Suppose that f:R->R is a continuous function such that f(x+y) = f(x) +f(y) for all x, y element of R. Prove that there exists k element of R such that f(x) = kx, for every x element of R
A beautiful beautiful problem. My very first question that I asked on this forum*. The necessary and sufficient conditions is for f to be continous at 0. Since f is continous everywhere it must be continous at 0. Follow the elegant solution hpe gave.
*)Actually it is find all such functions.
I believe this can be done through induction (if not, then someone correct me please).
First: check to see that it works for the first term: x = 1
f(1) = k
Second: assume that it works for some value: x = n
f(n) = kn
Third: show that it works for the subsequent term: x = n + 1
f(n + 1) = k(n + 1)
Since f(x + y) = f(x) + f(y), we have that
f(n + 1) = f(n) + f(1) = kn + k = k(n + 1)
This functional equation happens to be very old.
Cauchy Functional Equation