Find a continuous function f : [0, 1] → [0, 1] such that given any β ∈ [0, 1], we can find infinitely many α such that f(α) = β.
Solution
We use an adapted space-filling curve. Let g(x) be a piecewise linear function with period 2: g(x) = 0 on [0, 1/3], 1 on [2/3, 4/3] and 0 on [5/3, 2] with linear connecting pieces. For t any real in the range 0 ≤ t ≤ 1 let f(t) = g(t)/2 + g(9t)/4 + g(81t)/8 + g(729t)/16 + ... . The series is obviously uniformly convergent, so f is continuous.
Now suppose that in base 3 we have t = 0.c1c2c3 ... with all cn = 0 or 2. Then 9nt = even integer + 0.c2n+1c2n+2... . Now 0.c2n+1c2n+2... lies in [0, 1/3] if c2n+1 = 0 and in [2/3, 1] if c2n+1 = 2, so g(9nt) = 0 if c2n+1 = 0, 1 if c2n+1 = 2. Thus f(t) has the binary expansion 0.a1a2a3 ... , where an = c2n-1/2. So given any s in [0, 1] we can find infinitely many t with f(t) = s, for odd n pick cn to give the correct binary expansion and for even n pick cn arbitrarily.
© John Scholes
jscholes@kalva.demon.co.uk
15 Feb 2002