Problem 94 *

Find all strictly increasing functions f : \mathbb R \to \mathbb R such that f(x + f(y)) = f(x + y) + 1 for all real numbers x, y.

[originally posted by Mladenov on TSR]

Suppose that f(x) = f(y) for some x, y \in \mathbb{R}. Then certainly we cannot have that x < y as otherwise f(x) < f(y). Similarly we cannot have that x > y. Hence, x = y and so f is injective.

Now let x = y = 0 to obtain f(f(0)) = f(0) + 1.
Let y = -x to obtain f(x + f(-x)) = f(0) + 1 and thus:

f(x + f(-x)) = f(f(0)),

for all x \in \mathbb{R}. Since f is injective, it follows that

\begin{align*} x + f(-x) &= f(0) \\ f(x) &= x + f(0). \end{align*}

But by the original functional equation, x + y + 2f(0) = x + y + f(0) + 1 so f(0) = 1. Hence f(x) = x + 1 for all x \in \mathbb{R}. Clearly this also satisfies the given equation.

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