prove that, apart from y=a^x and y=0, there exist no other continuous functions such that for any real x and y the equation f(x+y) = f(x)f(y) holds.
To prove that apart from the two functions y = a^x and y = 0, no other continuous functions exist that satisfy the equation f(x + y) = f(x)f(y) for any real x and y, we can use a combination of proof by contradiction and a property of continuous functions.
First, let's assume that there exists another continuous function, f(x), that satisfies the equation f(x + y) = f(x)f(y) for any real x and y.
Proof by contradiction:
1. Suppose there exists a number b such that f(b) is not equal to 0 (f(b) ≠ 0).
2. Consider the equation f(x + b) = f(x)f(b). Since both x and b are real numbers, this equation holds for any real x.
3. Let's substitute x = 0 into the equation: f(0 + b) = f(0)f(b).
This simplifies to f(b) = f(0)f(b).
Since we assumed f(b) ≠ 0, we can divide both sides of the equation by f(b) to get:
1 = f(0).
4. Now consider the equation f(x + 0) = f(x)f(0). Again, since x is a real number, this equation holds for any real x.
Substituting x = 0, we get f(0) = f(0)f(0).
5. Since we've already established f(0) = 1 (from step 3), this simplifies to 1 = 1 * f(0), which is always true.
6. Therefore, f(x) = 1 for any real x.
We have now proved that if there exists a continuous function, f(x), satisfying the equation f(x + y) = f(x)f(y) for any real x and y, then f(x) must always equal 1.
However, the functions y = a^x and y = 0 also satisfy the equation f(x + y) = f(x)f(y) for any real x and y. These are not equivalent to the constant function f(x) = 1. Hence, apart from y = a^x and y = 0, no other continuous functions exist that satisfy the given equation.