Pure Maths---Function

11(a)
when n=1
g(x)=g(x)
So when n=1, the statement is true
Assume that P(k) is true
g(kx)=[g(x)]^k
when n=k+1
g[(k+1)x]=g(kx)g(x)=[g(x)]^kg(x)=[g(x)]^(k+1)
So when n=k+1, the statement is true
By MI for all positive integer n g(nx)=g(x)^n
if n is negative, then m=-n is positive
Using g(mx)=g(x)^m
g(-nx)=g(x)^(-n)
g(ny)=g(-y)^(-n)
g(ny)=g(y)^n
So when n is negative, the statement is also true
(ii)
For any real number y, we can find a value x∈(0,1) such that y=nx
So g(y)=g(nx)=g(x)^n
If g is also bound on R, then that means g(x)^n bound on R
Since n can be arbitrary large (both positive or negative)
We deduce that g(x)=1
b(i)
g(x) is a non-zero function , so f is a non-zero function
f(x+y)
=g(x+y)/g(1)^(x+y)
=g(x+y)/g(1)^xg(1)^y
=g(x)g(y)/g(1)^xg(1)^y
=[g(x)/g(1)^x][g(y)/g(1)^y]
=f(x)f(y)
(ii)
f(x+1)
=g(x+1)/g(1)^(x+1)
=g(x)g(1)/g(1)^xg(1)
=g(x)/g(1)^x
=f(x)
Show that f is a periodic function of period 1
Since all periodic function is a bounded function, f is bounded on R
(iii)
from part (a) f(x)=1
Then g(x)/g(1)^x=1
g(x)=g(1)^x

2008-01-09 21:26:32 補充：
for (ii) the correct argument isFirst, consider, for x∈(0,1) f(x) is bounded, (since g(x) and g(1) is bounded)Using f(x) is a periodic function , we deduce that f(x) is bounded for x∈R