Let f be any function defined on R and let it satisfy the condition : |f(x) - f(y)| ≤ |(x - y)²|, ∀(

Q: Wait, why can we divide by $|x - y|$? What if $x = y$?

That is a fantastic question! You are absolutely right to be cautious. We are interested in the behavior of the function as $x$ approaches $y$, so we consider the case where $x \neq y$. When $x = y$, the inequality becomes $|f(x) - f(x)| \le |(x - x)^2|$, which simplifies to $0 \le 0$, a trivial statement that doesn't help us find the derivative. By assuming $x \neq y$, we can safely divide to build the difference quotient, and the limit process then handles the behavior as $x$ gets arbitrarily close to $y$.

Q: How do we know the derivative is zero? Could it be something else?

Great intuition! Let's look at the result of our limit: $|f'(y)| \le \lim_{x \to y} |x - y|$. As $x$ approaches $y$, the distance $|x - y|$ shrinks to zero. This forces $|f'(y)| \le 0$. Since the absolute value of any real number must be non-negative ($|f'(y)| \ge 0$), the only possible value that satisfies both $|f'(y)| \le 0$ and $|f'(y)| \ge 0$ is $|f'(y)| = 0$. Therefore, $f'(y)$ must be zero.

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

This is a fundamental result from the Mean Value Theorem. If the derivative of a function is zero for all points in an interval, the function cannot increase or decrease. Imagine a path where your instantaneous velocity is always zero; you simply cannot move from your starting position. Thus, $f(x)$ must remain at its initial value, $f(0)$, for all $x$.

Question

Let $f$ be any function defined on $R$ and let it satisfy the condition : $|f(x) - f(y)| \le |(x - y)^2|, \forall(x, y) \in R$. If $f(0) = 1$, then:

Options:

(A) $f(x) < 0, \forall x \in R$

(B) $f(x)$ can take any value in $R$

(C) $f(x) = 0, \forall x \in R$

(D) $f(x) > 0, \forall x \in R$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast, flat landscape. You are given a mysterious function $f$ defined on all real numbers, and you are told it obeys a very strict rule:

$$|f(x) - f(y)| \le |(x - y)^2|$$

At first glance, this looks like a simple inequality, but it is actually a powerful constraint that dictates the entire behavior of the function. Let us embark on a journey to uncover its true identity.

### The Path to the Derivative

To understand how $f(x)$ behaves, we need to look at its rate of change. In calculus, we define the derivative as the limit of the difference quotient:

$$f'(y) = \lim_{x 	o y} \frac{f(x) - f(y)}{x - y}$$

Our given inequality holds for all $x$ and $y$. To bridge the gap between this inequality and the derivative, we need to create that difference quotient.

Let us assume $x 
eq y$ and divide both sides of the inequality by $|x - y|$. Because $|x - y|$ is always positive for $x 
eq y$, the inequality sign remains unchanged:

$$\frac{|f(x) - f(y)|}{|x - y|} \le \frac{|(x - y)^2|}{|x - y|}$$

### The Beauty of Simplification

Now, let us clean up this expression. On the left side, we can combine the absolute values:

$$\left| \frac{f(x) - f(y)}{x - y} ight|$$

On the right side, we have the absolute value of $(x - y)^2$ divided by the absolute value of $(x - y)$. One power of $(x - y)$ cancels out, leaving us with a much simpler expression:

$$\left| \frac{f(x) - f(y)}{x - y} ight| \le |x - y|$$

This is the moment of clarity. We have successfully isolated the difference quotient inside an absolute value sign, bounded by the distance between $x$ and $y$.

### The Calculus Magic

Now, we apply the limit as $x$ approaches $y$. As $x$ gets infinitely close to $y$, the left side becomes the formal definition of the absolute value of the derivative, $|f'(y)|$.

On the right side, as $x 	o y$, the term $|x - y|$ approaches zero. This leads us to the striking conclusion:

$$|f'(y)| \le 0$$

We know that the absolute value of any real number must be greater than or equal to zero. If it is also less than or equal to zero, it must be exactly zero. Thus, $f'(y) = 0$ for all $y \in \mathbb{R}$.

### The Final Revelation

If the derivative of a function is zero everywhere, the function has no slope. It is perfectly horizontal. This means $f(x)$ must be a constant function, $f(x) = C$.

We are given the initial condition $f(0) = 1$. Substituting $x = 0$ into our constant function, we find $f(0) = C$, which implies $C = 1$.

Therefore, the function is simply **$f(x) = 1$** for all $x$. Since $1 > 0$, we have proven that $f(x) > 0$ for all $x \in \mathbb{R}$.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f$ be any function defined on $R$ and let it satisfy the condition : $∣ f (x) - f (y) ∣ \leq ∣ (x - y)^{2} ∣, \forall (x, y) \in R$ . If $f (0) = 1$ , then:

Select Answer:

Visualized Solution

The Given Inequality

Preparing for the Derivative

Dividing by $∣ x - y ∣$

Simplifying the Right Hand Side

Applying the Limit

Evaluating the Limits

Deducing the Derivative

The Constant Function

Using the Initial Condition

Finding the Constant

Final Conclusion

The Sigma Insight: Differentiability of a Function

Analyzing the Setup

The Path to the Derivative

The Beauty of Simplification

The Calculus Magic

The Final Revelation

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