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Let $f\colon (0,\alpha)\to \def\R{\mathbf R}\R$ satisfy $f(x + y)=f(x)+f(y)$ for all $x,y,x + y \in (0,\alpha)$, where $\alpha$ is a positive real number. Show that there exists an additive function $A \colon \R \to \R$ such that $A(x) = f(x)$ for all $x \in (0, \alpha)$. Simply I want to define a function A In specific form as an extension of the function f wich is additive functional equation. I tried to define the function A .

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Let $x > 0$. Choose $n \in \def\N{\mathbf N}\N$ with $\frac xn < \alpha$. Define $A(x) := nf(\frac xn)$. Note that this is well-defnined: If $m\in \N$ is another natural number such that $\frac xm < \alpha$, we have \begin{align*} mf\left(\frac xm\right) &= mf\left(\sum_{k=1}^n \frac x{mn}\right)\\ &= m\sum_{k=1}^n f\left(\frac x{mn}\right)\\ &= \sum_{l=1}^m n f\left(\frac x{mn}\right)\\ &= nf\left(\sum_{l=1}^m \frac x{mn}\right)\\ &= nf\left(\frac x{n}\right). \end{align*} For $x < 0$ choose $n \in \N$ with $\frac xn > -\alpha$ and define $A(x) := -nf(-\frac xn)$, finally, let $A(0) = 0$. Then $A$ is an extension of $f$, to show that it is additive, let $x,y \in \def\R{\mathbf R}\R$. Choose $n \in \N$ such that $\frac xn, \frac yn, \frac{x+y}n \in (-\alpha, \alpha)$. We have if $x,y \ge 0$: \begin{align*} A(x+y) &= nf\left(\frac{x+y}n\right)\\ &= nf\left(\frac xn\right) + nf\left(\frac yn\right)\\ &= A(x) + A(y) \end{align*} If both $x,y \le 0$, we argue along the same lines. Now suppose $x \ge 0$, $y \le 0$, $x+y \ge 0$. We have $A(y) = -A(-y)$ be definition of $A$. Hence \begin{align*} -A(y) + A(x+y) &= A(-y) + A(x+y)\\ &= A(-y+x+y)\\ &= A(x). \end{align*} If $x \ge 0$, $y \le 0$, $x+y \le 0$, we have $-x \le 0$ and \begin{align*} -A(x) + A(x+y) &= A(-x) + A(x+y)\\ &= A(y) \end{align*}

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