let f:R->R such that for all x,y∈ R |f(x)-f(y)|ⁿ where n∈ N and n>1.Prove that f(x) is a constant function.

To show that the function \( f: \mathbb{R} \to \mathbb{R} \) is constant under the given condition, we need to analyze the implications of the inequality \( |f(x) - f(y)| \leq n \) for all \( x, y \in \mathbb{R} \) and for any natural number \( n > 1 \). This condition suggests that the difference in function values is bounded by a fixed positive number, regardless of how far apart \( x \) and \( y \) are.

Understanding the Implications of the Condition

The statement \( |f(x) - f(y)| \leq n \) means that for any two real numbers \( x \) and \( y \), the absolute difference between their function values is at most \( n \). This is a crucial point because it indicates that the function does not vary too much between any two points in its domain.

Exploring the Consequences of Bounded Differences

Let’s consider what happens as we choose different pairs of points in \( \mathbb{R} \). For instance, if we take \( y = x + k \) for some large \( k \), the inequality still holds:

• For \( x \) and \( y = x + k \), we have \( |f(x) - f(x + k)| \leq n \).

This means that no matter how far apart \( x \) and \( x + k \) are, the values of \( f \) at these points can only differ by at most \( n \). Now, if we let \( k \) grow larger and larger, we can see that the function values remain constrained within a fixed range.

Using the Density of Real Numbers

Since the real numbers are dense, we can choose \( k \) to be any positive number. For example, if we let \( k \) approach infinity, we can find points \( x \) and \( y \) such that \( |f(x) - f(y)| \) remains bounded. This leads us to consider the implications of this boundedness:

• If \( |f(x) - f(y)| \leq n \) for all \( x, y \), then as \( k \) increases, the function values cannot diverge significantly.
• In fact, if we take any two points \( x_1 \) and \( x_2 \) in \( \mathbb{R} \), we can find a sequence of intermediate points that keeps the function values within the bounds set by \( n \).

Concluding the Argument

To formalize this, we can argue that if \( f \) were not constant, there would exist at least two points \( a \) and \( b \) such that \( f(a) \neq f(b) \). However, the condition \( |f(a) - f(b)| \leq n \) implies that the function cannot take on values that differ by more than \( n \), which contradicts the assumption of non-constancy as we can always find points that are arbitrarily far apart.

Thus, the only way to satisfy the condition for all \( x, y \in \mathbb{R} \) is for \( f \) to be constant across its entire domain. Therefore, we conclude that \( f(x) \) must indeed be a constant function.