f is a continuous real-valued function on the reals such that for some 0 < a, b < 1/2, we have f(f(x)) = a f(x) + b x for all x. Show that for some constant k, f(x) = k x for all x.

**Solution**

Notice first that if f(x) = kx, then f(f(x)) = k^{2}x, so k^{2}x = akx + bx and hence k is a root of the quadratic t^{2} - at - b = 0.

If f(x) = f(y), then f(f(x)) = f(f(y)), so x = ( f(f(x)) - a f(x) )/b = ( f(f(y)) - a f(y) )/b = y. In other words, f is one-to-one. Since it is continuous, it must be strictly increasing or strictly decreasing. It cannot be bounded as x tends to plus infinity, otherwise f(f(x)) would be bounded, but a f(x) + b x would be unbounded. Similarly as x tends to minus infinity. So f must be onto and hence invertible.

Let h, k be the roots of t^{2} = a t + b. Evidently one root, say h, is positive and the other negative. Moreover both must have absolute value less than 1. This is where we need 0 < a, b < 1/2. Write h = (a + √(a^{2} + 4b) )/2, k = (a - √(a^{2} + 4b) )/2, and it follows immediately that 0 < h < 1, 0 > k > -h.

Now given x, take p, q so that x = p + q, f(x) = ph + qk. It is easy to see that this can be done: p = (f(x) - kx)/(h - k), q = (hx - f(x) )/(h - k). It is also easy to see that denoting f^{2}(x) = f(f(x)), f^{3}(x) = f(f^{2}(x)) etc, we have f^{n}(x) = ph^{n} + qk^{n}. Indeed this must also be true for n negative, where f^{-1} denotes the inverse of f.

Now suppose f is increasing. Suppose for some particular x we have q non-zero. Then for n sufficiently large, qk^{-n} dominates ph^{-n}, so f^{ -n}(x) alternate in sign but increases in absolute value as n increases. For suitable N we will have f^{ -N-2}(x) > f^{ -N}(x) > 0, but f^{ -N-1}(x) < f^{ -N+1}(x) < 0. In other words A > B, but f(A) < f(B), contradicting the fact that f is increasing. So we must have q zero for *all* x. Hence f(x) = hx for all x.

Similarly, suppose f is decreasing. Suppose for some particular x we have p non-zero. Then for sufficiently large n, ph^{n} dominates qk^{n}, so f^{ n}(x) has the same sign, but decreases in absolute value as n increases. So either we have N with f^{ N-1}(x) > f^{ N}(x) > f^{ N+1}(x) > f^{ N+2}(x) > 0, in which case A > B but f(A) > f(B) (with A = f^{ N-1}(x), B = f^{ N+1}(x) ), or we have N with f^{ N-1}(x) < f^{ N}(x) < f^{ N+1}(x) < f^{ N+2}(x) < 0, in which case A < B but f(A) < f(B) (with A = f^{ N-1}(x), B = f^{ N+1}(x) ). In either case we contradict f decreasing. So if f is decreasing we must have p zero for *all* x and hence f(x) = kx for all x.

© John Scholes

jscholes@kalva.demon.co.uk

16 Dec 2001