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1987 IMO Longlists
7
There exists a function u
There exists a function u
Source:
September 5, 2010
function
algebra proposed
algebra
Problem Statement
Let
f
:
(
0
,
+
∞
)
→
R
f : (0,+\infty) \to \mathbb R
f
:
(
0
,
+
∞
)
→
R
be a function having the property that
f
(
x
)
=
f
(
1
x
)
f(x) = f\left(\frac{1}{x}\right)
f
(
x
)
=
f
(
x
1
)
for all
x
>
0.
x > 0.
x
>
0.
Prove that there exists a function
u
:
[
1
,
+
∞
)
→
R
u : [1,+\infty) \to \mathbb R
u
:
[
1
,
+
∞
)
→
R
satisfying
u
(
x
+
1
x
2
)
=
f
(
x
)
u\left(\frac{x+\frac 1x }{2} \right) = f(x)
u
(
2
x
+
x
1
)
=
f
(
x
)
for all
x
>
0.
x > 0.
x
>
0.
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