Wednesday, 9 March 2016

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




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



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.


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