Solution of a functional equation $f(\alpha x+\beta)=\left( f(\alpha+\beta)-f(\beta) \right) f(x) + f(\beta)$ I am looking for an accessible proof of the following claim (which comes up in social choice theory). The proof is supposed to follow from a theorem on page 150 of Aczel's 1966 Lectures on functional equations and their applications. But, sadly, I find Aczel's discussion very hard to understand.
Suppose that
$$f(\alpha x+\beta)=\left( f(\alpha+\beta)-f(\beta) \right) f(x) + f(\beta)$$
for all real numbers $x$, all positive real numbers $\alpha$, all real numbers $\beta$, and where $f$ is real-valued, continuous and strictly increasing.
Then
$$f(x) = x$$
for all real numbers $x$.
(Or, at least, $f(x)=x$ for all real numbers if we normalize $f$ so that $f(0)=0$ and $f(1)=1$.)
Edit: Thanks for all the suggestions so far! I now know that if we set $\beta=0$, we get
$$f(\alpha x)=f(\alpha)f(x)$$
for all positive real $\alpha$ and all real $x$, which is a version of the Cauchy equation.
And if we set $\alpha=1$, we get
$$f(x+\beta)=\left( f(1+\beta)-f(\beta) \right) f(x) + f(\beta)$$
for all real $\beta$ and $x$.
And in the meantime I found out that this last thing is an instance of a functional equation from Aczel (1966, p. 150):
$$f(x+y)=f(x)h(y)+k(y)$$
to which he gives a number of possible solutions.
But I still have no clue how he arrives at these solutions. And I am still unsure why the result (that $f(x)=x$) follows if we put all these pieces together. Any help would be appreciated.
 A: Let $P(\alpha,x,\beta)$ denote the statement that $f(\alpha x+\beta)=\left( f(\alpha+\beta)-f(\beta) \right) f(x) + f(\beta)$ where $x,\beta \in \Bbb R$ and $\alpha \in \Bbb R^+$.Then note that $f(\alpha +\beta) -f(\beta )$ is always positive.Plugging in
$x=0$ we get:
$$
P(\alpha,0,\beta) \implies f(\beta)= (f(\alpha +\beta) -f(\beta ))f(0) +f(\beta) \\
\implies f(0)=0
$$
Plugging in $\beta =0$, we get $f(\alpha x)=f(\alpha )f(x)\ \forall x\in \Bbb R \ and \ \forall \alpha \in \Bbb R^+$.
Note that as $f$ is strictly increasing so $f(x)>0 \quad \forall x>0$ and $f(x)<0 \quad \forall x<0.$ Now,
$$
P(\alpha,x,-\alpha) \implies f(\alpha (x-1))=-f(-\alpha)f(x)+f(-\alpha)\\
\implies f(\alpha)f(x-1)=f(-\alpha)(1-f(x))\\
\implies f(\alpha)f(x-1)=f(-1)f(\alpha)(1-f(x))\\
\implies f(x-1)=f(-1)(1-f(x))\\
\text{Plugging in x=0.5, we get}\\
f(-0.5)=f(-1)(1-f(0.5))\\
\implies f(-1)f(0.5)=f(-1)(1-f(0.5))\\
\implies f(0.5)=0.5\\
as\ f(2*0.5)=f(2)f(0.5)\ ,so\ f(2)=2\\
\text{Plugging in $\alpha=\beta=1$, we get}\\
f(x+1)=f(x)+1\\
$$
As $f(x+1)=f(x)+1$ , so $f(-1)=-1$.
From induction we can now easily deduce $f(n)=n$ , as $f(-1)=-1$ ,  we get $f(-n)=-n$.
Hence, $f(z)=z$ $\forall z\in \Bbb Z$.
$$
As\ f(\frac mn .n)=m \\
\implies f(\frac mn)=\frac mn \\
\implies f(q)=q \qquad \forall q\in \Bbb Q
$$
As $f(x)$ is identity function on all rationals, it follows easily (from continuity or strictly increasing nature of the function) that $f(x)=x \quad \forall x\in \Bbb R$
A: As pointed out in the comments, it follows immediately that $f(0) = 0,$ and $f(1) = 1.$ This implies that
$$f(\alpha) f(x) = f(\alpha x),$$
(setting $\beta = 0$)
Now, since $f(0) = 0,$ it follows that $f(x) > 0,$ when $x>0$ (by strict monotonicity). Now, consider the composition $h=\log \circ f \circ \exp.$ The function $h$ satisfies $h(a + b) = h(a) + h(b)$ (and is continuous and monotone). It is a simple argument that this function is multiplication by a constant (clearly $f(n x) = n f(x),$ thus $f(q x) = q f(x)$ for $q \in mathbb{Q},$ thus by continuity we get $f(a x) = a f(x)$). From this the result follows.
