Autor |
Beitrag |
Son Wa
| Veröffentlicht am Dienstag, den 11. Dezember, 2001 - 23:36: |
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Hello guys. Sorry that I start writing in English, but unfortunately my german is not as perfect as it should be and I am going spare you all my spelling mistakes. I study computer science in Frankfurt and Analysis is really giving me the creeps. I actually would be happy if you could give me some basic approaches, but I wouldn`t mind detailed explenations either. Here we go: Sei f: (0;¥)->R eine stetige Funktion mit f(xy)=f(x)+f(y). Dann gilt entweder i) f(x)=0 für alle x aus (0;¥) , oder ii) f ist bijektiv unf f(x)=loga(x) mit f(a)=1, wobei loga die Umkehrfunktion von expa ist (Logarithmus zur Basis a). You really do not need to answer in English, I have my dictionary just in front of me ... Thanks a lot! |
Orion (Orion)
| Veröffentlicht am Mittwoch, den 12. Dezember, 2001 - 09:27: |
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Hallo Son Wa : A possible way to do this is the following. Suppose f is not the zero function.Then there exists a > 0 such that f(a) <> 0. Replacing f(x) by f(x)/f(a) if necessary we may assume w.l.o.g. that f(a) = 1. The function exp_a : x -> a^x ; x in |R is continous and maps |R onto (0,oo) one-to-one. Hence the function g:|R ->|R with g(x) := f(a^x) is a continous solution of the functional equation (*) g(x+y) = g(x) + g(y). It is well known that g(x) = x*g(1) is the unique continous solution of (*). To prove this we deduce the following facts from (*) proceeding step by step : g(0) = 0 , g(-x) = - g(x) , g(m) = m*g(1) for m in Z, g(1/n) = (1/n)*g(1) for n in |N , g(m/n) = (m/n)*g(1) for m in Z, n in |N <==> g(r) = r*g(1) for all rational r. Since every irrational x is the limit of a suitable sequence (r_n) of rationals it follows from the contiouity of g that g(x) = x*g(1) for all x in |R. Hence f(a^x) = x*f(a) = x for all x in |R Replacing x by log_a (x) we have f(x) = log_a (x) as the unique continous solution of the given functional equation. Best wishes Orion |
Son Wa
| Veröffentlicht am Donnerstag, den 13. Dezember, 2001 - 10:21: |
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Hello Orion! This is really awesome ... I am pretty impressed by your english skills! I owe you for this and hope you enjoy your weekend. Kind regards. Son Wa |
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