Determine-all-functions-f-R-R-such-that-af-b-bf-a-a-b-f-ab-ab-gt-0-

Question Number 169497 by Huy last updated on 01/May/22

Determine all functions f : R \to R such that

a f (b) + b f (a) = (a + b) f (\sqrt{ab}) \forall ab > 0

Answered by aleks041103 last updated on 02/May/22

l e t a = x, b = x + 2 d x

\sqrt{a b} = \sqrt{x (x + 2 d x)} = x {(1 + 2 \frac{d x}{x})}^{1 / 2} = x + d x

x f (x + 2 d x) + (x + 2 d x) f (x) = 2 (x + d x) f (x + d x)

f (x + d x) = f + f^{'} d x + \frac{f^{″}}{2} {d x}^{2}

f (x + 2 d x) = f + 2 f^{'} d x + 2 f^{″} {d x}^{2}

x (f + 2 f^{'} d x + 2 f^{″} {d x}^{2}) + x f + 2 f d x =

= 2 (x + d x) (f + f^{'} d x + \frac{f^{″}}{2} {d x}^{2})

x f + {x f}^{'} d x + f d x + {x f}^{″} {d x}^{2} =

= x f + {x f}^{'} d x + f d x + \frac{1}{2} {x f}^{″} {d x}^{2} + f^{'} {d x}^{2} + \frac{1}{2} f^{″} {d x}^{3}

\Rightarrow {x f}^{″} {d x}^{2} = (\frac{1}{2} {x f}^{″} + f^{'}) {d x}^{2}

\frac{1}{2} {x f}^{″} = f^{'}

l e t f^{'} = g

\Rightarrow \frac{d g}{d x} = \frac{2 g}{x}

\Rightarrow \frac{d g}{g} = \frac{2 d x}{x}

\Rightarrow d (l n (g) - 2 l n (x)) = 0

\Rightarrow g = 3 {c x}^{2}

\Rightarrow f = {c x}^{3} + s

\Rightarrow a ({c b}^{3} + s) + b ({c a}^{3} + s) = (a + b) (c {(a b)}^{3 / 2} + s)

\Rightarrow (a + b) (a^{2} + b^{2}) = (a + b) {(a b)}^{3 / 2}

\Rightarrow {(a^{2} + b^{2})}^{2} = a^{3} b^{3}

w h i c h i s g e n e r a l l y n o t t r u e!

i t w o u l d b e i f f c = 0

\Rightarrow f (x) = c o n s t . i s t h e o n l y s o l u t i o n .

N o t e!

T h i s a p p l i e s o n l y f o r c o n t i n u o u s f u n c t i o n s .