Let $M \in \mathbb{R}_+^* \cup \{+\infty\}$ and $f : {]-\infty, M[} \rightarrow \mathbb{R}$ be a continuous function such that $$\forall (x, y) \in {\left]-\infty, \frac{M}{2}\right[}^2, \quad 2f(x+y) = f(2x) + f(2y) \tag{IV.1}$$ Let $\alpha$ be a number strictly less than $\frac{M}{2}$ and $F$ be the antiderivative of $f$ vanishing at $\alpha$. Show that for all $x$ and $y$ in $]-\infty, \frac{M}{2}[$, with $y \neq \alpha$, we have: $$f(2x) = 2\frac{F(x+y) - F(x+\alpha) - \frac{1}{4}F(2y) + \frac{1}{4}F(2\alpha)}{y - \alpha}$$
Let $M \in \mathbb{R}_+^* \cup \{+\infty\}$ and $f : {]-\infty, M[} \rightarrow \mathbb{R}$ be a continuous function such that
$$\forall (x, y) \in {\left]-\infty, \frac{M}{2}\right[}^2, \quad 2f(x+y) = f(2x) + f(2y) \tag{IV.1}$$
Let $\alpha$ be a number strictly less than $\frac{M}{2}$ and $F$ be the antiderivative of $f$ vanishing at $\alpha$. Show that for all $x$ and $y$ in $]-\infty, \frac{M}{2}[$, with $y \neq \alpha$, we have:
$$f(2x) = 2\frac{F(x+y) - F(x+\alpha) - \frac{1}{4}F(2y) + \frac{1}{4}F(2\alpha)}{y - \alpha}$$