Almost A Linear Functional

Putting $x=y=0$ tells us that $f(0) = f(0) + f(0)$ , so that $f(0) = 0$ .

If $\alpha = 0$ then $f(x) + f(y) = f(0) = 0$ for all $x,y$ , so $2f(x) = f(x) + f(x) = 0$ for all $x$ , so that $f \equiv 0$ is constant.

If $\alpha \neq 0$ then since $x \,=\, \alpha\big(x + (\alpha^{-1}-1)x\big)$ , we deduce that $f(x) \; = \; f(x) + f\big((\alpha^{-1}-1)x\big)$ for all $x$ , so that $f\big((\alpha^{-1}-1)x\big) = 0$ for all $x$ . If $\alpha \neq 1$ , this implies that $f \equiv 0$ is constant.

Thus the only possible value of $\alpha$ is $1$ , which admits lots of nonconstant $f$ , such as $f(x) = x$ .

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An even quicker argument would be to observe that (for any real $x$ and any $\alpha \neq 1$ ): $\frac{\alpha x}{1-\alpha} \; = \; \alpha\left(\frac{\alpha x}{1-\alpha} + x\right)$ so that $f\left(\frac{\alpha x}{1-\alpha}\right) \; = \; f\left(\frac{\alpha x}{1-\alpha}\right) + f(x)$ and hence $f \equiv 0$ if $\alpha \neq 1$ .

[This is my scratch work. This is not a solution.]

For $\alpha = 1$ , there are numerous solutions like $f(x) = C x$ .

For $\alpha = 0$ ,
set $x=0, y = 0$ we get that $f(0) = f(0) + f(0) \Rightarrow f(0) = 0$ .
Set $x = x, y = 0$ , we get that $f(0) = f(x)$ .
Hence, we have the constant function $f(x) = 0$ .

For $\alpha = -1$ ,
set $x = 0, y = 0$ gives $f(0) = 0$ .
Set $x=x, y = 0$ gives $f(-x) = f(x)$ .
Set $x = x, y =-x$ gives $0 = f(0) = f(x) + f(-x) = 2f(x)$ .
Hence, we have the constant function $f(x) = 0$ .

Henceforth, $\alpha \neq 0, 1, -1$ .

For all other $\alpha$ ,
Set $x = 0, y = 0$ gives $f(0) = 0$ . Set $x = x, y = 0$ gives $f( \alpha x) = f(x)$ . By induction $f( \alpha^n x) = f(x)$ .
Using the above, $f(\alpha (x+y) ) = f(x) + f(y) = f( \alpha x) + f(\alpha y)$ .

Hence, $f(x)$ is a rational-linear functional. All solutions can be described as
1. Pick a Q basis over R, $\{ e_1, e_2, \ldots \}$
2. Set $f(e_i) = c_i$ .
3. Then $f( \sum b_i e_i ) = \sum b_i c_i$

Why must we have $c_i = 0$ ? Why does $f( \alpha x ) = f(x)$ force this constraint?

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If $\alpha$ is rational, then $\alpha f(x) = f(\alpha x ) = f(x)$ hence $f(x) = 0$ .
If $\alpha$ is algebraic, then let $\sum c_i \alpha^i = 0$ be it's minimial polynomial. We know that $\sum c_i \neq 0$ (since we can factor out $(x-1)$ otherwise ), and so $f(0) = f( x \times \sum c_i \alpha^i = \sum c_i f( x \times \alpha_i) \sum c_i f(x)$ . Hence $f(x) = 0$ .
What about transcedentals?