Consider the function f:R→R f(x+y)=f(x)+f(y)
which is known as Cauchy's functional equation. I know that if f is monotonic, continuous at one point, bounded, then the only solutions are f(x)=cx, ∀c But once I successfully derived Cauchy's functional equation from a problem, and I know it's an involution, bijective, as well as an odd function. How can I derive (if possible) that f(x)=x is the only solution?
Saturday, 21 November 2015
Cauchy's functional equation -- additional condition
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