Question

Let \( R \) be the set of all real numbers. Let \( f: R \to R \) be a function such that \( |f(x) - f(y)|^2 \le |x - y|^3, \forall x, y \in R \). Then \( f'(x) = \)

• A. \( f(x) \)
• B. \( 1 \)
• C. \( 0 \)
• D. \( x^2 \)
• E. \( x \)

Hint:

Whenever you see $|f(x) - f(y)| \le |x - y|^k$ with $k > 1$, the function is always a constant. Its "slope" is so restricted that it cannot change at all.

Answer 1

The Correct Option is C

Concept: We use the definition of the derivative: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] The given inequality allows us to bound the absolute value of this difference quotient.

Step 1: Rewrite the given inequality.
\[ |f(x) - f(y)|^2 \le |x - y|^3 \] Take the square root of both sides: \[ |f(x) - f(y)| \le |x - y|^{3/2} \]

Step 2: Form the difference quotient.
Let \( y = x + h \), so \( x - y = -h \). \[ |f(x+h) - f(x)| \le |h|^{3/2} \] Divide both sides by \( |h| \): \[ \left| \frac{f(x+h) - f(x)}{h} \right| \le \frac{|h|^{3/2}}{|h|} = |h|^{1/2} \]

Step 3: Apply the limit.
As \( h \to 0 \), the term \( |h|^{1/2} \to 0 \). By the Squeeze Theorem: \[ \lim_{h \to 0} \left| \frac{f(x+h) - f(x)}{h} \right| = 0 \] Since the derivative is 0 everywhere, \( f(x) \) is a constant function, and \( f'(x) = 0 \).