pure maths ...help

(a) f(x + y) = g(x + y)/[g(1)]x + y

= g(x)g(y)/[g(1)]x + y

= {g(x)/[g(1)]x} {g(y)/[g(1)]y}

= f(x) f(y)

(b) f(x + 1) = f(x) f(1) (By (a))

= f(x) {g(1)/[g(1)]1}

= f(x)

So f is a periodinc function of period 1.

Now, taking 0 <= x < 1:

|f(x)| = | g(x)/[g(1)]x |

= | g(x) |/| [g(1)]x |

With 0 < g(1) < 1, we have 0 < [g(1)]x < 1 for all 0 <= x < 1 and hence 1/| [g(1)]x | is a definite value.

So to speak, we can find a constant K > 0 such that:

1/| [g(1)]x | < K

Therefore:

|f(x)| = | g(x) |/| [g(1)]x | < KM

So, f(x) is bounded for all 0 <= x < 1.

Since f is periodic with period = 1, f is bounded for all real x.

2010-03-13 22:59:19 補充：
Special thanks to STEVIE-G™.

Also I share my answer...
http://i707.photobucket.com/albums/ww74/stevieg90/11-13.jpg