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Fie \( f:\mathbb{R}\rightarrow\mathbb{R} \) cu proprietatile:
(*) \( x(f(x+1)-f(x))=f(x) \), \( \forall x\in \mathbb{R} \)
(**) \( |f(x)-f(y)|\leq|x-y| \), \( \forall x,y\in\mathbb{R} \).
Aflati \( f \).

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\( \frac{f(x+1)}{x+1} = \frac{f(x)}{x} \) , \( \forall x \text{ diferit de } 0 \text{ si } -1 \).
Daca punem \( f(1) = p \), atunci se demonstreaza prin inductie ca \( f(n)=n \cdot p , \forall n \in \mathbb{Z} \).
Tot din prima relatie rezulta ca \( \frac{f(x+n)}{x+n} = \frac{f(x)}{x}, \forall x \in \mathbb{R} - \mathbb{Z} \) si \( n \in \mathbb{Z} \).
Atunci \( | f(x+n) - f(y+n) | = |\frac{x+n}{x}f(x) - \frac{y+n}{y}f(y)| \leq |x-y| \), oricare ar fi \( x,y \in \mathbb{R}- \mathbb{Z} \).
De aici \( \left| n \left( \frac{f(x)}{x} - \frac{f(y)}{y} \right) \right| - |f(x) - f(y)| \leq \left| f(x)- f(y) + n \left( \frac{f(x)}{x} - \frac{f(y)}{y} \right) \right| \leq |x-y| \).
Cum \( n \in \mathbb{N} \) l-am ales arbitrar, rezulta ca \( \frac{f(x)}{x}- \frac{f(y)}{y} = 0 \).
Atunci rezulta ca \( f= k \cdot x, \forall x \in \mathbb{R}- \mathbb{Z} \).
Daca \( x \in \mathbb{Z}^*, y \in \mathbb{N}^* \), avem din ipoteza ca\( \left | f \left(x+ \frac{1}{n} \right) \right| \leq \frac{1}{n} \), de unde\( | kx + \frac{k}{n} - px| \leq \frac{1}{n} \). Rezulta ca \( k=p \), deci \( f(x) = kx \) oricare ar fi \( x \in \mathbb{R} \).