f(x+y+z) = f(x) f(y) f(z)
Suppose there exists an x such that f(x) is non-zero (It does exist according to the question. For instance f(2) = 4). Then substitute z=-y.
f(x+y-y) = f(x) = f(x) f(y) f(-y)
Since f(x) is non-zero, f(x) may be eliminated. Then:
f(y) f(-y) = 1
For every y. This also means that f(y) is nonzero for every y. Since f(y) is differentiable, we can deduce that f(y) is either positive or negative for every y. And, since we know that f(2)=4, hence f(x) is positive.
Since f(x) is positive it may be written as:
f(x) = exp[ g(x) ]
Now recall that f(x+y+z) = f(x) f(y) f(z):
exp[ g(x+y+z) ] = exp[ g(x) ] exp[ g(y) ] exp[ g(z) ] = exp[ g(x) + g(y) + g(z) ]
g(x+y+z) = g(x) + g(y) + g(z)
g(0) = g(0) *3. Hence g(0) = 0
g(x+y) = g(x) + g(y). Hence g(x) is linear.
The solution for g(x) is: g(x) = kx.
Substitute:
- f(x) = exp[ kx ]
- f'(x) = exp[ kx ] * k
Substitute:
This gives:
- exp[ 2k ] = 4
- exp[ 0 ] * k = 3
Which does not comply
Hence - there is no such a function.