# Answer to Question #476 in Calculus for Anthony

Question #476

Is there a bounded function f: R → R that f(1) > f(0) and ∀ x, y ∈ R: f(x + y) ≥ f²(x) + 2f(xy) + f²(y)?

Expert's answer

Take any x1 ≠ 0 and y1 = 1/x1. Then f²(x1 + y1) ≥ f²(x1) + 2f(1) + f²(y1) ≥ f²(x1) + a, where a = 2f(1) > 0. Then take xn = x(n-1) + y(n-1), yn = 1/xn, n ≥ 2. Then f²(xn + yn) ≥ f²(xn) + a = f²(x(n-1) + y(n-1)) + a ≥ f²(x(n-1)) + 2a ≥ ⋯ ≥≥ f²(x1) + na. It is clear that the sequence f(x1), f(x2), … , f(xn), … is an unlimited sequence.

Answer. No, there is not

Answer. No, there is not

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