Let f be any function defined on R and let it satisfy the condition :
`| f (x) - f(x) | le ( x-y)^2, AA (x,y) in R ` if ` f(0) =1, ` then :

Options

A. f(x) can take any value in R
B. `f(x) lt 0, AA x in R`
C. `f(x) =0 ,AA x in R`
D. `f(x) gt 0 , AA x in R `

Answer

To solve the problem, we start with the given condition:

\[ |f(x) - f(y)| \leq (x - y)^2 \quad \forall (x, y) \in \mathbb{R} \]

and the initial condition \( f(0) = 1 \).

### Step 1: Understanding the condition

The condition \( |f(x) - f(y)| \leq (x - y)^2 \) suggests that the function \( f \) is Lipschitz continuous with a Lipschitz constant that depends on the square of the distance between \( x \) and \( y \). This implies that as \( x \) approaches \( y \), the difference \( |f(x) - f(y)| \) approaches zero faster than the linear term \( |x - y| \).

### Step 2: Setting up the limit

To analyze the behavior of \( f \), we can rewrite the condition by dividing both sides by \( |x - y| \) (assuming \( x \neq y \)):

\[ \frac{|f(x) - f(y)|}{|x - y|} \leq |x - y| \]

Let \( y = x + h \) where \( h \) approaches \( 0 \). Then, we can rewrite the inequality as:

\[ \frac{|f(x + h) - f(x)|}{|h|} \leq |h| \]

### Step 3: Taking the limit

Now, we take the limit as \( h \to 0 \):

\[ \lim_{h \to 0} \frac{|f(x + h) - f(x)|}{|h|} \leq \lim_{h \to 0} |h| = 0 \]

This implies that:

\[ |f'(x)| \leq 0 \]

### Step 4: Conclusion about the function

Since the derivative \( f'(x) \) is less than or equal to zero, it indicates that \( f \) is a non-increasing function. The only way for the derivative to be non-positive everywhere and still satisfy the condition given is if \( f \) is a constant function.

### Step 5: Determining the constant

Given that \( f(0) = 1 \), we conclude that:

\[ f(x) = 1 \quad \forall x \in \mathbb{R} \]

### Final Answer

Thus, the function \( f(x) \) is:

\[ f(x) = 1 \]