Let f:R→R be a solution of the additive Cauchy functional equation satisfying the condition
f(x)=x2f(1/x)∀x∈R∖{0}.
Then show that f(x)=cx, where c is an arbitrary constant.
Answer
Let F(x)=f(x)−xf(1)
For some x≠0, F(1x)=f(1x)−1xf(1).
Hence for x≠0 x2F(1x)=x2f(1x)−xf(1)=f(x)−xf(1)=F(x)
and of course F(1)=0 and F is additive.
Let us prove that ∀x∈R,F(x)=−F(−x)
Indeed, 0=F(1)=F(x+1−x)=F(x)+F(1)+F(−x)=F(x)+F(−x)
Also, for some x≠−1,
F(x)=F(x+1)=(x+1)2F(1x+1)=(x+1)2F(1−xx+1)=−(x+1)2F(xx+1)=−(x+1)2(xx+1)2F(x+1x)=−x2F(1+1x)=−x2F(1x)=−F(x)
This also holds for x=−1 since F(−1)=−F(1).
Hence 2F=0.
Hence F=0 and we're done.
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