Question

$Sa~se~determine~functiile~monotone~f:\mathbb{R}\rightarrow \mathbb{R}~cu~proprietatea~ca~f(1)=1~si\\ f(\frac{1918a+100b}{2018})=f(\frac{1918}{2018}\cdot a)+f(\frac{100}{2018}\cdot b), \forall a,b\in \mathbb{Q}.$

o idee de a mea a fost inductia matematica...insa nu stiu in principiu cum sa abordez in aceasta situatie..ma puteti ajuta?

Answer 1

dacă ar fi sigur cu inducție matematică eu la asta m-am gândit

Answer 2

$\displaystyle Cand~a~si~b~parcurg~\mathbb{Q},~\frac{1918}{2018}a~si~\frac{100}{2018}b~parcurg~\mathbb{Q}. \\ \\ Deci~notand~u= \frac{1918}{2018}a~si~v= \frac{100}{2018}b,~ecuatia~functionala~devine \\ \\ f(u)+f(v)=f(u+v)~\forall~u,v \in \mathbb{Q}. \\ \\ Desigur,~nu~avem~nicio~informatie~despre~numerele~irationale, \\ \\ dar~vom~vedea~ca~monotonia~de~va~fi~de~ajutor.$

$\displaystyle Relatia~f(u)+f(v)=f(u+v)~\forall~u,v \in \mathbb{Q}~se~extinde~la \\ \\ f(u)+f(v)+f(w)=f(u+v+w)~\forall~u,v,w \in \mathbb{Q}~observand~ca \\ \\ f(u+v+w)=f(u+v)+f(w)=f(u)+f(v)+f(w)~si,~in~fine, \\ \\ inductiv~deducem~ca \\ \\ f(x_1)+f(x_2)+...+f(x_n)=f(x_1+x_2+...+x_n) ~\forall~x_1,x_2,...,x_n \in \mathbb{Q}.$

$\displaystyle Punem~x_1=x_2=...=x_n=x~si~obtinem~f(nx)=nf(x)~\forall~x \in \mathbb{Q}. \\ \\ In~ particular~f(n)=nf(1)=n~\forall~n \in \mathbb{Z}. \\ \\ Punem~x= \frac{m}{n},~cu~m,n \in \mathbb{Z},~n \neq 0~si~rezulta~f(m)=n f \left( \frac{m}{n} \right)~\forall \\ \\ m,n \in \mathbb{Z},~n \neq 0.\\ \\ Deci~f \left( \frac{m}{n} \right)= \frac{f(m)}{n}= \frac{m}{n}~\forall~m,n \in \mathbb{Z},~n \neq 0. \\ \\ Cu~alte~cuvinte~f(x)=x~\forall~x \in \mathbb{Q}.$

$\displaystyle Vom~demonstra~ca~acest~rezultat~este~valabil~pe~\mathbb{R}. \\ \\ Din~ce~am~demonstrat~pana~acum~rezulta~f-crescatoare. \\ \\ Presupunem~ca~exista~r \in \mathbb{R}- \mathbb{Q}~a.i.~f(r) \neq r. \\ \\ \bullet ~Daca~f(r)<r,~atunci~exista~s>0~a.i.~f(r)=r-s. \\ \\ In~intervalul~(r-s,r)~exista~o~infinitate~de~numere~rationale.~Fie\\ \\ q~un~numar~din~(r-s,r) \cap \mathbb{Q}. \\ \\ Avem~q<r,~si~totusi~f(q)=q>r-s=f(r),~contradictie~cu~faptul \\ \\ ca~ f~este~crescatoare.$

$\dsplaystyle \bullet ~Daca~f(r)>r,~atunci~exista~s>0~a.i.~f(r)=r+s. \\ \\ In~intervalul~(r,r+s)~exista~o~infinitate~de~numere~rationale.~Fie\\ \\ q~un~numar~din~(r,r+s) \cap \mathbb{Q}. \\ \\ Avem~q>r,~si~totusi~f(q)=q<r+s=f(r),~contradictie~cu~faptul \\ \\ ca~ f~este~crescatoare. \\ \\ In~concluzie~f(x)=x~este~valabil~si~pentru~numerele~irationale, \\ \\ si,~deci~pentru~orice~x \in \mathbb{R}.$