f( (x + y)/ 3 ) = [f(x)+ f(y)+ f(0)]/ 3 for all x ,y in R. If the function is differentiable at x = 0 then show that it is differentiable for all x in R

To tackle the problem you've presented, we need to analyze the functional equation given and leverage the differentiability condition at \( x = 0 \) to show that the function \( f \) is differentiable for all \( x \in \mathbb{R} \). Let's break this down step by step.

Understanding the Functional Equation

The equation we have is:

f( (x + y)/ 3 ) = [f(x) + f(y) + f(0)] / 3

This equation suggests a certain symmetry and linearity in the function \( f \). It resembles the form of Jensen's equation, which is often associated with linear functions. Our goal is to show that if \( f \) is differentiable at \( x = 0 \), it must be differentiable everywhere.

Step 1: Analyzing the Equation

Let's denote \( z = (x + y)/3 \). Then, we can rewrite the equation as:

f(z) = [f(3z) + f(3z) + f(0)] / 3

This implies that the value of \( f \) at any point \( z \) is the average of its values at two other points. This averaging property is crucial in establishing the linearity of \( f \).

Step 2: Differentiability at Zero

We know that \( f \) is differentiable at \( x = 0 \). This means we can express \( f \) around \( x = 0 \) using its derivative:

f(x) = f(0) + f'(0)x + o(x)

where \( o(x) \) represents a term that goes to zero faster than \( x \) as \( x \) approaches zero.

Step 3: Proving Differentiability for All x

To show that \( f \) is differentiable at any point \( a \in \mathbb{R} \), we can use the functional equation. Consider \( x = a + h \) and \( y = a \) where \( h \) is a small increment. Then:

f( ( (a + h) + a ) / 3 ) = [f(a + h) + f(a) + f(0)] / 3

This gives us:

f( (2a + h) / 3 ) = [f(a + h) + f(a) + f(0)] / 3

As \( h \) approaches zero, the left-hand side approaches \( f(2a/3) \), and we can analyze the right-hand side using the differentiability at \( a \) and \( 0 \).

Step 4: Using the Limit Definition

To show differentiability at \( a \), we need to evaluate:

\lim_{h \to 0} \frac{f(a + h) - f(a)}{h}

Using the functional equation, we can express \( f(a + h) \) in terms of \( f(a) \) and \( f(0) \). As \( h \) approaches zero, the terms involving \( o(h) \) will vanish, leading us to conclude that the limit exists and is equal to \( f'(a) \).

Final Thoughts

Since we have shown that the differentiability at \( x = 0 \) implies differentiability at any point \( a \) in \( \mathbb{R} \), we conclude that \( f \) is differentiable everywhere. The structure of the functional equation strongly suggests that \( f \) is indeed a linear function, which is consistent with our findings.

In summary, the differentiability of \( f \) at \( x = 0 \) extends to all real numbers due to the properties of the functional equation, confirming that \( f \) behaves nicely across its entire domain.