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Since we have solution for Cauchy functional equation, $$f(xy)= f(x)+f(y)$$ which is $C\log(x)$. However, if we have $$2f(xy)=f(x)+f(y)$$ type of functional equation, I found that its solutions are any arbitrary constant $C$, but failed to prove.

How do I prove the above claim?

Mohsen Shahriari
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1 Answers1

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For $y = 1$ we obtain $f(x) = f(1)$ so $f$ is constant.

Jakobian
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  • Why do you choose only y=1 and why not y any real number? If we let x= y then we have f(x^2)= f(x) still the solution is same but how is constant function I'm not getting ? – 1729 Jun 11 '21 at 22:24
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    @LeeLee1729 The $f(xy)$ term looks complicated. So let's substitute something so that it isn't complicated anymore. For example, $y = 1$. – Jakobian Jun 11 '21 at 22:47