# Find all functions satisfying the functional equation

• Sep 10th 2010, 11:20 PM
PaulinaAnna
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:

$\displaystyle f(\sqrt{\frac{x^2+xy+y^2}{3}}) = \frac{f(x) + f(y)}{2}$
• Nov 19th 2010, 09:30 AM
helgamauer
It is quite interesting question, isn't it? Did someone find the answer?
• Dec 4th 2010, 01:53 AM
Glyper
Can't manage to solve this interesting problem either.. Any ideas?
• Dec 5th 2010, 10:36 AM
zzzoak
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

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

$\displaystyle \displaystyle { f(x) =const. }$
• Dec 5th 2010, 10:45 AM
Glyper
Hmm.. That's the full solution?
• Dec 8th 2010, 02:15 PM
zzzoak
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
$\displaystyle g(x)=g(y)=0.$