a continuous function, satisfying $f(α) = f(β) +f(α −β)$ for any $α, β ∈ \mathbb{R}$

Question

Hi need some help with this problem:

Assume $f : \mathbb{R} → \mathbb{R}$ is a continuous function, satisfying $f(α) = f(β) +f(α −β)$ for any $α, β ∈ \mathbb{R}$, and $f(0) = 0$. Then $f(α) = α f(1)$.

any hints, thank you.

Answer 1

Hint: Show that $f(\alpha)=\alpha f(1)$ for all $\alpha\in\mathbb{Z}$. Then show that $f(\alpha)=\alpha f(1)$ for all $\alpha\in\mathbb{Q}$. Then use continuity.

Answer 2

If you can show that the function must be differentiable at $0$, we have $$\frac{f(a)-f(b)}{a-b}=\frac{f(a-b)}{a-b}$$ Taking limits as $a\to b$ gives $$f'(b)=f'(0)$$ Therefore, because $b$ is arbitrary we must have $$f'(x)=c\in\mathbb{R}$$ so $$f(x)=cx=f(1)x$$

Answer 3

Note that $f(-\beta)=f(0-\beta)=f(0)-f(\beta)=-f(\beta)$ then $f(\alpha-\beta)=f(\alpha)+f(-\beta)$ and also $f(\alpha+\beta)=f(\alpha-(-\beta))=f(\alpha)+f(-(-\beta))=f(\alpha)+f(\beta)$ and this is a "Cauchy functional equation" conditon of continuity.

See the hint of @Nate. or see the book "Kaczor and Nowak - Problems in Mathematical Analysis II (2000)" 1.6.2 exercise and then conclude your proof.