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Find all function $f,g$ that satisfy: $$g(x)-g(y)=1/6 (x-y)(f(x)+f((x+y)/2)+f(y))$$ For $y=0$ we have an equation in $f$: $$4(x-y)(f(x/2)-f(x/2+y/2))=xf(y)-yf(x)$$ How can i do it?

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Could you kindly typeset your question to make it more readable? meta.math.stackexchange.com/questions/5020/… –  user17762 Nov 22 '12 at 21:38
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Why does $y$ appear in the equation with $y=0$ and why does $g$ not appear? –  Hagen von Eitzen Nov 22 '12 at 22:03
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If we additionally assume that $f$ is continuous, then by letting $y\to x$ we find that necessarily $g'(x)=\frac12 f(x)$. This immediately gives us a few solutions:

  • $g(x)=1, f(x)=0$
  • $g(x)=x, f(x)=2$
  • $g(x)=x^2, f(x)=4x$

and since the solutions form a vector space, any linear combination thereof. However, already with $g(x)=x^3, f(x)=6x$ the pattern breaks and the functional equation does not hold throughout!

Note that $g(x+h)-g(x-h) = \frac13 h(f(x+h)+f(x)+f(x-h)$ but also $g(x+h)-g(x-h) = g(x+h)-g(x)+g(x)-g(x-h)= \frac16 h(f(x+h)+f(x+h/2)+f(x))+\frac16h(f(x)+f(x-h/2)+f(x-h))$, hence $ f(x+h)+f(x-h)=f(x+h/2)+f(x-h/2)$ and by continuity at $x$, we conclude that $ f(x+h)+f(x-h)=2f(x)$ for all $h$. Then the right hand side of the f.e. is simply $\frac12(x-y)f(\frac{x+y}2)$. From this we have $g(x+h)-g(x-h)=hf(x)$. By subtracting a multiple of $x$ from $g$ and a corresponding multiple of $2$ from $f$, we may assume wlog. that $f(0)=0$, hence $g(x)=g(-x)$.

(to be continued)

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