Let R be the set of real numbers and `f: R to R` suchthat for all x and y in R.
`|f(x) -f(y)| le |x-y|^(3)`, then `f(x)` is a constant.

Answer

To show that the function \( f: \mathbb{R} \to \mathbb{R} \) defined by the condition

\[ |f(x) - f(y)| \leq |x - y|^3 \]

for all \( x, y \in \mathbb{R} \) is a constant function, we can follow these steps:

### Step 1: Analyze the given condition
We start with the inequality:

\[ |f(x) - f(y)| \leq |x - y|^3 \]

This tells us that the difference in the function values \( f(x) \) and \( f(y) \) is bounded by the cube of the difference in \( x \) and \( y \).

### Step 2: Choose \( y = x + h \)
Let’s set \( y = x + h \) where \( h \) is a small increment. Then we can rewrite the inequality as:

\[ |f(x) - f(x + h)| \leq |h|^3 \]

### Step 3: Divide by \( |h| \)
Now, we divide both sides of the inequality by \( |h| \) (assuming \( h \neq 0 \)):

\[ \frac{|f(x) - f(x + h)|}{|h|} \leq |h|^2 \]

### Step 4: Take the limit as \( h \to 0 \)
Next, we take the limit as \( h \) approaches \( 0 \):

\[ \lim_{h \to 0} \frac{|f(x) - f(x + h)|}{|h|} \leq \lim_{h \to 0} |h|^2 \]

The right-hand side approaches \( 0 \) because \( |h|^2 \) goes to \( 0 \) as \( h \to 0 \).

### Step 5: Recognize the left-hand side as the definition of the derivative
The left-hand side is the definition of the derivative of \( f \) at point \( x \):

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

### Step 6: Conclude that \( f'(x) = 0 \)
Since the absolute value of the derivative is non-negative, the only possibility is:

\[ f'(x) = 0 \]

for all \( x \in \mathbb{R} \).

### Step 7: Integrate to find \( f(x) \)
If the derivative \( f'(x) = 0 \), then \( f(x) \) must be a constant function. Thus, we can write:

\[ f(x) = k \]

where \( k \) is a constant.

### Conclusion
Therefore, we conclude that \( f(x) \) is a constant function.

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