f(x) is a real valued function on the reals, and has a continuous derivative. f '(x)^{2} + f(x)^{3} → 0 as x → ∞. Show that f(x) and f '(x) → 0 as x → ∞.

**Solution**

The key to getting started is to notice that if f ' = 0 for arbitrarily large values of x then the result is certainly true. Suppose f '(x_{n}) = 0 and x_{n} → ∞. Then since f '^{2} + f^{3} → 0, we have f(x_{n}) → 0. But f is monotonic on the interval [x_{n}, x_{n+1}} since its derivative does not change sign, hence f → 0. Hence also f ' → 0. So we may assume that for sufficiently large x, f ' does not change sign.

Now suppose f tends to a limit as x → ∞. Then f ' must also tend to a limit. If that limit is non-zero, then f increases or decreases faster than some non-constant linear function for sufficiently large x and so cannot tend to a limit. Hence f ' must tend to zero. Hence f also.

So we may assume that either (1) for sufficiently large x, f is strictly monotonic increasing and tends to infinity, or (2) for sufficiently large x, f is strictly monotonic decreasing and tends to minus infinity.

The first case is impossible, because then f^{3} and hence also f^{3} + f '^{2} would tend to infinity.

Showing that the second case is impossible needs a little more work. Suppose that for x ≥ X, we have f(x) < -1 and 1/4 > f(x)^{3} + f '(x)^{2} > -1/4. Then 1/2 f(x)^{3} < -1/2, so -1/2 f(x)^{3} > 1/2. Hence f '(x)^{2} + 1/2 f(x)^{3} > -1/4 - 1/2 f(x)^{3} > 1/4 > 0. So f '(x)^{2} > -1/2 f(x)^{3}. f '(x) is negative, so f '(x) < -1/2 |f(x)|^{3/2} (*).

Now define g(x) to satisfy g(X) = -1, g' = -1/2 |g|^{3/2}. Solving, we get g(x) = (1 - (x - X)/4 )^{-2} for x >= X. (*) shows that we must have f(x) < g(x) for x ≥ X. But g(x) → -∞ as x → 5X, so f must be discontinuous on the interval (X, 6X). Contradiction.

*Comment. I found this hard. Note that the exact form of the relation is important. If, for example, it was replaced by f '(x) ^{3} + f(x)^{2} → 0 as x → ∞, then the result would not be true - we could have f(x) = -x^{3}/27, in which case f '^{3} + f^{2} is identically zero.
*

© John Scholes

jscholes@kalva.demon.co.uk

18 Aug 2001