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Consider the functional equation $$2f\left(\frac{x+2y}{2}\right)+2f\left(\frac{x-2y}{2}\right)=f(x)+4f(y)\text.\tag1\label1$$ I noticed that if $f(ax)=a^2f(x)$, then \eqref{1} reduces to the quadratic functional equation $$f(x+y)+f(x-y)=2f(x)+2f(y)\text.\tag2\label2$$ My question is "Can we reduce \eqref{1} into \eqref{2} without the assumed condition on $f$?"

Thanks!

Mohsen Shahriari
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mark haokip
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1 Answers1

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Can we reduce $(1)$ into $(2)$ without the assumed condition on $f$?

  • $\;x=y=0\,$: $\quad 4f(0)=5f(0) \implies f(0)=0$

  • $\;y=0\,$: $\quad\displaystyle 4 f\left(\frac{x}{2}\right) = f(x) +4 f(0)=f(x)\implies f(2x)=4 f(x)$

  • $\;x \mapsto 2x\,$: $\quad 2f(x+y)+2f(x-y)=f(2x)+4f(y) = 4f(x)+4f(y)\\ \implies f(x+y)+f(x-y) = 2f(x)+2f(y)$

dxiv
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