Let $c > 1$ be a real number. A function $f: [0 ,1 ] \to R$ is called c-friendly if $f(0) = 0, f(1) = 1$ and $|f(x) -f(y)| \le c|x - y|$ for all the numbers $x ,y \in [0,1]$. Find the maximum of the expression $|f(x) - f(y)|$ for all [i]c-friendly[/i] functions  $f$ and for all the numbers $x,y \in [0,1]$. 
1. **Define the problem and constraints:**
   We are given a function \( f: [0, 1] \to \mathbb{R} \) that is \( c \)-friendly, meaning:
   - \( f(0) = 0 \)
   - \( f(1) = 1 \)
   - \( |f(x) - f(y)| \le c|x - y| \) for all \( x, y \in [0, 1] \)

2. **Establish the goal:**
   We need to find the maximum value of \( |f(x) - f(y)| \) for all \( c \)-friendly functions \( f \) and for all \( x, y \in [0, 1] \).

3. **Use the given constraints:**
   For any \( 0 \leq x < y \leq 1 \), we have:
   \[
   |f(x) - f(y)| \leq c(y - x)
   \]
   This is directly from the definition of \( c \)-friendliness.

4. **Consider the triangle inequality:**
   We can also use the triangle inequality to bound \( |f(x) - f(y)| \):
   \[
   |f(x) - f(y)| \leq |f(x) - f(0)| + |f(0) - f(1)| + |f(1) - f(y)|
   \]
   Given \( f(0) = 0 \) and \( f(1) = 1 \), this simplifies to:
   \[
   |f(x) - f(y)| \leq |f(x)| + 1 + |1 - f(y)|
   \]

5. **Apply the \( c \)-friendly condition:**
   Using the \( c \)-friendly condition, we have:
   \[
   |f(x)| \leq cx \quad \text{and} \quad |1 - f(y)| \leq c(1 - y)
   \]
   Therefore:
   \[
   |f(x) - f(y)| \leq cx + 1 + c(1 - y) = c(x - y) + c + 1
   \]

6. **Combine the inequalities:**
   We now have two inequalities:
   \[
   |f(x) - f(y)| \leq c(y - x)
   \]
   \[
   |f(x) - f(y)| \leq c(x - y) + c + 1
   \]
   Adding these two inequalities, we get:
   \[
   2|f(x) - f(y)| \leq c + 1
   \]
   Therefore:
   \[
   |f(x) - f(y)| \leq \frac{c + 1}{2}
   \]

7. **Construct a function to achieve the maximum:**
   Consider the piecewise linear function \( f \) defined by:
   \[
   f(x) = \begin{cases}
   -cx & \text{if } x \leq \frac{c-1}{4c} \\
   cx - \frac{c-1}{2} & \text{if } \frac{c-1}{4c} < x < 1 - \frac{c-1}{4c} \\
   -cx + c + 1 & \text{if } x \geq 1 - \frac{c-1}{4c}
   \end{cases}
   \]
   For the points \( x = \frac{c-1}{4c} \) and \( y = 1 - \frac{c-1}{4c} \), we can verify that:
   \[
   |f(x) - f(y)| = \frac{c + 1}{2}
   \]

Conclusion:
\[
\boxed{\frac{c + 1}{2}}
\]