1f ( x ) = 1 .
Full answer on request : )
Find all continuous functions f:\mathbb{R}\to [1,\infty) for which there exist an a\in\mathbb{R} and k, a positive integer, such that
f(x)f(2x)\cdots f(nx)\le an^k
for every real number x and positive integer n.
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6 Answers
341Consider g(x) = \ln f(x) which is defined and continuous and also g(x)≥0 for all x. Further g(x) is integrable
Now, we are given that
f \left(\frac{x}{n} \right)f \left(\frac{2x}{n} \right)...f \left(\frac{(n-1)x}{n} \right)f \left(\frac{nx}{n} \right) \le an^k
or by taking logarithms on both sides
\sum_{r=0}^n g \left(\frac{rx}{n} \right)\frac{1}{n} \le \frac{\ln a }{n}+ k \frac{\ln n}{n}
Now letting n→∞, and recognising the riemann sum on the LHS, and evaluating the limit on RHS we have
0 \le \int_0^{x} g(t) \ dt \le 0 \Rightarrow \int_0^{x} g(t) =0 for all x.
Hence it follows that g(x) = 0 for all x.
And hence f(x) =1 for all x
341@nasiko: I hate you for diverting my attention from work for the good part of an hour :D
21good in fact great
i ws trying elementary solutions...din notice 1,2,3...n that way
thank u
ha ha
341i had proved f(0)=1, by setting x =0 and n→∞.
So, i was trying to use this somehow so i put it in that form. That's when i realized that integration was called for.
