Key Concepts and Formulas

• Definition of the Derivative: The derivative of a function $f(x)$ at a point $x$ is defined as: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ or equivalently, $f'(x) = \lim_{y \to x} \frac{f(y) - f(x)}{y - x}$
• Properties of Absolute Value: For any real number $A$, $|A| \ge 0$. If $|A| \le 0$, then $A = 0$.
• Fundamental Theorem of Calculus: If $f'(x) = 0$ for all $x$ in an interval, then $f(x)$ is a constant function on that interval.
• Definite Integration: $\int_a^b c \, dx = c(b-a)$ for a constant $c$.

Step-by-Step Solution

Step 1: Analyze the Given Inequality and Relate it to the Derivative We are given that $f$ is a differentiable function such that for all $x, y \in \mathbb{R}$, $|f(x) - f(y)| \le 2|x - y|^{3/2}$. Since $f$ is differentiable, its derivative $f'(x)$ exists for all $x$. To find information about $f'(x)$, we will manipulate the given inequality to resemble the definition of the derivative.

Step 2: Form the Difference Quotient For $x \neq y$, we can divide both sides of the inequality by $|x - y|$: $\frac{|f(x) - f(y)|}{|x - y|} \le \frac{2|x - y|^{3/2}}{|x - y|}$ This simplifies to: $\left| \frac{f(x) - f(y)}{x - y} \right| \le 2|x - y|^{3/2 - 1}$ $\left| \frac{f(x) - f(y)}{x - y} \right| \le 2|x - y|^{1/2}$ Reasoning: We divide by $|x-y|$ to isolate the difference quotient, which is the core of the derivative's definition.

Step 3: Apply the Limit as $y \to x$ Now, we take the limit as $y \to x$ on both sides of the inequality. $\mathop {\lim }\limits_{y \to x} \left| {\frac{{f\left( x \right) - f\left( y \right)}}{{x - y}}} \right| \le \mathop {\lim }\limits_{y \to x} 2{\left| {x - y} \right|^{{1 \over 2}}}$ Reasoning: We take the limit to transition from an inequality holding for all pairs of points to a statement about the instantaneous rate of change at a single point.

Step 4: Evaluate the Limits

• The left-hand side becomes: $\mathop {\lim }\limits_{y \to x} \left| {\frac{{f\left( x \right) - f\left( y \right)}}{{x - y}}} \right| = \left| {\mathop {\lim }\limits_{y \to x} \frac{{f\left( x \right) - f\left( y \right)}}{{x - y}}} \right| = |f'(x)|$ (The absolute value can be moved inside the limit because the absolute value function is continuous).
• The right-hand side becomes: $\mathop {\lim }\limits_{y \to x} 2{\left| {x - y} \right|^{{1 \over 2}}} = 2|0|^{1/2} = 0$ Reasoning: The left limit directly applies the definition of the derivative. The right limit simplifies because $|x-y|$ approaches 0.

Step 5: Deduce the Value of the Derivative Combining the results from Step 4, we have: $|f'(x)| \le 0$ Since the absolute value of any real number is always non-negative ($|f'(x)| \ge 0$), the only way for $|f'(x)| \le 0$ to be true is if $|f'(x)| = 0$. Therefore, $f'(x) = 0$ for all $x \in \mathbb{R}$. Reasoning: This is a crucial deduction. The combination of the inequality and the properties of absolute values forces the derivative to be zero everywhere.

Step 6: Determine the Function $f(x)$ Since $f'(x) = 0$ for all $x$, the function $f(x)$ must be a constant function. We can find $f(x)$ by integrating $f'(x)$: $f(x) = \int f'(x) \, dx = \int 0 \, dx = C$ where $C$ is the constant of integration. We are given the initial condition $f(0) = 1$. Substituting this into our constant function: $f(0) = C \Rightarrow 1 = C$ Thus, the function is $f(x) = 1$ for all $x \in \mathbb{R}$. Reasoning: A zero derivative implies a constant function. The initial condition allows us to uniquely determine that constant.

Step 7: Evaluate the Definite Integral We need to compute $\int_0^1 f^2(x) \, dx$. Substituting $f(x) = 1$: $\int_0^1 f^2(x) \, dx = \int_0^1 (1)^2 \, dx = \int_0^1 1 \, dx$ Evaluating the integral: $\int_0^1 1 \, dx = [x]_0^1 = 1 - 0 = 1$ Reasoning: With the function $f(x)$ determined, the integral becomes a simple calculation.

Common Mistakes & Tips

• Exponent Value: The exponent $3/2$ is greater than 1. This is critical. If the exponent were 1, the inequality would be $|f(x) - f(y)| \le K|x - y|$, implying Lipschitz continuity, not necessarily $f'(x)=0$.
• Absolute Value Logic: Remember that $|A| \ge 0$. The deduction that $|f'(x)| \le 0$ implies $f'(x) = 0$ relies heavily on this property.
• Differentiability Assumption: The problem explicitly states $f$ is differentiable. This is essential for forming the limit of the difference quotient and concluding $f'(x)$ exists.

Summary The given inequality $|f(x) - f(y)| \le 2|x - y|^{3/2}$ implies that the derivative of $f(x)$ must be zero everywhere. By forming the difference quotient, taking the limit as $y \to x$, and using the properties of absolute values, we showed that $|f'(x)| \le 0$, which means $f'(x) = 0$. A function with a zero derivative is a constant. Using the initial condition $f(0) = 1$, we determined that $f(x) = 1$. Finally, we integrated $f^2(x) = 1^2 = 1$ from 0 to 1, which yielded the result 1.

The final answer is \boxed{1}.