To prove that [tex]\( f \)[/tex] is a constant function given that [tex]\( f(x + y) = f(xy) \)[/tex] for all [tex]\( x, y \in \mathbb{R} \)[/tex], follow these steps:
1. Initial Observation:
First, we need to investigate how the given functional equation constrains [tex]\( f \)[/tex]. Let's assume [tex]\( f \)[/tex] is differentiable.
2. Differentiate with respect to [tex]\( x \)[/tex]:
Consider the equation [tex]\( f(x + y) = f(xy) \)[/tex]. Differentiating both sides with respect to [tex]\( x \)[/tex] using the chain rule,
[tex]\[ \frac{d}{dx}[f(x + y)] = \frac{d}{dx}[f(xy)]. \][/tex]
3. Applying the Chain Rule:
For the left side, we have,
[tex]\[ \frac{d}{dx}[f(x + y)] = f'(x + y). \][/tex]
For the right side,
[tex]\[ \frac{d}{dx}[f(xy)] = y \cdot f'(xy), \][/tex]
where the factor [tex]\( y \)[/tex] arises from differentiating the inner function [tex]\( xy \)[/tex] with respect to [tex]\( x \)[/tex].
4. Equate the Derivatives:
Equating these derivatives,
[tex]\[ f'(x + y) = y \cdot f'(xy). \][/tex]
5. Substitute [tex]\( x = 0 \)[/tex]:
To simplify, substitute [tex]\( x = 0 \)[/tex]. We get,
[tex]\[ f'(y) = y \cdot f'(0). \][/tex]
6. Analyze the Result:
Rearrange this equation to solve for [tex]\( f'(y) \)[/tex],
[tex]\[ f'(y) = y \cdot C, \][/tex]
where [tex]\( C \)[/tex] is a constant ([tex]\( C = f'(0) \)[/tex]).
7. Implication for [tex]\( f \)[/tex]:
The equation [tex]\( f'(y) = y \cdot C \)[/tex] suggests that the derivative of [tex]\( f \)[/tex] is proportional to [tex]\( y \)[/tex]. This form is problematic unless [tex]\( C = 0 \)[/tex] because it would imply that [tex]\( f' \)[/tex] depends linearly on [tex]\( y \)[/tex], contradicting the requirement for it to reduce to a linear term guaranteeing [tex]\( f' \)[/tex] itself is constant indicating non-dependence on [tex]\( y \)[/tex].
8. Conclude [tex]\( f' \)[/tex] is Zero:
Thus, the only consistent solution is [tex]\( C = 0 \)[/tex], implying [tex]\( f'(y) = 0 \)[/tex] for all [tex]\( y \)[/tex].
9. Integrate [tex]\( f' \)[/tex] to Find [tex]\( f: \
If \( f'(y) = 0 \)[/tex], integrating with respect to [tex]\( y \)[/tex] implies that [tex]\( f(y) \)[/tex] must be a constant function:
[tex]\[ f(y) = k, \][/tex]
where [tex]\( k \)[/tex] is some constant.
10. Conclusion:
Thus, we have shown that the function [tex]\( f \)[/tex] must be constant. Specifically, for any inputs [tex]\( x \)[/tex] and [tex]\( y \)[/tex], [tex]\( f(x + y) = f(xy) \)[/tex] holds true if and only if [tex]\( f \)[/tex] does not change with [tex]\( x \)[/tex] and [tex]\( y \)[/tex], therefore, [tex]\( f \)[/tex] is a constant function.
Therefore, [tex]\( \boxed{f \text{ is a constant function}} \)[/tex].
