Find all functions satisfying the functional equation

**PaulinaAnna** · September 10th 2010, 11:20 PM

Find all functions $f$ defined on the set of positive real numbers and having real values, that for any real $x, y$ there is satisfied the following equasion:

$f(\sqrt{\frac{x^2+xy+y^2}{3}}) = \frac{f(x) + f(y)}{2}$

**helgamauer** · November 19th 2010, 09:30 AM

It is quite interesting question, isn't it? Did someone find the answer?

**Glyper** · December 4th 2010, 01:53 AM

Can't manage to solve this interesting problem either.. Any ideas?

**zzzoak** · December 5th 2010, 10:36 AM

May be this helps

$ z=\sqrt{\frac{x^2+xy+y^2}{3}} $

$ f(z)=(f(x)+f(y))/2 $

$ f'_z \; z'_x \; = \; f'_z \; \frac{2x+y}{2 \sqrt{3(x^2+xy+y^2})}} \; = \; f'_x/2 $

$ f'_z \; z'_y \; = \; f'_z \; \frac{2y+x}{2 \sqrt{3(x^2+xy+y^2})}} \; = \; f'_y/2 $

$ \displaystyle { \frac{2x+y}{2y+x}=\frac{f'_x}{f'_y}. } $

One solution

$ \displaystyle { f'_x =f'_y=0 } $

$ \displaystyle { f(x) =const. } $

**Glyper** · December 5th 2010, 10:45 AM

Hmm.. That's the full solution?

**zzzoak** · December 8th 2010, 02:15 PM

If

$ { \displaystyle { f'_x =g(x) } } $

$ { \displaystyle { f'_y =g(y) } } $

$ (2x+y)g(y)=(2y+x)g(x). $

We get a system

$ 2xg(y)=xg(x) $

$ yg(y)=2yg(x) $

which has a solution
$ g(x)=g(y)=0. $

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