Additive function a: R → R is an increasing function .it is continus.why?
If your notation is correct, then the range of the function is the Real Number set and is increasing. This suggests the graph of the function is smooth. As long as there are no breaks in the graph and it is smooth on its entire domain (or whatever interval you're considering) it is considered 'continuous'.
Any additive increasing function α: R → R is linear of the form α(x) = α(1)x, hence is continuous. Proof. Suppose that (1) α(x + y) = α(x) + α(y), for any x, y in R, and (2) α(x) ≤ α(y), for any x, y in R, x ≤ y. Applying (1) we have: α(0) = α(0+0) = α(0)+α(0) ==> α(0) = 0 0 = α(0) = α(x+(–x)) = α(x) + α(–x) ==> α(–x) = – α(x), for any x in R α(x – y) = α(x + (–y)) = α(x) + α(–y) = α(x) – α(y), for any x, y in R If n is a positive integer then, for any x in R α(nx) = α(x+x+...+x) = α(x)+α(x)+...+α(x) = nα(x) If n is a positive integer then, for any x in R α(x) = α(n(x/n)) = nα(x/n) ==> α(x/n) = (1/n)α(x) If m, n are positive integers then, for any x in R α((n/m)x) = α(n(x/m)) = nα(x/m) = n((1/m)α(x)) = (n/m)α(x) Collecting the previous results, we have α(qx) = qα(x), for any x in R, q in Q In particular, α(q) = α(1)q, for any q in Q Next I apply (2) to show that α(x) = α(1)x, for any x in R. Fix x in R. Since Q is dense in R, given ε > 0 there are p, q in Q such that p < x < q an q–p < ε. Then α(1)p = α(p) ≤ α(x) ≤ α(q) = α(1)q, so –α(1)(x–p) ≤ α(x) – α(1)x ≤ α(1)(q–x) –|α(1)|ε ≤ α(x) – α(1)x ≤ |α(1)|ε The previous inequality holds for any ε > 0, hence α(x) – α(1)x = 0 bye
