Analysis II Home Assignment 4 Subhadip...
Transcript of Analysis II Home Assignment 4 Subhadip...
Analysis IIHome Assignment 4
Subhadip Chowdhury
Problem 4.1
f ∈ Lp(RN) ⇐⇒∫RN
∣∣(1 + |x|α)−1(1 + | log |x||β)−1∣∣p <∞
Problem 4.2
∫|f |p ≤
(∫(|f |p)
qp
) pq(∫
(1)1/(1− pq
)dµ
)(1− pq
)
= ‖f‖pq .|Ω|1−p/q
⇒‖f‖p ≤ |Ω|1p− 1q ‖f‖q
Thus f ∈ Lq ⇒ f ∈ Lp. Also ‖f − g‖p ≤ ε
|Ω|1p−
1q⇒ ‖f − g‖q ≤ ε. So the Lp(Ω) ⊆ Lq(Ω) with continuous
injection.
Problem 4.3
4.3.1 Note that
h = maxf, g =|f − g|+ f + g
2
Now Lp is a vector space. So f + g ∈ Lp. Then by definition, |f + g| ∈ Lp. So h ∈ Lp.
4.3.2 Note that
hn =|fn − gn|+ fn + gn
2
Now fn → f in Lp(Ω) and gn → g in Lp(Ω) imply that |fn − gn| → |f − g| in Lp(Ω). Thus hn → h inLp(Ω).
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Analysis Subhadip Chowdhury Assignment 4
4.3.3 Note that
|fngn − fg|p ≤ c (|fn − f |p|gn|p + |gn − g|p|f |p) (*)
for some constant c depending only on p. Now gpn → gp a.e. and (gn) is bounded in L∞, suppose gn ≤ kfor all n. Hence by DCT, we get
∫|gn − g|p → 0. Thus taking n→∞ in (∗), we get∫
|fngn − fg|p ≤ ckp∫|fn − f |p + |f |p
∫|gn − g|p → 0
So fngn → fg in Lp.
Problem 4.4
4.4.1 We prove the result by inducting on k. For the base case, k = 2. Then for 1p
= 1p1
+ 1p2
, we
have
‖f1f2‖p =
(∫|f1f2|p
) 1p
≤
((∫|f1|p1
) pp1
.
(∫|f2|p.
p1p1−p
)1− pp1
) 1p
= ‖f1‖p1 .‖f2‖ pp1p1−p
Note that 1p
= 1p1
+ 1p2⇒ 1
p2= p1−p
pp1⇒ pp1
p1−p = p2.
Thus ‖f2‖ pp1p1−p
= ‖f2‖p2 ⇒‖f1f2‖p ≤ ‖f1‖p1‖f2‖p2
Suppose for k = m ∈ N, we have ‖g‖p ≤∏m
i=1 ‖gi‖pi whenever gi ∈ Lpi and 1p
=∑m
i=11pi
.
Then let 1q
=∑m+1
i=11qi≤ 1 and fi ∈ Lqi . We have, 1
q= 1
qm+1+∑m
i=11qi
. Then by induction hypothesis,
(f1f2 . . . fm) ∈ L1∑m
i=11qi
and
‖f‖q ≤ ‖fm+1‖qm+1 .‖f1f2 . . . fm‖ 1∑mi=1
1qi
≤ ‖fm+1‖qm+1 .m∏i=1
‖fi‖qi =m+1∏i=1
‖fi‖pi
Thus by induction principle the result is true for all k ∈ N.
4.4.2 Suppose α 6= 0, 1. Put p1 = pα, p2 = q
1−α . Then note that f ∈ Lp ⇒ fα ∈ L pα . Similarly,
f 1−α ∈ Lq
1−α . So applying 4.4.1 with p1, p2 defined as above and noting that 1r
= 1p1
+ 1p2≤ 1, We have
‖f‖r ≤ ‖fα‖p1 .‖f 1−α‖p2 =
(∫|fα|p/α
)αp
.
(∫|f 1−α|
q1−α
) 1−αq
= ‖f‖αp‖f‖1−αq
For the cases α = 0, 1, the inequality is trivially true.
Problem 4.5
4.5.2 We call the set C. Let fn be a sequence in C which converges to f in Lp(Ω). We want to
prove that f ∈ C.Now fn → f in Lp(Ω)⇒ ∃ a subsequence (fnk) and a function h ∈ Lp such that
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Analysis Subhadip Chowdhury Assignment 4
(a) fnk(x)→ f(x) a.e. on Ω
(b) |fnk(x)| ≤ h(x) ∀k, a.e. on Ω
Then applying Fatou’s lemma to (|fnk |q), we get,∫|f |q ≤ 1 ⇒ f ∈ C. Thus C contains all its limit
points. hence C is closed.
4.5.3 By 4.4.2, we have
‖fn − f‖r ≤ ‖fn − f‖αp‖fn − f‖1−αq
where α is defined as in 4.4.2. Note that by part 2, we have ‖f‖q ≤ C. Thus ‖fn − f‖q ≤ 2C. Hence
‖fn − f‖r ≤ ‖fn − f‖αp (2C)1−α → 0
SO fn → f in Lr(Ω).
Problem 4.6
4.6.1 Taking p→∞ in problem 4.2, we get that
lim supp→∞
‖f‖p ≤ ‖f‖∞ (1)
We know that there exist a sequence Cn → ‖f‖∞ such that |f(x)| ≤ Cn a.e on Ω for all n. Then letΩn = x ∈ Ω : |f(x)| > Cn. We have
‖f‖pp ≥∫
Ωn
|f |p ≥ µ(Ωn)(Cn)p ⇒ ‖f‖p ≥ Cn(µ(Ω))1/p
Taking p→∞, we then have lim infp→∞
‖f‖p ≥ Cn for all n. Taking n→∞ we then have
lim infp→∞
‖f‖p ≥ ‖f‖∞ (2)
Combining above inequalities (1) and (2), we get our result.
4.6.2 Fix a k > C. Consider the set S = x ∈ Ω : |f(x)| > k. Suppose µ(S) = ε > 0. Then
C ≥ ‖f‖p ≥ kε1/p
Taing p→∞ we then have C ≥ k,contradiction!! So µ(S) = 0 proving that ‖f‖∞ ≤ C i.e. f ∈ L∞(Ω).
4.6.3 Consider f(x) = ln x for x ∈ (0, 1). Then we know that for 1 ≤ p <∞
limt0
(ln t)12p = lim
t→0+
1t
− 12pt−
12p−1
= 0
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Analysis Subhadip Chowdhury Assignment 4
⇒ ∃ε > 0 such that for 0 < x < ε, | lnx|p < 1
x12
. Then∫ 1
0
| lnx|pdx <∫ ε
0
1√xdx+
∫ 1
ε
| lnx|pdx
=ε12
−12
+ a finite number
<∞
Thus f ∈ Lp(Ω) for 1 ≤ p <∞. But clearly | lnx| is unbounded on (0, 1). So f 6∈ L∞(Ω).
Problem 4.7
Let (un) be sequence in Lp(Ω) converging to u such that aun → f in Lq(Ω). By theorem 4.9, by passingto a subsequence if necessary, we may assume un(x) → u(x) a.e. Then aun(x) → au(x) a.e. Hencef = au a.e. Thus by closed graph theorem, the operator T : Lp → Lq given by u 7→ au is continuous. So∃C > 0 such that
‖au‖q ≤ C‖u‖p (*)
Case 1: If p <∞, then ∫Ω
|au|q ≤ Cq
(∫Ω
|u|p) q
p
Putting u = v1q for some v ∈ L
pq , we get that∫Ω
|a|q|v| ≤ Cq
(∫Ω
|v|pq
) qp
= Cq‖v‖ pq
for all v ∈ Lpq . Thus the map S : L
pq → R given by v 7→
∫Ω|a|q|v| is a continuous linear map.
Hence S ∈(Lpq
)∗⇒ aq ∈ L
11− qp i.e. a ∈ L
qpp−q .
Case 2: If p =∞, from equation (∗) we have ‖a‖q ≤ C‖1‖∞ ⇒ a ∈ Lq. [Note that 1 ∈ L∞].
Problem 4.8
4.8.1 For every integer n ≥ 1, consider the sets
Xn = f ∈ X ∩ L1+ 1n (Ω) : ‖f‖1+ 1
n≤ n
Note that if f ∈ X then f ∈ Lp for some p > 1. But also f ∈ L1. Hence f ∈ Lr for all 1 ≤ r ≤ p. Notethat from 4.4.2, ‖f‖r ≤ K for some constant K for all 1 ≤ r ≤ p. Hence f ∈ Xn for some n ∈ N.Thus ⋃
n∈N
Xn = X
Then since each Xn is closed, by Baire category theorem, there exists N ∈ N such that Int(XN) 6= ∅.Let x0 ∈ Int(XN). Then there exists r > 0 such that x0 +B(0, r) ⊆ XN ⊆ L1+ 1
n . But x0 ∈ L1+ 1n .
Hence B(0, r) ⊆ L1+ 1n ⇒ λB(0, r) ⊆ L1+ 1
n for all λ ∈ R⇒ X ⊆⋃λ
λB(0, r) ⊆ Lq for q = 1 + 1n.
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Analysis Subhadip Chowdhury Assignment 4
4.8.2 Consider the inclusion map T : X → L1 given by Tu = u. Suppose we have a sequence (un)
converging to u in X (note that X is closed) such that Tun → v in L1. Then by theorem 4.9, there is asubsequence (unk) such that unk(x)→ u(x) a.e. Then Tunk(x)→ u(x) a.e. implying u = v a.e. So byclosed graph theorem, T is continuous. Hence there exist some C such that
‖f‖p ≤ C‖f‖1; ∀f ∈ X
Problem 4.9
We may assume WLOG |Ω| = 1, by dividing the measure by |Ω| if necessary. Then∫Ω
j(f)dµ =
∫Ω
supa,b∈R
at+b≤j(t)∀t∈D(j)
(af + b)dµ since j is convex l.s.c.
≥ supa,b∈R
at+b≤j(t)∀t∈D(j)
(b+ a
∫Ω
fdµ
)
= j
(∫Ω
fdµ
)
Problem 4.10
4.10.1 Since j is convex and integration is linear, we have J is convex.
Problem 4.11
4.11.2 We want to show that∫|u+ v|α ≤
∫(|u|+ |v|)α ≤
∫|u|α +
∫|v|α
We claim that (|u|+ |v|)α ≤ |u|α + |v|α. Then it is enough to show that for x ≥ 0,
(x+ 1)α ≤ xα + 1 (*)
Let f(x) = (x+1)α−xα. Then f ′(x) = α((x+1)α−1−xα−1) < 0 implying that f is a decreasing functionof x ∈ [0,∞). Hence f(x) ≤ f(0) = 1⇒ (∗).Putting x = |u|
|v| , we get our required result.
4.11.2 Let
f =
∫|u|α g =
∫|v|α
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Analysis Subhadip Chowdhury Assignment 4
Then We have
[u]α + [v]α = f1α + g
1α = f
1α−1
∫|u|α + g
1α−1
∫|v|α
=
∫f
1α−1|u|α +
∫g
1α−1|v|α
=
∫f
1α
(1−α)|u|α +
∫g
1α
(1−α)|v|α
≤∫ (
f1α
)1−α(|u+ v|α) +
∫ (g
1α
)1−α(|u+ v|α) since u, v ≥ 0
=
∫ (f
1α + g
1α
)1−α(|u+ v|α)
⇒ ([u]α + [v]α)α =(f
1α + g
1α
)α≤∫|u+ v|α = ([u+ v]α)α
⇒ [u]α + [v]α ≤ [u+ v]α
Thus Lα is a vector space but not a norm.
Problem 4.12
4.12.1 Since the inequality is homogeneous, we may divide by |b|p if necessary, to assume wlog
that b = 1. Let
f(a) =(|a|p + 1)1− p
2
(|a|p + 1− 2 |a+1|p
2p
) p2
|a− 1|p
be a function from a : |a+ 1| > 0 → R.
Problem 4.13
4.13.1
Case 1: |a+ b| ≥ |a| Then,
||a+ b| − |a| − |b|| ≤ ||a+ b| − |a||+ | − |b|| = |a+ b| − |a|+ |b| ≤ |b|+ |b| = 2|b|
Case 2: |a+ b| < |a| Then |a| − |a+ b| ≤ | − b| = |b| ⇒ ||a| − |a+ b|| ≤ |b|. Thus,
||a+ b| − |a| − |b|| = ||a| − |a+ b|+ |b|| ≤ |b|+ |b| = 2|b|
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Analysis Subhadip Chowdhury Assignment 4
4.13.2 Note that supn∫|fn| ≤M . Thus by Fatou’s lemma,
∫|f | ≤ lim
n→∞
∫|fn| ≤ ∞. So f ∈ L1(Ω).
Let a = fn − f, b = f . Consider the sequence ϕn = ||a+ b| − |a| − |b||. Then ϕn ≤ 2|b| = 2|f |.Also note that (|a+ b| − |a| − |b|)(x) = |fn(x)| − |f(x)|+ |fn(x)− f(x)| → 0.Then by dominated convergence theorem, ‖ϕn‖1 → 0. Now∫
(|fn| − |fn − f |)−∫|f | ≤
∫(||fn| − |fn − f | − |f ||) =
∫|ϕn| → 0
⇒∫
(|fn| − |fn − f |)→∫|f | (*)
4.13.3 Note that ‖fn‖ → ‖f | ⇒ ‖fn‖ < M for some constant M for all n. Then by part 2,
limn→∞
‖fn − f‖ = limn→∞
∫|fn − f |
by (∗)= lim
n→∞
∫|fn| −
∫|f | = lim
n→∞‖fn‖ − ‖f‖ = 0
Problem 4.14
4.14.1 Let us denote the measure by µ. See part 2.
4.14.2 Note that for a.e. x, fn(x)→ f(x)⇒ |fn(x)− f(x)| < α for all n bigger than a sufficiently
large N . Thus
x 6∈⋃k≥N
x : |fk(x)− f(x)| > α
for some N for a.e. x. Thus χSn(α)(x)→ 0 a.e.Also clearly |χSn(α)| < 1 ∈ L1(Ω) since µ(Ω) <∞. Thus by dominated convergence theorem,∫
|χSn(α)|n→∞−−−→ 0⇒ µ(Sn(α)) −−−→
n→∞0
Also µ[|fn − f | > α] ≤ µ(Sn(α))⇒ µ[|fn − f | > α] −−−→
n→∞0⇒ 4.14.1
4.14.3 Let α = 1m
for some fixed m ∈ N. Then by part 2,
µ(Sn(1
m))
n→∞−−−→ 0
Thus given δ > 0, there exists Nm ∈ N such that µ(Sn( 1m
)) < δ2m
for all n ≥ Nm.Now
x 6∈ SNm(1
m)⇒ |fk(x)− f(x)| ≤ 1
mfor all k ≥ Nm
Let Σm = SNm( 1m
). Let Σ =⋃m∈N
Σm. Then µ(Σ) < δ. Note that x 6∈ Σ⇒ x 6∈ Σm for all m.
Fix any ε > 0. Then there exists M ∈ N such that 1M< ε Then x ∈ Ω\Σ⇒ x 6∈ ΣM ⇒
|fk(x)− f(x)| ≤ 1
M< ε for all k ≥ NM
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Analysis Subhadip Chowdhury Assignment 4
i.e. fk → f uniformly on Ω\Σ. Taking A = Σ, we prove the theorem.
4.14.4 Fix an ε > 0. Then ∃δ > 0 such that
∫A
|fn|p < ε for all n and ∀A ⊆ Ω measurable with
µ(A) < δ.By part 3, there exists a certainAδ and an integerNδ > 0 such that µ(Aδ) < δ and |fn(x)−f(x)|p < ε
µ(Ω\Aδ)for all n ≥ Nδ for all x ∈ Ω\Aδ. So in particular,∫
Aδ
|fn|p < ε and
∫Ω\Aδ|fn(x)− f(x)|p < ε
for all n ≥ Nδ. Then ∫|fn − f |p =
(∫Aδ
|fn − f |p +
∫Ω\Aδ|fn − f |p
)≤∫Aδ
|f |p +
∫Aδ
|fn|p + ε
= 2ε+
∫Aδ
|f |p
But |fn|p ∈ L1(Ω) and supn∫Aδ|fn|p < ∞ . Also |fn(x)|p → |f(x)|p for a.e. x ∈ Aδ. So by Fatou’s
lemma,∫Aδ|f |p ≤ lim inf
n→∞
∫Aδ|fn|p ≤ ε. Thus∫
|fn − f |p ≤ 3ε for all n ≥ Nδ
Hence ‖fn − f‖pn→∞−−−→ 0 i.e. fn → f in Lp(Ω).
Note that (fn − f) and fn both are in Lp(Ω), imply that f ∈ Lp(Ω).
Problem 4.15
4.15.1
4.15.1.(i) For x 6= 0,
limn→∞
ne−nx = limn→∞
n
enx= lim
n→∞
11n
+ x+ nx2/2 + n2x3/6 + . . .= 0
Thus fn → 0 a.e.
4.15.1.(ii)
‖fn‖1 =
∫ 1
0
|fn| =∫ 1
0
ne−nxdx =
∫ n
0
e−zdz = 1− e−n ≤ 1;∀n ∈ N
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Analysis Subhadip Chowdhury Assignment 4
4.15.1.(iii)‖fn − 0‖1 = 1− e−n 6→ 0
4.15.2
4.15.2.(i) For x 6= 0,
limn→∞
n1p e−nx = lim
n→∞
n1p
enx= lim
n→∞
1
n−1p + n1− 1
px+ n2− 1px2/2 + n3− 1
px3/6 + . . .= 0
Thus gn → 0 a.e.
4.15.2.(ii)
‖gn‖p =
∫ 1
0
|gn|p =
∫ 1
0
ne−pnxdx =1
p
∫ np
0
e−zdz =1
p(1− e−np) ≤ 1
p;∀n ∈ N
4.15.2.(iii)
‖gn − 0‖p =1
p(1− e−np) 6→ 0
4.15.2.(iv) Observe that∫ 1
0
gn(x)xmdx = n1p
∫ 1
0
e−nxxm =n
1p
nm+1
∫ n
0
e−yymdy ≤ n1p
nm+1Γ(m+ 1)
n→∞−−−→ 0
Thus for any polynomial ϕ(x) ∈ R[x], we have∫ 1
0
gnϕ→ 0
Since the polynomials are dense in Lp′(Ω), we have∫ 1
0
gnh→ 0
for all h ∈ Lp′(Ω) i.e. gn 0.
Problem 4.16
4.16.1 Let
Kn = conv (∪∞i=nfi)We claim that ∩∞n=1Kn = f. Indeed since fn’s are bounded, by Cantor’s intersection theorem,∩∞n=1Kn 6= ∅. if g ∈ ∩∞n=1Kn, then g(x) ∈ conv (∪∞i=nfi(x)) for all n i.e. fn(x) → g(x) a.e. Thusf(x) = g(x) a.e. So g ≡ f in Lp(Ω) and ∩∞n=1Kn = f. Then by exercise 3.13.2, fn f weakly inσ(Lp, Lp
′).
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Analysis Subhadip Chowdhury Assignment 4
4.16.2 By theorem 4.9 applied to the sequence (fn − f) in Lp, we get that there is a subsequence
(fnk − f) such that fnk − f → 0 a.e. Then by part 1, fnk f weakly. Indeed, we can say that everysubsequence of (fn− f) has a further subsubsequence (fnki − f)such that fnki f weakly. Thus fn fweakly.
4.16.3 fix an ε > 0. By Egorov’s theorem, ∃A ⊆ Ω such that |A| < ε and fn → f uniformly on
Ω\A. Now can write ∫|fn − f |q =
∫A
|fn − f |q︸ ︷︷ ︸I
+
∫Ω\A|fn − f |q︸ ︷︷ ︸J
Note that by uniform convexity for sufficiently large n,
J ≤ |Ω\A|εq
And by problem 4.2, we have
I1q ≤ |A|
1q− 1p‖fn − f‖p <⇒ I ≤ ε1−
qp‖fn − f‖qp
But ‖fn − f‖p is bounded by some constant for all n since fn, f ∈ Lp(Ω)⇒ fn − f ∈ Lp(Ω) and fn is
bounded in Lp. Thus I + J < C1εq + C2ε
1− qp for constants Ci not depending on p. So,∫
|fn − f |q → 0⇒ ‖fn − f‖q → 0
Problem 4.17
4.17.1 Note that the inequality is homogeneous of degree p. WLOG, let |a| ≥ |b|. Then we can
divide both sides by |a|p. Let t = b/a. Then it is enough to show that
||1 + t|p − 1− |t|p| ≤ C(|t|+ |t|p−1
)for some constant C depending on p for all |t| ≤ 1. Note that
limt→0+
||1 + t|p − 1− |t|p|(|t|+ |t|p−1)
= limt→0+
|p(1 + t)p−1 − p(t)p−1|(1 + (p− 1)tp−2)
= p
and
limt→0−
||1 + t|p − 1− |t|p|(|t|+ |t|p−1)
= limt→0−
|p(1 + t)p−1 − p(−t)p−1|(1 + (p− 1)(−t)p−2)
= p
Let
g(t) =
||1+t|p−1−|t|p|
(|t|+|t|p−1)t 6= 0
p t = 0
Then clearly g is a continuous function on [−1, 1]. So it is bounded above since [−1, 1] is compact. Thus
sup|t|≤1
||1 + t|p − 1− |t|p|(|t|+ |t|p−1)
< C
for some C depending only on p which proves the given inequality.
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Analysis Subhadip Chowdhury Assignment 4
4.17.2 Note that by problem 4.16.1, we have fn − f 0 in Lp. Also f ∈ Lp ⇒ fp ∈ L1 ⇒ fp−1pp ∈
Lpp−1 ⇒ fp−1 ∈ Lp′ . So ∫
|fn − f ||f |p−1 → 0
Also note that ∫ (|fn − f |p−1
) 1
1− 1p =
∫|fn − f |p <∞
So by 4.16.1, (fn − f)p−1 0 in Lp′. Thus∫
|fn − f |p−1|f | → 0
Put a = fn − f and b = f in question 1. Then∫||fn|p − |fn − f |p − |f |p| <
∫C(|fn − f |p−1|f |+ |fn − f ||f |p−1
)→ 0
The assertion in part 2 follows.
4.17.3 By part 2,
limn→∞
∫|fn − f |p = 0
So ‖fn − f‖p → 0.
Problem 4.18
4.18.1 Take any step function g ∈ L∞(0, 1) such that g =∑l
i=1 aiχAi , where ai ∈ R and Ai is an
interval (bi, bi+1) with⊔Ai = (0, 1). We have∫ 1
0
ung =
∫ 1
0
f(nx)g(x)dx
=1
n
∫ n
0
f(y)g(y/n)dy
=1
n
(l∑
i=1
ainbi+1 − nbi
T
∫ T
0
f(y)dy
)+
1
n
l∑i=1
ai
∫ nbi+1
nb bi+1−biT
cT+nbi
f(y)dy
≤
(f
l∑i=1
ai
∫ bi+1
bi
dy
)+
1
n
l∑i=1
ai
∫ nbi+1
nb bi+1−biT
cT+nbi
|f(y)|dy
≤∫ 1
0
fg +1
n
l∑i=1
ai‖f‖1,[0,T ] [since f is periodic]
→∫ 1
0
fg
since f ∈ Lploc means the second term is of the form bounded termn
; and hence it goes to 0. Since functions
of above form are dense in Lp′(0, 1), we get that un f weakly.
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Analysis Subhadip Chowdhury Assignment 4
4.18.2 Let kT ≤ n ≤ (k + 1)T for k ∈ Z.∫ 1
0
|un − f |p =
∫ 1
0
∣∣f(nx)− f∣∣p dx
=1
n
∫ n
0
∣∣f(y)− f∣∣p dy
=1
n
(k
∫ T
0
∣∣f(y)− f∣∣p dy +
∫ n
kT
∣∣f(y)− f∣∣p dy)
≤ 1
T
∫ T
0
∣∣f(y)− f∣∣p dy +
1
n‖f − f‖pp,[0,T ]
⇒∫ 1
0
|un − f |p −1
T
∫ T
0
∣∣f(y)− f∣∣p dy → 0
So
limn→∞
‖un − f‖p =
(1
T
∫ T
0
∣∣f(y)− f∣∣p dy) 1
p
4.18.3
4.18.3.(i) un(x) = sin(nx)⇒ un ∈ Lp(0, 1) for all 1 ≤ p ≤ ∞.
By part 1, un 1
2π
∫ 2π
0sin ydy = 0 weakly in σ(Lp, Lp
′).
Also ‖un − sin‖p →(
12π
∫ 2π
0|sin y|p dy
) 1p
4.18.3.(ii)∫ 1
0|un|pdx = n
2(αp+βp) <∞ for all 1 ≤ p <∞. Hence un ∈ Lp(0, 1) for all 1 ≤ p <∞.
Also sup |un| = maxα, β. Hence un ∈ Lp(0, 1) for all 1 ≤ p ≤ ∞.un
α+β2
weakly in σ(Lp, Lp′).
Also ‖un − f‖p →(∫ 1
0|f |p dy
) 1p
=(|α|p+|β|p
2
) 1p
Problem 4.19
4.19.1 Since Lp is uniformly convex for 1 < p <∞, we have by prop. 3.32, fn → f strongly in Lp.
4.19.2 Consider fn(x) = sin(2πnx) + 1. Then given g ∈ L∞(0, 1), we have∫ 1
0
sin(2πnx)g(x)dx+
∫ 1
0
g =1
n
∫ n
0
sin(2πy)g(y)dy +
∫ 1
0
g ≤ 1
n‖g‖∞ ×+
∫ 1
0
g →∫ 1
0
g
So, fn 1.
12
Analysis Subhadip Chowdhury Assignment 4
Problem 4.20
4.20.1 Clearly by the given inequality, f ∈ Lp ⇒ Af ∈ Lq. Now take any sequence un → u
strongly in Lp. Then by theorem 4.9, we can get a subsequence unk and a function h ∈ Lp such thatunk(x)→ u(x) a.e. and |unk(x)| ≤ h(x) for all k, a.e. Then for some h′ ∈ Lq, we have Aunk(x)→ Au(x)a.e. and |Aunk(x)| ≤ h′(x) for all k, a.e. Thus by dominated convergence theorem applied to (Aunk)
q weget that ‖Aunk −Au‖q → 0 By uniqueness of limit we then get that Aun → Au strongly in Lq. Hence Ais continuous from Lp strong to Lq strong.
4.20.2 From 4.18.3(ii), we know that un(x) = f(nx) α+β2
where f is defined as in the problem
4.18.3(ii). Now by given hypothesis Aun → Au; i.e. a(un) a(α+β2
) weakly in σ(Lq, Lq′). Now
a(un(x)) = a(f(nx)) =
a(α) k
n< x <
k+ 12
n
a(β)k+ 1
2
n< x < k+1
n
Since 1 ∈ Lq′(0, 1), integrating Aun and Au against 1, we get
n1
2n(a(α) + a(β)) = a(
α + β
2)
i.e1
2(a(α) + a(β)) = a(
α + β
2)
for all α, β ∈ R. Clearly the only solution to the functional equation is a(x) = (a(1) − a(0))x + a(0)which is an affine function.
Problem 4.21
4.21.1 Take g = χ(a,b). Then∫Rung =
∫ b
a
u0(x+ n)dx =
∫ b+n
a+n
u0(x)dx =
∫ b+n
0
u0(x)dx−∫ a+n
0
u0(x)dx→ 0
as n → ∞. Thus for any step functions(with compact support) g, we have∫ung → 0. Since step
functions(with compact support) is dense in Lp(R). we get that un 0 weakly.
4.21.2 Take g = χI . Let I = (a, b). Fix a δ > 0. Let E = [|u0| > δ]. Then∫Rung =
∫I
u0(x+ n)dx =
∫I+n
u0(y)dy =
∫(I+n)∩E
u0(y)dy +
∫(I+n)∩Ec
u0(y)dy
Now since |E| <∞, there exists N such that for n > N , we have |(I + n) ∩ E| < δ. Then∣∣∣∣∫Rung
∣∣∣∣ ≤ ‖u0‖∞δ + δ|(I + n) ∩ Ec| ≤ ‖u0‖∞δ + δ|I|
Taking δ → 0, we get that∣∣∫
R ung∣∣→ 0. Thus for step function g with compact support,
∣∣∫R ung
∣∣→ 0.
Since step functions with compact support are dense in L1(R), we get that un∗ 0.
13
Analysis Subhadip Chowdhury Assignment 4
4.21.3 Suppose, by contradiction, there is a subsequence unk which weakly converge. Then for any
g ∈ L∞,∫unkg → 0. Take g ∈ L∞ defined by
g(x) =
(−1)i ni < x < ni+1
0 o.w.
Then∫unkg = (−1)k which do not converge as k →∞. Thus we have a contradiction! Hence such a
subsequence does not exist.
Problem 4.22
4.22.1 A⇒ B Since χE ∈ Lp′
for any measurable E with |E| <∞, this implication is clear by
definition of weak convergence. [or weak* convergence in case p =∞.]
B ⇒ A Note that, since simple functions are dense in Lp′(Ω), the vector space spanned by characteristic
functions χA of subsets A of Ω with |A| <∞ are dense in Lp′(Ω) for 1 ≤ p′ <∞. Now for any element
g of the vector space, we have∫
Ωfng →
∫Ωfg. Thus
∫Ωfnh→
∫Ωfh for all h ∈ Lp′(Ω) i.e. fn f .
4.22.2 Again the implication A⇒ B is trivial. For the other direction observe that every simple
function belongs to L∞(Ω) if |Ω| <∞. Hence the vector space spanned by characteristic functions χA ofsubsets A of Ω with |A| <∞ are dense in L∞(Ω). Thus A follows by the same argument as in part 1.
4.22.3 By proposition 3.5, and definition of weak convergence, A⇒ B.
To see that the opposite implication is not true consider Ω = R. Consider the sequence fn = χ(n,n+1).then clearly ‖fn‖1 = 1 <∞ and
∫Efn → 0 =
∫E
0 for any E ⊆ R with |E| <∞. But clearly by problem4.21.3 fn does not weakly converge.
4.22.4 Given ε > 0, we can find a subset ω ⊆ Ω such that∫ωcf < ε and |ω| <∞.
Now ∫ωcfn +
∫ω
fn =
∫Ω
fn →∫
Ω
f =
∫ωcf +
∫ω
f
⇒∫ωcfn −
∫ωcf →
∫ω
f −∫ω
fn → 0 (*)
since |ω| <∞. Also we have, for any F ⊆ Ω with |F | ≤ ∞,∫F
fn =
∫F∩ω
fn +
∫F∩ωc
fn →∫F∩ω
f +
∫F∩ωc
fn
⇒∫F
(fn − f)→∫F∩ωc
(fn − f) ≤∫ωcfn +
∫ωcf → 2
∫ωcf < 2ε
since fn, f ≥ 0. Thus we get∫F
(fn − f)→ 0 for F measurable subset of Ω, |F | ≤ ∞. Since vector spacespanned by χF for such F ’s is dense in L∞(Ω), we get that fn f .
14
Analysis Subhadip Chowdhury Assignment 4
Problem 4.23
4.23.1 Clearly by definition C is convex. Suppose (un) is a sequence in C which converge to u in
Lp. Then there is a subsequence (unk) such that unk(x)→ u(x) a.e.[since 1 ≤ p <∞] Thus u(x) ≥ f(x)a.e. Hence C is closed.Now C is convex and strongly closed ⇒ C is weakly closed.
4.23.2 Assume first that f ∈ L∞(Ω). Call the set mentioned in part 2 C ′. Then clearly, C ⊆ C ′.
Now suppose, u ∈ C ′. Thus∫fϕ ≥
∫uϕ ∀ϕ ∈ L1(Ω) with fϕ ∈ L1(Ω) and ϕ ≥ 0 a.e. Since Ω is
σ−finite we can find Ωn with ∪Ωn = Ω and |Ωn| <∞. Let Ω′n = Ωn ∩ |f | < n. Then⋃
Ω′n = Ω. LetA = f < u. Then choose ϕ = χA∩Ωn∫
A∩Ω′n
|f − u| ≤ 0⇒ |A ∩ Ω′n| = 0⇒ |A| = 0
So f ≥ ua.e. implying C ′ ⊆ C.
4.23.3 The second definition C clearly shows that C is closed in σ(L∞, L1) since it is the intersection
of all closed sets in σ(L∞, L1) of the formu ∈ L∞(ω)
∣∣∣∣∫ fϕ ≥∫uϕ and fϕ ∈ L1(Ω)
for a fixed 0 ≤ ϕ ∈ L1(Ω).
4.23.4 Note that a weak* closed and bounded set is compact in weak* topology, since we can scale
down the bounded set into a closed subset of the unit ball in weak* topology which is compact. Hencebeing a closed subset of a compact set, the scaled down set will be compact giving us that the originalset is compact. Now given set C is a weak*(σ((L1)∗, L1))closed and bounded set; hence it is compact.
Problem 4.24
4.24.1 Given ϕ ∈ L1(RN), we have by prop. 4.16,∫ϕ(ρn ? ζnu) =
∫ζnu(ρn ? ϕ)
Hence, ∫vnϕ−
∫vϕ =
∫ζnu(ρn ? ϕ)−
∫ζϕu
=
∫ζnu(ρn ? ϕ− ϕ) +
∫uϕ(ζn − ζ)
≤ ‖ζn‖∞‖u‖∞‖ρn ? ϕ− ϕ‖1 + ‖u‖∞‖(ζn − ζ)ϕ‖1
≤ ‖u‖∞‖ρn ? ϕ− ϕ‖1 + ‖u‖∞‖(ζn − ζ)ϕ‖1
→ 0
first term by thm 4.22. and second term by DCT.
15
Analysis Subhadip Chowdhury Assignment 4
4.24.2 Let B = B(x0, R) and let χ := χB(x0,R+1). Then let wn = ρn ? (ζnuχ). Note that wn = vnon B since by proposition 4.18,
supp(wn − vn) ⊆ B(0,1
n) +B(x0, R + 1)c)
Thus ∫B
|vn − v| =∫B
|wn − χv|
≤∫RN|ρn ? (ζnuχ)− χζu|
≤∫RN|ρn ? (ζnuχ)− ρn ? (χζu)|+
∫RN|ρn ? (χζu)− χζu|
≤∫RN|(ζnuχ)− (χζu)|+
∫RN|ρn ? (χζu)− χζu|
→ 0
Problem 4.25
4.25.1 Let
u′(x) =
u(x) x ∈ Ω
0 x ∈ Ωc
Then u′ ∈ L∞(RN). LetΩn = x ∈ Ω|dist(x,Ωc) > 2/n; |x| < n
Defineζn := χΩn , ζ := χΩ
Then note that ζn → ζ a.e. on RN and ‖ζn‖∞ ≤ 1 for all n. Define vn = ρn ? (ζnu′) and v = ζu′ = u′.
Then by problem 4.24, vn∗ v in L∞(RN). Thus vn
∗ u = u′ on L∞(Ω). Also vn ∈ C∞c (Ω). and∫
B|vn − v| =
∫B|vn − u′| → 0 for every ball B. Thus we may find an subsequence (vnk) which converges
a.e. to u′(x) on B. By a diagonal process, we can find a further subsequence (vnki ) which converges a.e.
to u′ on RN . Renaming this subsequence (un), we see that un satisfies all given properties.
4.25.2 By our construction u ≥ 0 a.e. ⇒ ρn ? (ζnu′) ≥ 0 a.e.
4.25.3 By part 1, C∞c (Ω) is dense in L∞(Ω).
16
Analysis Subhadip Chowdhury Assignment 4
Problem 4.26
4.26.1 If f ∈ L1(Ω) then A ≤ sup‖ϕ‖∞≤1
‖f‖1‖ϕ‖∞ ≤ ‖f‖1 <∞.
Conversely, if A <∞, Then for all ϕ ∈ Cc(Ω), we have∫f
ϕ
‖ϕ‖∞≤ A⇒
∫fϕ ≤ A‖ϕ‖∞
Fix any compact subset K of Ω. Let ψ ∈ Cc(Ω) be a function such that 0 ≤ ψ ≤ 1 and ψ ≡ 1 on K. Letu ∈ L∞(Ω) be any function. Then by problem 4.25 , we can find a sequence (un) in Cc(Ω) such that
‖un‖∞ ≤ ‖u‖∞ and un → u a.e. on Ω and un∗ u in σ(L∞, L1). Then taking ϕ = unψ we get∫
fψun ≤ A‖ψun‖∞ ≤ A‖u‖∞
By DCT, taking limit as n→∞, we then have∫fψu ≤ A‖u‖∞
Choose u = sign(f). Then ∫K
|f |.1 ≤∫f.u.ψ ≤ A
for all compact subset K of Ω. Thus ‖f‖1 ≤ A <∞ i.e. f ∈ L1(Ω).
In case f ∈ L1(Ω), by above proof we have A ≤ ‖f‖1 ≤ A⇒ ‖f‖1 = A.
4.26.2 If f+ ∈ L1(Ω) then B ≤ sup‖ϕ‖∞≤1ϕ≥0
∫(maxf, 0ϕ) ≤ sup
‖ϕ‖∞≤1ϕ≥0
‖f+‖1‖ϕ‖∞ ≤ ‖f+‖1 <∞.
Conversely if B <∞, we can proceed as in part 1, to get that∫fψu ≤ B‖u‖∞
for all u ∈ L∞(Ω) Put u = χf≥0∩K . Then∫Kf+ ≤ B for every compact subset K of Ω. Hence
‖f+‖1 ≤ B.
The proof clearly shows that ‖f+‖1 = B if f ∈ L1(Ω).
4.26.3 Note that the sequence constructed in problem 4.25 is actually in C∞c (Ω). Hence the proof
is same.
4.26.4 [∫fϕ = 0∀ϕ ∈ C∞c (Ω)
]⇒ A = ‖f‖1 = 0⇒ f = 0 a.e.[∫
fϕ ≥ 0∀ϕ ∈ C∞c (Ω), ϕ ≥ 0
]⇒ B = ‖f+‖1 ≥ 0⇒ f > 0 a.e.
17
Analysis Subhadip Chowdhury Assignment 4
Problem 4.27
Fix a ϕ0 ∈ C∞c (Ω) such that∫uϕ0 = 0. Fix any ε > 0. Then∫
u(ϕ− (ϕ0
∫(uϕ)) + εϕ0) = 0 + ε > 0⇒
∫v(ϕ− (ϕ0
∫(uϕ)) + εϕ0) ≥ 0
Taking ε→ 0, we then have ∫vϕ ≥
∫uϕ
∫vϕ0 = λ
∫uϕ
Conversely, taking ε < 0 and replacing ϕ − (ϕ0
∫(uϕ)) + εϕ0 by ϕ − (ϕ0
∫(uϕ)) − εϕ0 we get that as
ε→ 0, ∫vϕ ≤
∫uϕ
∫vϕ0 = λ
∫uϕ
. Thus ∫(v − λu)ϕ = 0
for all ϕ ∈ C∞c (Ω). Hence v = λu a.e. Clearly by construction λ ≥ 0.
Problem 4.28
We can approximate ρ by a sequence of C∞c (Ω) functions (ψi) such that ‖ρ − ψi‖1 → 0. Defineψin(x) = nNψi(nx). Note that
∫ψin =
∫RN n
Nψi(nx)dx1 . . . dxN =∫RN ψ
i(y)dy1 . . . dyN →∫ρ = 1. Fix
K ⊆ RN a compact set and ε > 0. Then there exist n large enough such that suppψin ⊆ B(0, δ) where δis such that |f(x− y)− f(y)| < ε for all x ∈ K and for all y ∈ B(0, δ). Then for n sufficiently large,∫
|(ψin ? f)(x)− f(x)|pdx =
∫ ∣∣∣∣∫ |f(x− y)− f(x)|ψin(y)dy
∣∣∣∣p dx≤∫ ∫
|f(x− y)− f(x)|pψin(y)dydx
=
∫ψin(y)dy
(∫|f(x− y)− f(x)|pdx
)≤∫B(0,δ)
εpψi ≤ εp∫ψi → εp
Hence taking ε→ 0, we get that ψin ? fn→∞−−−→ f in Lp. Taking limit i→∞, we then have ρn ? f → f in
Lp.
Problem 4.29
Let χn := χK+B(0, 12n
) and take un = ρ2n ? χn where ρ(x) =
e
1|x|2−1 |x| < 1
0 |x| > 1and ρn(x) = 1∫
ρnNρ(nx).
Then all it remains to check is that un satisfies (a), (b), (c) and (d). Note that
‖un‖∞ ≤ ‖χn‖∞‖ρ2n‖1 = 1
18
Analysis Subhadip Chowdhury Assignment 4
So (a) is true. Next on K,
ρ2n ? χn =
∫ρ2n = 1
So (b) holds. Also
supp(un) ⊆ K +B(0,1
2n) +B(0,
1
2n) ⊆ K +B(0,
1
n)
So (c) holds. Lastly,|Dαun(x)| = |Dαρ2n(x) ? χn| ≤ ‖Dαρ2n‖1
NowDα(CnNρ(2nx)) = CnN(2n)|α|(Dαρ)(2nx)
So
‖Dαρ2n‖1 = C(2n)|α|∫|(Dαρ)(2nx)|nNdx = C(2n)|α|
∫|(Dαρ)(2y)|dy = Cαn
|α|
So (d) follows.
Problem 4.30
4.30.2 Let 1p
+ 1p′
= 1 = 1q
+ 1q′
. Then 1r
= 1 − 1p′− 1
q′. Then set α = p
q′and β = q
p′. Note that
(1 − α)r = (1p− 1
q′)rp = 1
rrp = p, and similarly (1 − β)r = q. Then ϕ1(y) = f(x − y)α ∈ Lq′ , ϕ2(y) =
g(y)β ∈ Lp′ . Also for ϕ3(y) = f(x− y)1−αg(y)1−β, we have ϕr3(y) = f(x− y)pg(y)q ∈ L1 ⇒ ϕ3 ∈ Lr. Wehave,
|(f ? g)(x)| =∣∣∣∣∫ f(x− y)g(y)dy
∣∣∣∣ ≤ ∫ |f(x− y)g(y)| dy
=
∫|f(x− y)|α|g(y)|β
(|f(x− y)|1−α|g(y)|1−β
)By Holder’s inequality y 7→ f(x− y)g(y) is integrable. In fact,
|(f ? g)(x)| ≤ ‖ϕ1‖q′‖ϕ2‖p′‖ϕ3‖r = ‖f‖αp‖g‖βq ‖ϕ2‖r
⇒∫|f ? g|r ≤ ‖f‖αrp ‖g‖βrq
∫|f(x− y)(1−α)rg(y)(1−β)r|dy
= ‖f‖αrp ‖g‖βrq∫|f(x− y)pg(y)q|dy
≤ ‖f‖αrp ‖g‖βrq ‖f‖pp‖g‖qq= ‖f‖rp‖g‖rq
⇒ ‖f ? g‖r ≤ ‖f‖p‖g‖q
4.30.3 By previous part f ? g ∈ L∞(RN ) if r =∞. Also we know that convolution of a Lp function
and a Lp′
function is continuous. Hence f ? g ∈ C(RN) ∩ L∞(RN).
For 1 < p, q <∞, note that we can construct sequences (fn), (gn) in Cc(Ω) so that fn → f in Lp andgn → g in Lq. But clearly, supp(fn ? gn) ⊆ supp(fn) + supp(gn) ⇒ (fn ? gn)(x) = 0 as |x| → ∞. But‖fn ? gn − f ? g‖1 → 0. Hence (f ? g)(x)→ 0 as |x| → ∞.
19
Analysis Subhadip Chowdhury Assignment 4
Problem 4.31
4.31.1 Let χr :=χB(0,r)
|B(0,r)| . Then
fr(x) =
∫χr(x− y)f(y)dy = (χr ? f)(x)
Now f ∈ Lp and χr ∈ L1. Hence f ? χr ∈ Lp. Also χr ∈ Lp′ ⇒ f ? χr ∈ C(RN) by problem 4.30. Hence
fr ∈ C(RN) ∩ Lp(RN). Also by 4.30, fr(x)→ 0 as |x| → ∞ for 1 < p <∞.
4.31.2 Note that∫χr = 1, and supp(χr) = B(0, r) Hebce χ 1
nis a sequence of mollifiers. Hence
f1/n → f in Lp as n→∞⇒ fr → f in Lp as r → 0.
Problem 4.32
4.32.1 By definition of convolution it is immediate that f ? g = g ? f . Next note that
(f ? g) ? h(u) =
∫(f ? g)(x)h(u− x)dx
=
∫ (∫f(y)g(x− y)dy
)h(u− x)dx
=
∫ ∫f(y)g(x− y)h(u− x)dydx
=
∫ ∫f(y)g(x− y)h(u− x)dxdy
=
∫f(y)
(∫g(x− y)h(u− x)dx
)dy
=
∫f(y)
(∫g(x+ y − y)h(u− x− y)dx
)dy
=
∫f(y)(g ? h)(u− y)dy
= (f ? (g ? h))(u)
Thus (f ? g) ? h = f ? (g ? h).
Problem 4.33
4.33.1 Note that compactly supported continuous functions are uniformly continuous. Hence
given ε > 0, there exist δ > 0 such that |h| < δ implies |ϕ(h + x) − ϕ(x)| < ε for any x ∈ R. Thusin fact |ϕ(x + n + h) − ϕ(x + n)| < ε for any x ∈ R and any n ∈ N. So for |h| < δ, we have‖τhϕn − ϕn‖p < ε.|supp(ϕ)|1/p for any n. Hence ∀ε > 0;∃δ > 0 such that ‖τhf − f‖p < ε ∀f ∈ F and∀h ∈ R with |h| < δ.
20
Analysis Subhadip Chowdhury Assignment 4
4.33.2 Fix ε = 13. Consider the open balls B(f, ε) ⊆ Lp(R) for each f ∈ F . Then clearly they
form an open cover of the closure of F . If the closure is compact, this cover has a finite subcover.
Let it bek⋃i=1
B(ϕni , ε). Take N >> maxn1 . . . , nk so that supp(ϕn)⋂( k⋃
i=1
supp(ϕni)
)= ∅. Then
‖ϕN − ϕni‖p =(2∫ϕp) 1p = C, some constant. Hence if ε < C/2, ϕN 6∈
k⋃i=1
B(ϕni , ε). Contradiction!
Hence the closure is not compact.
Problem 4.34
Fix an ε > 0. Then there is a finite covering of F by open ε−balls. Let F ⊆n⋃i=1
B(fi, ε).
4.34.1 Given any f ∈ F , f ∈ B(fi, ε) for some i ∈ 1, . . . , n. Hence ‖f‖p ≤ ‖fi‖p+ε ≤ max‖fi‖p :
1 ≤ i ≤ n+ ε. Hence F is bounded.
4.34.2 By lemma 4.3, for each i ∈ 1, . . . , n;
limh→0‖τhfi − fi‖p = 0
Hence for the same ε as above, there exists δi > 0 such that
‖τhfi − fi‖p < ε ∀i and ∀h ∈ RN with |h| < δi
Put δ = minδi : 1 ≤ i ≤ n. Take any f ∈ F , suppose f ∈ B(fk, ε). Then
‖τhf − f‖p ≤ ‖τhf − τhfk‖p + ‖τhfk − fk‖p + ‖fk − f‖p = 2‖fk − f‖p + ‖τhfk − fk‖p ≤ 3ε
for all |h| < δ.
4.34.3 Note that fi ∈ Lp(RN)⇒ ∃Ωi ⊂ RN bounded,open, such that ‖fi‖Lp(RN )\Ωi < ε where ε is
as above. Let Ω =n⋃i=1
Ωi. Take any f ∈ F and let f ∈ B(fk, ε). Then
‖f‖Lp(RN\Ω) ≤ ‖fk‖Lp(RN\Ω) + ‖f − fk‖Lp(RN\Ω) ≤ ‖fk‖Lp(RN\Ωk) + ‖f − fk‖Lp(RN ) = 2ε
Comparing with corollary 4.27, we find that this problem is the converse of the corollary, and togetherthey characterise all compact sets in Lp(RN).
Problem 4.35
Clearly F|Ω is bounded in Lp(Ω). Now take f ∈ F and write f = G ? u for an u ∈ B. Then
‖τhf − f‖p = ‖(τhG−G) ? u‖p ≤ C‖τhG−G‖p → 0
by lemma 4.3. Hence by theorem 4.26, F|Ω has compact closure in Lp(Ω) for any measurable Ω withfinite measure.
21
Analysis Subhadip Chowdhury Assignment 4
Problem 4.36
4.36.1 [(d) + (e)]⇒ [(a) + (b) + (c)]
Suppose (d) and (e) hold. Then given ε > 0, there exists N > 0 such that
supf∈F
∫[|f |>t]
|f | < ε ∀t ≥ N
⇒∫
[|f |>t]|f | < ε ∀f ∈ F , ∀t ≥ N
Now given E measurable with |E| < εN
, we have∫E
|f | =∫E∩[|f |>N ]
|f |+∫E∩[|f |≤N ]
|f | ≤∫
[|f |>N ]
|f |+N |E| ≤ 2ε
Thus given ε > 0, there exists δ > 0 such that (b) holds.
We also know that given ε > 0, there exists N ∈ N such that
supf∈F
∫Ω\Ωn
|f | < ε ∀n ≥ N
⇒∫
Ω\Ωn|f | < ε ∀f ∈ F ∀n ≥ N
So we may take ω = ΩN and thus (c) holds.
Clearly (d) implies (a) is true.
[(a) + (b) + (c)]⇒ [(d) + (e)]
Suppose (a), (b) and (c) hold. Then (a) implies there is an M ∈ N such that∫|f | < M . Fix an ε > 0.
Then (b) gives us an δ > 0. Let E = [|f | > Mδ
] Then |E| < δ since otherwise,∫E|f | > M >
∫|f | which
is not possible. Thus
(b)⇒ supf∈F
∫[|f |>M
δ]
|f | < ε
Thus given ε > 0, there exists a N > 0 such that supf∈F
∫[|f |>t] |f | < ε for all t > N implying
limt→∞
supf∈F
∫[|f |>t]
|f | = 0
By (c) we can produce an ω satisfying properties in (c). Then note that since |ω| <∞, and Ωn arenondecreasing with union the whole of Ω; we can find a N ∈ N such that |ω∆Ωn| < ε/M for all n > N .Then ∫
Ω\Ωn|f | ≤
∫Ω\ω|f |+
∫Ωn∆ω
|f | ≤∫
Ω\ω|f |+ ε ≤ 2ε ∀f ∈ F
Thus we have (e) i.e.
limn→∞
supf∈F
∫Ω\Ωn
|f | = 0
22
Analysis Subhadip Chowdhury Assignment 4
Problem 4.37
4.37.1 We have∫I
un(x)ϕ(x)dx =
∫I
nf(nx)ϕ(x)dx
=
∫ n
−nf(t)ϕ(t/n)dt
=
∫ n
−nf(t) (ϕ(t/n)− ϕ(0)) dt+ ϕ(0)
∫ n
−nf(t)dt
But the first term converges to 0 by Lebesgue’s theorem and the second term goes to 0 by given condition:∫ +∞
−∞f = 0
4.37.2 ∫I
|un(x)|dx =
∫ n
−n|f(t)|dt ≤
∫ +∞
−∞|f | <∞
since f ∈ L1(R). Hence (un) is bounded.
Note that for all δ > 0, ∫ δ
0
|un| =∫ nδ
0
|f | →∫ ∞
0
|f | > 0
Hence there exist∫∞
0|f | > ε > 0 such that for all δ > 0, there exists E = (0, δ) such that
∫E|un| > ε for
sufficiently large n. Hence no subsequence of (un) is equi-integrable.
4.37.3 Suppose there exists such a function u ∈ L1(I). Then by part 1,∫I
uϕ = 0 ∀ϕ ∈ C([−1,+1])
Then by corollary 4.24, u ≡ 0 a.e. But for ϕ = χ(0,1), we have∫unϕ =
∫ n
0
f →∫ ∞
0
f > 0
Contradiction! Hence u does not exist.
4.37.4 By Dunford-Pettis thm, we have
[ No subsequence of (un) is equi-integrable]⇒ [no subseqence of (un) weakly converges]
which agrees with the fact that part 2 and part 3 hold simultaneously. The thm states that part 3 is adirect implication of part 2.
23
Analysis Subhadip Chowdhury Assignment 4
4.37.5 ∫I
|un(x)|dx =
∫[−n−
12 ,n−
12 ]
|un(x)|dx+
∫[n−
12<|x|<1]
|un(x)|dx
=
∫ n12
−n12
|f(t)|dt+
∫[n
12<|t|<n]
|f(t)|dt
→ 0 + 0
since f ∈ L1. Thus un → 0 in L1. Hence there is a subsequence (unk) such that unk(x)→ 0 a.e.
Problem 4.38
4.38.1 By definition |supp(un)| =∑n−1
0 ( 1n2 ) = 1
n. Also∫
I
un =n−1∑j=0
∫ jn
+ 1n2
jn
n =n−1∑j=0
n
n2= 1⇒ ‖un‖1 = 1
4.38.2 Suppose ϕ ∈ C1([0, 1]). Then
limn→∞
∫I
unϕ−∫I
ϕ = limn→∞
n−1∑j=0
n
∫ jn
+ 1n2
jn
ϕ− limn→∞
n−1∑j=0
1
nϕ(xj) where xj ∈ (j/n, (j + 1)/n)
= limn→∞
n−1∑j=0
1
n
∫ nj+1
nj
(ϕ
(t
n2
)− ϕ
( yjn2
))dt where yj ∈ (nj, n(j + 1))
= limn→∞
n−1∑j=0
1
n3
∫ nj+1
nj
((t− yj)ϕ′(zj)) dt where zj ∈ (t, yj)
≤ limn→∞
n−1∑j=0
1
n3
∫ nj+1
nj
(|n(j + 1)− nj|ϕ′(zj)) dt
= limn→∞
O(n−1∑j=0
1
n3n)
= limn→∞
O(1
n)
= 0
But C1([0, 1]) is dense in C([0, 1]). Hence limn→∞
∫Iunϕ =
∫Iϕ for all ϕ ∈ C([0, 1]).
4.38.3 The sequence (un) is not equi-integrable because for En = supp(un); |En| → 0 but
1 =
∫I
un =
∫En
|un|
So (b) does not hold in definition of equi-integrability.
24
Analysis Subhadip Chowdhury Assignment 4
4.38.4 If (unk) u, we have by part 2 and corollary 4.24, u ≡ 1 a.e. Choose a further subsequence
(un′k) such that∑
k |supp(un′k)| < 1. Let ϕ = χA where
A = I\
(⋃k
supp(un′k)
)
so that |A| > 0. We have∫Iun′kϕ = 0 for all k and hence 0 =
∫Iϕ = |A|. Contradiction!
4.38.5 Consider a subsequence (unk) such that∑k
|supp(unk)| <∞
Let Bk =⋃j≥k(supp(unj)) and B =
⋃k Bk. Clearly |Bk| =
∑∞j=k
1nj→ 0 as k →∞, and thus |B| = 0.
If x 6∈ B, there exists some k0 such that unk(x) = 0 for all k ≥ k0. Thus unk(x)→ 0 a.e. as k →∞.
25