To prove that $f$ is constant, we will leverage the given inequality:

$|f(x) - f(y)| \leq K|x - y|^\alpha \quad \text{for all } x, y \in (a, b),$

where $\alpha > 1$ and $K \geq 0$.

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Step 1: Rewrite the inequality

Divide both sides of the inequality by $|x - y|$ (assuming $x \neq y$):

$\frac{|f(x) - f(y)|}{|x - y|} \leq K|x - y|^{\alpha - 1}.$

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Step 2: Behavior as $|x - y| \to 0$

As $|x - y| \to 0$, the term $|x - y|^{\alpha - 1}$ approaches 0 because $\alpha - 1 > 0$ (since $\alpha > 1$).

Thus, the inequality implies:

$\frac{|f(x) - f(y)|}{|x - y|} \to 0 \quad \text{as } |x - y| \to 0.$

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Step 3: Relate this to the definition of the derivative

The left-hand side $\frac{|f(x) - f(y)|}{|x - y|}$ is the definition of the difference quotient for $f$. In the limit as $x \to y$, this quantity is related to the derivative $f'(x)$. Specifically:

$\lim_{y \to x} \frac{f(x) - f(y)}{x - y} = f'(x).$

From Step 2, we observe that:

$f'(x) = 0 \quad \text{for all } x \in (a, b).$

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Step 4: The Fundamental Theorem of Calculus

If the derivative of a function $f$ is 0 everywhere on an interval $(a, b)$, then $f$ is constant on $(a, b)$.

Thus, $f$ is constant on $(a, b)$.

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Conclusion

Using the inequality and the property $\alpha > 1$, we have shown that the derivative of $f$ must be zero everywhere on $(a, b)$. Therefore, $f$ is constant.

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Supporting Concepts Used:

1. Inequality Analysis: Exploiting $|x - y|^{\alpha - 1} \to 0$ as $|x - y| \to 0$.
2. Limit Definition of Derivative: Connecting the difference quotient to the derivative.
3. Fundamental Theorem of Calculus: A function with zero derivative everywhere on an interval is constant.

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Would you like further explanation or examples of these principles? 😊

Related Questions:

1. What does the derivative of a function represent geometrically?
2. Can this proof extend to cases where $\alpha = 1$?
3. What if $f$ were not continuous—would this proof still hold?
4. How does the Fundamental Theorem of Calculus apply to other intervals or cases?
5. Could we generalize this proof for multivariable functions?

Tip:

Always check the differentiability and continuity conditions when reasoning about derivatives and their implications!