Let f be a differentiable function from R to R such that |f(x)-f(y)| ≤ 2|x-y|^(3/2), for all x, y ∈

Q: Wait, why can we just divide by $|x - y|$? Isn't that undefined when $x = y$?

That is a brilliant observation! You are absolutely right that division by zero is undefined. However, in the context of limits, we are interested in the behavior of the function as $x$ gets arbitrarily close to $y$, but not equal to it. By considering the limit as $x \to y$, we are essentially looking at the behavior of the difference quotient in the neighborhood of $y$, which is exactly how we define the derivative.

Q: What if the exponent was 1 instead of 3/2? Would the function still be constant?

Not necessarily! If the exponent were 1, the condition would be $|f(x) - f(y)| \le 2|x - y|$, which is the definition of a Lipschitz continuous function with a Lipschitz constant of 2. Such functions don't have to be constant; for example, $f(x) = 2x$ satisfies this. The fact that our exponent is strictly greater than 1 (3/2 > 1) is the 'magic' that forces the derivative to be zero.

Q: Why does $|f'(y)| = 0$ imply that $f(x)$ is a constant function?

Think about the geometric meaning of the derivative. The derivative $f'(y)$ represents the slope of the tangent line to the curve at point $y$. If the slope is zero everywhere, the tangent line is always horizontal. A function whose slope is zero at every point must be a flat, horizontal line, which is mathematically defined as a constant function, $f(x) = C$.

Question

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

Options:

(A) 0

(B) 1/2

(C) 2

(D) 1

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast, flat plain. You are looking for a function $f(x)$ that is so incredibly smooth, so perfectly restricted, that it cannot move.

We are given a differentiable function $f: R 	o R$ that satisfies the inequality $|f(x) - f(y)| \le 2|x - y|^{3/2}$.

At first glance, this looks like a standard calculus problem, but it is actually a profound statement about the "wiggliness" of a function. The exponent $3/2$ is the key to the entire puzzle.

### The Mathematical Surgery

To understand why this function is so restricted, we need to perform a bit of mathematical surgery. We want to see the derivative hidden inside this inequality.

Recall the definition of the derivative: 
$$f'(y) = \lim_{x 	o y} \frac{f(x) - f(y)}{x - y}$$

Our inequality involves the absolute difference $|f(x) - f(y)|$. If we divide both sides by $|x - y|$, we get: 
$$\frac{|f(x) - f(y)|}{|x - y|} \le \frac{2|x - y|^{3/2}}{|x - y|}$$

### The Master Equation

Now, look at the right-hand side. Using the laws of exponents, we have $|x - y|^{3/2} / |x - y|^1 = |x - y|^{1/2}$.

So, our inequality simplifies to: 
$$\left|\frac{f(x) - f(y)}{x - y}ight| \le 2|x - y|^{1/2}$$

This is where the magic happens. As we let $x$ approach $y$, the left side becomes the absolute value of the derivative, $|f'(y)|$.

On the right side, as $x 	o y$, the term $2|x - y|^{1/2}$ approaches zero. Thus, we are left with the condition $|f'(y)| \le 0$.

### The Conclusion

Since the absolute value of any real number must be non-negative ($|f'(y)| \ge 0$), the only way for $|f'(y)|$ to be less than or equal to zero is if it is exactly zero.

Therefore, $f'(y) = 0$ for all $y \in R$. A function whose derivative is zero everywhere is a constant function.

So, $f(x) = C$ for some constant $C$. We are given $f(0) = 1$, which immediately tells us that $C = 1$. Our mysterious function is simply $f(x) = 1$.

### The Final Integral

With the function identified as $f(x) = 1$, the integral becomes trivial: 
$$\int_0^1 f^2(x) dx = \int_0^1 (1)^2 dx = \int_0^1 1 dx$$

This is simply the area of a rectangle with height 1 and width 1. The final result is **1**.

### The Golden Rule of Lipschitz Continuity

This problem illustrates a beautiful general principle. If you encounter a condition $|f(x) - f(y)| \le K|x - y|^\alpha$ where $\alpha > 1$, the function is forced to be constant.

This is a specific case of a broader concept in analysis related to Lipschitz continuity. When the exponent $\alpha$ is greater than 1, the function's rate of change is so constrained that it cannot change at all.

Keep this "Golden Rule" in your toolkit—it is a powerful weapon for JEE Advanced problems!

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f$ be a differentiable function from $R$ to $R$ such that $∣ f (x) - f (y) ∣ \leq 2∣ x - y ∣^{3/2}$ , for all $x, y \in R$ . If $f (0) = 1$ then $\int_{0}^{1} f^{2} (x) d x$ is equal to

Select Answer:

Visualized Solution

Given Condition

Rearranging the Inequality

Simplifying the RHS

Applying the Limit

Evaluating the Limits

The Derivative is Zero

Identifying the Function

Finding the Constant

Setting up the Integral

Final Calculation

Summary and Takeaway

The Sigma Insight: Differentiability of a Function

Analyzing the Setup

The Mathematical Surgery

The Master Equation

The Conclusion

The Final Integral

The Golden Rule of Lipschitz Continuity

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