Real Analysis Final Exam, 200818
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1. (10 points) Let
E
⊂
R
n
be a measurable set and
q
(
x
)
≥
0
be a nonnegative measurable function on
E
. Show that for any constant
α>
0
we have the inequality
Z
E
qdx
·
Z
E
q
α
dx
≤

E
·
Z
E
q
1+
α
dx.
solution:
By Hölder inequality
Z
E
≤
μZ
E
q
1+
α
dx
¶
1
1+
α
μZ
E
dx
¶
α
1+
α
,
Z
E
q
α
dx
≤
μZ
E
(
q
α
)
1+
α
α
dx
¶
α
1+
α
μZ
E
dx
¶
1
1+
α
the proof is done.
¤
2. (10 points) Assume

E

<
∞
and
u
is a measurable function on
E,
which is everywhere positive. For
0
<p<
∞
,
set
Φ
p
(
u
):=
μ
1

E

Z
E

u
(
x
)

p
dx
¶
1
/p
=
μ
1

E

Z
E
u
p
(
x
)
dx
¶
1
/p
.
(0.1)
Assume that
⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
(1)
.u
p
(
x
)
→
1
uniformly on
E
as
p
→
0
+
.
(2)
.
d
dp
¡R
E
u
p
(
x
)
dx
¢
=
R
E
d
dp
(
u
p
(
x
))
dx.
(3)
.

log
u
(
x
)

≤
C
for all
x
∈
for some constant
C>
0
.
Evaluate the limit
lim
p
→
0
+
Φ
p
(
u
)
in terms of an integral involving
u.
(Compare with the limit
lim
p
→∞
Φ
p
(
u
)=
k
u
k
∞
.
)
solution:
As
p
→
0
+
,
Φ
p
(
u
)
has the form
1
∞
.
Hence we can use
Lopital rule
.Notethat
log
Φ
p
(
u
log
³
1

E

R
E
u
p
(
x
)
dx
´
p
μ
0
0
form
¶
and so
lim
p
→
0
+
log
Φ
p
(
u
)= lim
p
→
0
+
d
dp
log
³
1

E

R
E
u
p
(
x
)
dx
´
1
=l
im
p
→
0
+
1

E

d
dp
R
E
u
p
(
x
)
dx
1

E

R
E
u
p
(
x
)
dx
p
→
0
+
1

E

R
E
d
dp
(
u
p
(
x
))
dx
1

E

R
E
u
p
(
x
)
dx
p
→
0
+
1

E

R
E
u
p
(
x
)log
u
(
x
)
dx
1

E

R
E
u
p
(
x
)
dx
=
1

E

Z
E
log
u
(
x
)
dx.
Hencewehavetheformu
la
lim
p
→
0
+
Φ
p
(
u
)=exp
μ
1

E

Z
E
log
u
(
x
)
dx
¶
.
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 Spring '08
 Akhmedov,A
 dx, lim sup

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