# Thread: Find all functions satisfying the functional equation

1. ## Find all functions satisfying the functional equation

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

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

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

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

4. May be this helps

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

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

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

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

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

One solution

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

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

5. Hmm.. That's the full solution?

6. If

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

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

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

We get a system

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

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

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

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