1. ## Functional equation

Hi, I need help on these two problems dealing with functional equation

Let f be a function defined everywhere on the real axis. Suppose also that f satisfies the functional equation

f(x+y) = f(x)f(y) for all x and y.

A) Using only the functional equation, prove that f(0) is either 0 or 1. Also, prove that if f(0) then f(x) is not equal to 0 for all x.

Thanks

2. Originally Posted by BrianW
Hi, I need help on these two problems dealing with functional equation

Let f be a function defined everywhere on the real axis. Suppose also that f satisfies the functional equation

f(x+y) = f(x)f(y) for all x and y.

A) Using only the functional equation, prove that f(0) is either 0 or 1.
f(0) = f(0+0) = f(0)*f(0) = [f(0)]^2

so we have f(0) = [f(0)]^2
=> [f(0)]^2 - f(0) = 0
=> f(0)[f(0) - 1] = 0
=> f(0) = 0 or f(0) - 1 = 0
=> f(0) = 0 or f(0) = 1

Also, prove that if f(0) then f(x) is not equal to 0 for all x.
i believe a part of this question is missing. "if f(0) is what? then f(x) is not equal to 0 for all x."

3. oh sorry, I forgot that part.

It should be: "Also, prove that if f(0) is not equal to 0, then f(x) is not equal to 0 for all x"

4. Originally Posted by BrianW
oh sorry, I forgot that part.

It should be: "Also, prove that if f(0) is not equal to 0, then f(x) is not equal to 0 for all x"
From above, we see if f(0) is not zero, then f(0) = 1

now recall f(x + y) = f(x)f(y)
=> f(x) = f(x + y)/f(y)
if f(x + y) = f(0) we have:
f(x) = f(0)/f(y)
=> f(x) = 1/f(y) if f(0) is not 0
but 1/f(y) is never zero. thus we have f(x) is never zero for any x

5. Originally Posted by BrianW
oh sorry, I forgot that part.

It should be: "Also, prove that if f(0) is not equal to 0, then f(x) is not equal to 0 for all x"
I do not like what Jhevon did because he did not show that f(y)!=0.

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We are given that f(0) is non-zero.
Note, for any real x we have that,
f(0) = f[(x)+(-x)] =f(x)*f(-x)
Now since the left hand side is non-zero implies f(x) is non-zero.

6. Originally Posted by ThePerfectHacker
I do not like what Jhevon did because he did not show that f(y)!=0.
i did not think that was necessary (f(x) won't be zero no matter what f(y) is). but ok, i guess your way leaves no uncertainty