A P -function is a differentiable function f :R →R with a continuous derivative f ′ on R such that f (x + f ′ (x)) = f (x) for all x in R .

8.4.4. A P -function is a differentiable function f :R →R with a continuous derivative f ′ on R such that f (x + f ′ (x)) = f (x) for all x in R .

(i) Prove that the derivative of a P -function has at least one zero. (ii) Provide an example of a nonconstant P -function. (iii) Prove that a P -function whose derivative has at least two distinct zeros is constant.

D. Andrica and M. Piticari

8.4 Qualitative Results 321 Solution. (i) If f is a P-function, and f ′

theorem shows that f ′ vanishes at some point ξ between x and x +f ′ (x): 0 = f (x +

f ′ (x)) − f (x) = f ′ (x) f ′ ( ξ ). (ii) Try a nonconstant polynomial function f . Identification of coefficients forces

f (x) = −x 2 + px + q, where p and q are two arbitrarily fixed real numbers. This is not at all accidental. As shown in the comment that follows the solu- tion, every nonconstant P-function whose derivative vanishes at a single point is of this form.

(iii) Let f be a P-function. By (a), the set Z = {x : x ∈ R and f ′ (x) = 0} has at least one element. We now show that if it has more than one element, then it must be all of R. The conclusion will follow. The proof is broken into three steps.

1. If f ′ vanishes at some point a, then f ′ (x) ≥ 0 for x ≤ a, and f ′ (x) ≤ 0 for x ≥ a. The argument is essentially the same in both cases, so we deal only with the first one. We argue by reductio ad absurdum. Suppose f ′ (x 0 ) < 0 for some x 0 < a and let α = inf{x : x > x 0 and f ′ (x) = 0}; clearly, this infimum exists. By continuity of f ′ ,f ′ ( α ) = 0 and f ′ (x) < 0 for x 0 <x< α ; in partic- ular, f is strictly monotonic (decreasing) on (x 0 , α ). Consider further the contin- uous real-valued function g : x

S TEP

′ (x), x ∈ R, and note that g(x) < x for x 0 <x< α , and g ( α )= α . Since g ( α )= α >x 0 and g is continuous, g (x) > x 0

for x in (x 0 , α ), sufficiently close to α . Consequently, for any such x, x 0 < g(x) < x < α , and f (g(x)) = f (x), which contradicts the strict monotonicity of f on (x 0 , α ). S TEP

2. If f ′ vanishes at two points a and b, a < b, then f is constant on [a, b]. By Step 1, f ′ (x) ≥ 0 for x ≤ b and f ′ (x) ≤ 0 for x ≥ a, so f ′ vanishes identically on [a, b]. Consequently, f is constant on [a, b].

We are now in a position to conclude the proof. S TEP

3. If the set Z = {x : x ∈ R and f ′ (x) = 0} has more than one element, then Z is all of R. By Step 2, Z is a nondegenerate interval, and f is constant on Z :f (x) = c for all x in Z. We show that α = inf Z = −∞ and β = sup Z = +∞. Suppose, if possible, that α > −∞. Then α is a member of Z, by continuity of f ′ . Recall the function g from Step 1. By Step 1, f ′ (x) > 0 for x < α , so g (x) > x,

f (x) is strictly monotonic (increasing), and f (x) < c for x < α . Since f (x) is strictly monotonic for x < α , the conditions f (g(x)) = f (x) and g(x) > x force x < α < g(x). Since g( α )= α < β , and g is continuous, it follows that g (x) < β for x < α , sufficiently close to α . Finally, take any such x and recall that Z is an interval to conclude that g (x) ∈ Z, so f (x) = f (g(x)) = c, in contradiction to

f (x) < c established above. Consequently, α = −∞. A similar argument shows that β = +∞.

The antiderivative test for series, which we now state and prove, follows from the mean value theorem and the fact that a series of positive terms converges if and only if its sequence of partial sums is bounded above. ⊓ ⊔

322 8 Antiderivatives