1. The problem statement, all variables and given/known data Prove that if f(x) satisfies the functional equation f(x+y) = f(x) + f(y) and if f is continous then f(x) = cx for some constant c. 2. Relevant equations N/A 3. The attempt at a solution Assume |f(a)| > |ca| for some a in the domain of f. Since f is continuous at every point then for every e > 0 we can find a d > 0 such that |f(x) - f(x+a)| < e whenever |x - (x+a)| < d. We have |f(x) - (f(x) + f(a)| = |-f(a)| = |f(a)| < e whenever |a| < d. Since |ca| < f(a)| < e then |ca| < e, implying that |a| < e/|c|. Thus we can take d = e/|c|, implying that |c|d = e. Now, since d would be points on the x axis and e points on the y axis, then we have y = |c|x, or f(x) = cx for all x in the domain of f. However, we said that |f(a)| > |ca|. So this is a contradiction. If we assume |f(a)| < |ca| for some a in the domain of f, then we have again |f(a)| < e whenever |a| < d. However, we don't know if |f(a)| < |ca| < e or |f(a)| < e < |ca|. If it's |f(a)| < |ca| < e then we can continuous with our proof using the same argument as before. If it's |f(a)| < e < |ca| then we can't. However, the function g(x) = cx is also continuous for all x and so we have |f(a)| < |ca| < e'. Then |a| < e'/|c| and we can choose d = e'|c|, implying that |c|d = e', implying that |c|x = y = f(x). This is again a contradiction, so therefore |f(x)| = |cx|. So if f(x) > 0 then f(x) = cx. And if f(x) < 0 then -f(x) = -(cx), implying that f(x) = cx. Therefore f(x) = cx. QED Is this a valid proof? I found the question difficult so I want to make sure that the proof is correct.