A function `f: I to I, f(x + y) = f(x) + f(y) - 1/2 AAx, y in I`, where `f(1) >` 0, then f(x) is

Options

A. one-one and decreasing when `x in I^(-)`
B. many-one when `inl` and decreasing when `x in I^(-)`
C. one-one and increasing when `x in l`
D. many-one when `x in I` and decreasing when `x in I^(-)`

Answer

To solve the problem, we start with the functional equation given:

\[ f(x + y) = f(x) + f(y) - \frac{1}{2} \]

for all \( x, y \in I \), where \( I \) is the set of integers, and we know that \( f(1) > 0 \).

### Step 1: Find \( f(0) \)

We can find \( f(0) \) by substituting \( x = 0 \) and \( y = 0 \) into the functional equation:

\[ f(0 + 0) = f(0) + f(0) - \frac{1}{2} \]

This simplifies to:

\[ f(0) = 2f(0) - \frac{1}{2} \]

Rearranging gives:

\[ f(0) - 2f(0) = -\frac{1}{2} \implies -f(0) = -\frac{1}{2} \implies f(0) = \frac{1}{2} \]

### Step 2: Differentiate the function

Assuming \( f \) is differentiable, we can express the derivative \( f'(x) \) using the limit definition:

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

Substituting \( y = h \) into the functional equation gives:

\[ f(x + h) = f(x) + f(h) - \frac{1}{2} \]

Thus,

\[ f'(x) = \lim_{h \to 0} \frac{f(x) + f(h) - \frac{1}{2} - f(x)}{h} = \lim_{h \to 0} \frac{f(h) - \frac{1}{2}}{h} \]

### Step 3: Evaluate \( f(h) \) as \( h \to 0 \)

As \( h \to 0 \), we know \( f(0) = \frac{1}{2} \). Therefore:

\[ f'(x) = \lim_{h \to 0} \frac{f(h) - \frac{1}{2}}{h} \]

This implies that \( f'(0) = 0 \).

### Step 4: Integrate to find \( f(x) \)

Since \( f'(x) = 0 \) implies that \( f(x) \) is a constant function, we can write:

\[ f(x) = C \]

for some constant \( C \). We know \( f(0) = \frac{1}{2} \), so \( C = \frac{1}{2} \).

### Step 5: Check the condition \( f(1) > 0 \)

We have \( f(x) = \frac{1}{2} \) for all \( x \). Thus:

\[ f(1) = \frac{1}{2} > 0 \]

This satisfies the condition given in the problem.

### Conclusion

The function \( f(x) \) is constant and equal to \( \frac{1}{2} \) for all integers \( x \).