why is g cintinous?
Asked by Rajat Somani
| 21st Jul, 2011,
12:00: AM
Expert Answer:
g(x+y) = g(x) g(y)
at x=0, g(0+y) = g(y) = g(0) g(y)
so g(0) = 1
If g is continuous at x=0, then g(0+y) = g(y) is continuous for all y.
Since any y can be decomposed into y=a+b, where a,b are real numbers, then g(y) = g(a) g(b) must be continuous, so g must be continuous everywhere.
If g(a) = 0 for some a in R, then write a = x+y where x,y are real numbers, then
g(a) = g(x+y) = g(x) g(y) = 0
then either g(x) = 0 or g(y) = 0 (or both).
that implies that g(x) = 0 for all x in R.
at x=0, g(0+y) = g(y) = g(0) g(y)
so g(0) = 1
If g is continuous at x=0, then g(0+y) = g(y) is continuous for all y.
Since any y can be decomposed into y=a+b, where a,b are real numbers, then g(y) = g(a) g(b) must be continuous, so g must be continuous everywhere.
If g(a) = 0 for some a in R, then write a = x+y where x,y are real numbers, then
g(a) = g(x+y) = g(x) g(y) = 0
then either g(x) = 0 or g(y) = 0 (or both).
that implies that g(x) = 0 for all x in R.
Answered by
| 21st Jul, 2011,
08:06: AM
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