Determinati toate functiile \( f : (0,\infty) \rightarrow (0,\infty) \) cu proprietatea : \( f(x^2f(y))=xyf(f(x)) \forall x,y\in{(0,\infty)} \)
M. Opincariu G.M.B. 2008
Ecuatie functionala simpla
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Pentru \( x=1 \) obtinem \( f(f(y))=yf(f(1))\ \ (1) \) si cum \( f(x)\ >\ 0,\ \forall x\in(0,\ \infty) \) rezulta ca functia \( f \) este injectiva.
Din enunt si din relatia \( (1) \) putem scrie \( f(x^2f(y))=x^2yf(f(1)),\ \forall x,y\ >\ 0\ (2) \).
In relatia \( (2) \) luam \( x=\frac{1}{\sqrt{y}} \) si obtinem \( f\left\(\frac{f(y)}{y}\right\)=f(f(1))\ \Longrightarrow f(y)=yf(1) \).
Asadar \( f(x)=\alpha x,\ \alpha\ >\ 0 \), care verifica enuntul.
Din enunt si din relatia \( (1) \) putem scrie \( f(x^2f(y))=x^2yf(f(1)),\ \forall x,y\ >\ 0\ (2) \).
In relatia \( (2) \) luam \( x=\frac{1}{\sqrt{y}} \) si obtinem \( f\left\(\frac{f(y)}{y}\right\)=f(f(1))\ \Longrightarrow f(y)=yf(1) \).
Asadar \( f(x)=\alpha x,\ \alpha\ >\ 0 \), care verifica enuntul.