Given a comtnuous function from F:[0,1]→[0,∞) that satisfy the follow:
f(12x+12)=f(x)+1f\left( \frac{1}{2}x+\frac{1}{2} \right)=f\left( x \right)+1f(21x+21)=f(x)+1
f(1−x)=1f(x)f\left( 1-x \right)=\dfrac{1}{f\left( x \right)}f(1−x)=f(x)1
for all x∈(0,1)x\in (0,1)x∈(0,1)
Determine value of ∫01f(x)dx\int\limits_{0}^{1}{f\left( x \right)dx}0∫1f(x)dx
Note by Idham Muqoddas 7 years, 9 months ago
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You need to clarify your definitions a bit. If fff is continuous on [0,1][0,1][0,1] (including 000 and 111) and f(12x+12)=f(x)+10<x<1 f\big(\tfrac12x+\tfrac12\big) = f(x) + 1 \qquad 0 < x < 1 f(21x+21)=f(x)+10<x<1 then this identity is also true when x=0,1x=0,1x=0,1, since both sides of this identity are continuous functions on [0,1][0,1][0,1]. In particular, putting x=1x=1x=1, f(1)=f(1)+1f(1) = f(1) + 1f(1)=f(1)+1, which is not possible.
Do you mean that fff is only to be defined and continuous on (0,1)(0,1)(0,1)? Then fff must be unbounded, tending to ∞\infty∞ as x→∞x\to\inftyx→∞, so we are looking at either an improper Riemann or a Lebesgue integral.
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Comments
You need to clarify your definitions a bit. If f is continuous on [0,1] (including 0 and 1) and f(21x+21)=f(x)+10<x<1 then this identity is also true when x=0,1, since both sides of this identity are continuous functions on [0,1]. In particular, putting x=1, f(1)=f(1)+1, which is not possible.
Do you mean that f is only to be defined and continuous on (0,1)? Then f must be unbounded, tending to ∞ as x→∞, so we are looking at either an improper Riemann or a Lebesgue integral.