x)|p dn x
=
Rn
Z
|f (x − h) − g(x − h)|p dn x
=
Rn
Z
=
Rn
|f (u) − g(u)|p dn u = ||f − g||pp
so we also have ||τh f − τh g||pp < /3. Since g is uniformly continuous on a
compact set, we clearly have
Z
||τh g − g|| =
|g(x − h) − g(x)|p dn x < /3
Rn
for sufficiently small h and thus
||τh f − f ||p
= ||τh f − τh g + τh g − g + g − f ||p
≤
||τh f − τh g||p + ||τh g − g||p + ||g − f ||p
+ + = .
<
3 3 3
This proof fails for p = ∞ because the set of continuous functions of
compact support is not dense in L∞ .
Exercise 12.17 Prove that the unit ball in Lp ([0, 1]), where 1 ≤ p ≤ ∞ is not
strongly compact.
Proof: We need to find a sequence fn in Lp ([0, 1]) such that ||fn ||p = 1
and fn does not have a convergent subsequence.
n
Let In = 21n , 2 21n and fn = 2 p χIn then for finite p
Z
||fn ||p
1
n
|2 p χIn |p dµ
=
0
Z
2−n+1
2n dµ
=
2−n
=
2n · (2
=
2n ·
6
1
1
− n)
2n
2
1
=1
2n
and for n 6= m we note that χIm χIn = 0 and
1
Z
||fn − fm ||pp
m
n
|2 p χIn − 2 p χIm |p dµ
=
0
1
Z
=
n
m
2 p χIn + 2 p χIm dµ
0
= ||fm ||p + ||fn ||p
=
1 + 1 = 2.
Thus, the distance between any two points is a constant and thus we
can have no Cauchy sequence and no convergent subsequence. If p = ∞,
let fn = χIn then clearly ||fn ||p = 1 and ||fn − fm ||∞ = 1 since we have
disjoint characteristic functions. From here- we repeat the same argument
to conclude we have no convergent subsqeuences.
Exercise 12.18 Give an example of bounded sequence in L1 ([0, 1]) that does
not have a weakly convergent subsequence. Why doesn’t this contradict
the Banach-Alaoglu theorem.
Proof: We use the same constructions from the previous problem but
with p = 1. The sequence fn converges weakly if for any element of
ρ(L1 )∗ then ρ(fn ) → ρ(f ). If we take
1
Z
ρ(f ) =
f
0
∞
X
(−1)k χInk dµ
k=1
then
Z
ρ(fnj )
1
=
f nk
0
=
2
nj
2 nj
Z
0
=
(−1)k χInk dµ
k=1
Z
1
χInj
0
=
∞
X
∞
X
(−1)k χInk dµ
k=1
1
χInj (−1)j dµ
(−1)j 2nj
Z
2−nj +1
dµ
2−nj
=
(−1)j 2nj
1
= (−1)j .
2nj
so fnj does not converge weakly. This does not contradict the BanachAlaoglu theorem because L1 is not reflexive and weak-∗ and weak convergence are not the same.
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