Fallbrook Epsilon Academy - Monthly Problem 11402

11402. Let f:[0,1] -> [0,∞) be a continuous function such that f(0) = f(1) = 0 and f(x) > 0 for 0 < x < 1. Show that there exists a square with two vertices in the interval (0,1) on the x-axis and the other two vertices on the graph of f.

Solution by the Fallbrook Epsilon Academy Problem Group, Fallbrook, CA. Extend f continuously to [0,∞) by defining f(x) = 0 for x > 1. Consider the functions 


	g(x) = x + f(x)

	h(x) = f(x) - f(g(x)) [ = f(x) - f(x + f(x))]



both also defined and continuous on [0,∞).  We seek a point s in (0,1) with h(s) = 0, for then the vertices


(s,0)

(s+f(s),0)

(s+f(s),f(s+f(s)))

(s,f(s))



form the desired square, with all sides equal to f(s).  Note that the square's right side x-coordinates are
still in (0,1) because f(x) = 0 and not f(s) > 0 beyond this interval.




Since f is continuous on [0,1] it takes its (positive) maximum there at some point M which is in (0,1)
because of the conditions on f.  Since this is also a maximum of the extended f we have f(M) ≥ f(M+f(M)),
so h(M) ≥ 0.
On the other hand, since g(0) = 0, g(1) = 1 and g is continuous, the connected set g([0,1]) contains
(at least) the interval [0,1],
so there is an m (necessarily < M) in (0,1) with g(m) = M.  Then f(m) ≤ f(M) = f(g(m)),
so h(m) ≤ 0.  By the Intermediate Value Theorem, h(s) = 0 for some 0 < m ≤ s ≤ M < 1.