Question

Let $f:\mathbb{R} \to \mathbb{R}$ be a differentiable function such that $|f(x)-f(y)| \le 2|x-y|^{3/2} \ \forall x,y \in \mathbb{R}$. If $f(0)=1$, then $\int_0^1 f^2(x)dx=$

• A. $-2$
• B. $\frac{1}{2}$
• C. 0
• D. 1

Hint:

Implications of H\"older-Type Bounds.

• If $|f(x)-f(y)| \le C|x-y|^\alpha$ with $\alpha>1$ and $f$ differentiable, then $f'(x) = 0$.
• The function must be constant, reducing the integral to a basic one.

Answer 1

The Correct Option is D

Given: $|f(x)-f(y)| \le 2|x-y|^{3/2}$ and $f$ is differentiable. Divide both sides by $|x-y|$: \[ \left|\frac{f(x)-f(y)}{x-y}\right| \le 2|x-y|^{1/2} \Rightarrow \lim_{y \to x} \left|\frac{f(x)-f(y)}{x-y}\right| = 0 \] \[ \Rightarrow f'(x) = 0 \text{ for all } x. \text{ So } f \text{ is a constant function.} \] Given \( f(0) = 1 \Rightarrow f(x) = 1 \). \[ \int_0^1 f^2(x)dx = \int_0^1 1^2 dx = 1 \]

Answer 2

Step 1: Understand the given condition
The function \( f:\mathbb{R} \to \mathbb{R} \) is differentiable and satisfies:
\[ |f(x) - f(y)| \le 2 |x - y|^{3/2} \quad \forall x,y \in \mathbb{R} \]
This means the difference in function values grows at most as the power \(3/2\) of the distance between \(x\) and \(y\).

Step 2: Interpret the inequality for differentiability
Since \(f\) is differentiable, by definition:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
Given the inequality:
\[ |f(x+h) - f(x)| \le 2 |h|^{3/2} \]
Dividing both sides by \(|h|\) (assuming \(h \neq 0\)):
\[ \left|\frac{f(x+h) - f(x)}{h}\right| \le 2 |h|^{1/2} \]
Taking the limit as \(h \to 0\):
\[ |f'(x)| = \lim_{h \to 0} \left|\frac{f(x+h) - f(x)}{h}\right| \le \lim_{h \to 0} 2 |h|^{1/2} = 0 \]

Step 3: Deduce the derivative
Since the derivative magnitude tends to zero everywhere:
\[ f'(x) = 0 \quad \forall x \in \mathbb{R} \]

Step 4: Find the form of \(f\)
If \(f'(x) = 0\) for all \(x\), then \(f\) is a constant function:
\[ f(x) = c \]
Given \( f(0) = 1 \), we get:
\[ c = 1 \]
So:
\[ f(x) = 1 \quad \forall x \]

Step 5: Calculate the integral
\[ \int_0^1 f^2(x) \, dx = \int_0^1 1^2 \, dx = \int_0^1 1 \, dx = 1 \]

Final Answer:
\[ \boxed{1} \]