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Miklós Schweitzer
1992 Miklós Schweitzer
6
analysis
analysis
Source: miklos schweitzer 1992 q6
October 24, 2021
real analysis
Problem Statement
Let
E
⊂
[
0
,
1
]
E \subset [0,1]
E
⊂
[
0
,
1
]
be a Lebesgue measurable set having Lebesgue measure
∣
E
∣
<
1
2
| E |<\frac{1}{2}
∣
E
∣
<
2
1
. Let
h
(
s
)
=
∫
E
‾
d
t
(
s
−
t
)
2
h (s) = \int _ {\overline {E}} \frac{dt}{{(s-t)}^2}
h
(
s
)
=
∫
E
(
s
−
t
)
2
d
t
where
E
‾
=
[
0
,
1
]
\
E
\overline {E} = [0,1] \backslash E
E
=
[
0
,
1
]
\
E
. Prove that there is one
t
∈
E
‾
t \in \overline {E}
t
∈
E
for which
∫
E
d
s
h
(
s
)
(
s
−
t
)
2
≤
c
∣
E
∣
2
\int_E \frac {ds} {h (s) {(s-t)} ^ 2} \leq c {| E |} ^ 2
∫
E
h
(
s
)
(
s
−
t
)
2
d
s
≤
c
∣
E
∣
2
with some absolute constant c .
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