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

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Question Number 169497 by Huy last updated on 01/May/22
Determine all functions f:R→R such that af(b)+bf(a)=(a+b)f((√(ab))) ∀ab>0
$$\mathrm{Determine}\:\mathrm{all}\:\mathrm{functions}\:{f}:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{a}{f}\left(\mathrm{b}\right)+\mathrm{b}{f}\left(\mathrm{a}\right)=\left(\mathrm{a}+\mathrm{b}\right){f}\left(\sqrt{\mathrm{ab}}\right)\:\forall\mathrm{ab}>\mathrm{0} \\ $$
Answered by aleks041103 last updated on 02/May/22
let a=x, b=x+2dx (√(ab))=(√(x(x+2dx)))=x(1+2(dx/x))^(1/2) =x+dx xf(x+2dx)+(x+2dx)f(x)=2(x+dx)f(x+dx) f(x+dx)=f+f ′dx+((f′′)/2)dx^2 f(x+2dx)=f+2f ′dx+2f ′′dx^2 x(f+2f ′dx+2f ′′dx^2 )+xf+2fdx= =2(x+dx)(f+f ′dx+((f′′)/2)dx^2 ) xf+xf′dx+fdx+xf′′dx^2 = =xf+xf′dx+fdx+(1/2)xf′′dx^2 +f′dx^2 +(1/2)f′′dx^3 ⇒xf′′dx^2 =((1/2)xf′′+f′)dx^2 (1/2)xf′′=f′ let f′=g ⇒(dg/dx)=((2g)/x) ⇒(dg/g)=((2dx)/x) ⇒d(ln(g)−2ln(x))=0 ⇒g=3cx^2 ⇒f=cx^3 +s ⇒a(cb^3 +s)+b(ca^3 +s)=(a+b)(c(ab)^(3/2) +s) ⇒(a+b)(a^2 +b^2 )=(a+b)(ab)^(3/2) ⇒(a^2 +b^2 )^2 =a^3 b^3 which is generally not true! it would be iff c=0 ⇒f(x)=const. is the only solution. Note! This applies only for continuous functions.
$${let}\:{a}={x},\:{b}={x}+\mathrm{2}{dx} \\ $$$$\sqrt{{ab}}=\sqrt{{x}\left({x}+\mathrm{2}{dx}\right)}={x}\left(\mathrm{1}+\mathrm{2}\frac{{dx}}{{x}}\right)^{\mathrm{1}/\mathrm{2}} ={x}+{dx} \\ $$$${xf}\left({x}+\mathrm{2}{dx}\right)+\left({x}+\mathrm{2}{dx}\right){f}\left({x}\right)=\mathrm{2}\left({x}+{dx}\right){f}\left({x}+{dx}\right) \\ $$$${f}\left({x}+{dx}\right)={f}+{f}\:'{dx}+\frac{{f}''}{\mathrm{2}}{dx}^{\mathrm{2}} \\ $$$${f}\left({x}+\mathrm{2}{dx}\right)={f}+\mathrm{2}{f}\:'{dx}+\mathrm{2}{f}\:''{dx}^{\mathrm{2}} \\ $$$${x}\left({f}+\mathrm{2}{f}\:'{dx}+\mathrm{2}{f}\:''{dx}^{\mathrm{2}} \right)+{xf}+\mathrm{2}{fdx}= \\ $$$$=\mathrm{2}\left({x}+{dx}\right)\left({f}+{f}\:'{dx}+\frac{{f}''}{\mathrm{2}}{dx}^{\mathrm{2}} \right) \\ $$$${xf}+{xf}'{dx}+{fdx}+{xf}''{dx}^{\mathrm{2}} = \\ $$$$={xf}+{xf}'{dx}+{fdx}+\frac{\mathrm{1}}{\mathrm{2}}{xf}''{dx}^{\mathrm{2}} +{f}'{dx}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{f}''{dx}^{\mathrm{3}} \\ $$$$\Rightarrow{xf}''{dx}^{\mathrm{2}} =\left(\frac{\mathrm{1}}{\mathrm{2}}{xf}''+{f}'\right){dx}^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{xf}''={f}' \\ $$$${let}\:{f}'={g} \\ $$$$\Rightarrow\frac{{dg}}{{dx}}=\frac{\mathrm{2}{g}}{{x}} \\ $$$$\Rightarrow\frac{{dg}}{{g}}=\frac{\mathrm{2}{dx}}{{x}} \\ $$$$\Rightarrow{d}\left({ln}\left({g}\right)−\mathrm{2}{ln}\left({x}\right)\right)=\mathrm{0} \\ $$$$\Rightarrow{g}=\mathrm{3}{cx}^{\mathrm{2}} \\ $$$$\Rightarrow{f}={cx}^{\mathrm{3}} +{s} \\ $$$$\Rightarrow{a}\left({cb}^{\mathrm{3}} +{s}\right)+{b}\left({ca}^{\mathrm{3}} +{s}\right)=\left({a}+{b}\right)\left({c}\left({ab}\right)^{\mathrm{3}/\mathrm{2}} +{s}\right) \\ $$$$\Rightarrow\left({a}+{b}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)=\left({a}+{b}\right)\left({ab}\right)^{\mathrm{3}/\mathrm{2}} \\ $$$$\Rightarrow\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} ={a}^{\mathrm{3}} {b}^{\mathrm{3}} \\ $$$${which}\:{is}\:{generally}\:{not}\:{true}! \\ $$$${it}\:{would}\:{be}\:{iff}\:{c}=\mathrm{0} \\ $$$$\Rightarrow{f}\left({x}\right)={const}.\:{is}\:{the}\:{only}\:{solution}. \\ $$$$ \\ $$$${Note}! \\ $$$${This}\:{applies}\:{only}\:{for}\:{continuous}\:{functions}. \\ $$

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