Question

Let \(f(x)\) be a real valued function defined for all \(x \in \mathbb{R}\), such that \(|f(x)-f(y)| \le (x-y)^2\), for all \(x,y \in \mathbb{R}\), then

• A. \(f(x)\) is nowhere differentiable
• B. \(f(x)\) is a constant function
• C. \(f(x)\) is strictly increasing function in the interval [0,1]
• D. \(f(x)\) is strictly increasing function for all \(x \in \mathbb{R}\)

Hint:

A condition of the form \(|f(x)-f(y)| \le K|x-y|^\alpha\) is a Holder condition. When \(\alpha>1\), as in this problem (\(\alpha=2\)), it implies that the function's derivative is zero everywhere, meaning the function is constant. This is a useful result to remember.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The given inequality is a form of a Lipschitz condition. It places a strong constraint on how fast the function can change. We can use this condition to find the derivative of the function.

Step 2: Key Formula or Approach:
We use the definition of the derivative: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] The given condition is \( |f(a) - f(b)| \le (a-b)^2 \). Let's apply this to the definition of the derivative.

Step 3: Detailed Explanation:
Let \( y = x+h \). Then \( y-x = h \). The given condition becomes \( |f(x+h) - f(x)| \le h^2 \). Now let's look at the expression for the derivative's magnitude. For \( h \neq 0 \): \[ \left| \frac{f(x+h) - f(x)}{h} \right| = \frac{|f(x+h) - f(x)|}{|h|} \] Using the given inequality: \[ \frac{|f(x+h) - f(x)|}{|h|} \le \frac{h^2}{|h|} = |h| \] So, we have \( \left| \frac{f(x+h) - f(x)}{h} \right| \le |h| \). Now, take the limit as \( h \to 0 \) to find the derivative: \[ |f'(x)| = \left| \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \right| = \lim_{h \to 0} \left| \frac{f(x+h) - f(x)}{h} \right| \] Using the squeeze theorem with the inequality we derived: \[ 0 \le \lim_{h \to 0} \left| \frac{f(x+h) - f(x)}{h} \right| \le \lim_{h \to 0} |h| = 0 \] This implies that \( |f'(x)| = 0 \), which means \( f'(x) = 0 \) for all \( x \in \mathbb{R} \). If the derivative of a function is zero everywhere, the function must be a constant function. \[ f(x) = C \] This means that options A, C, and D are incorrect.

Step 4: Final Answer:
The function \(f(x)\) must be a constant function.