Finding continuous functions

Consider 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

@nasiko: I hate you for diverting my attention from work for the good part of an hour :D

good in fact great

i ws trying elementary solutions...din notice 1,2,3...n that way

thank u

ha ha

i 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.