Let f : [0, ∞)→R be a function such that (i) f (x) + f (y) ≤ f (x + y) , for all x ,y≥0 ;

3.3.4. Let f : [0, ∞)→R be a function such that (i) f (x) + f (y) ≤ f (x + y) , for all x ,y≥0 ;

(ii) there exists M >0 such that | f (x)| ≤ Mx , for any x ≥0 . Prove that the following properties hold: (a) if there exists a sequence (x n ) such that lim n → ∞ x n = +∞ and lim n → ∞

f (x n )/x n =ℓ , then f (x) ≤ ℓx , for all x >0 ;

(b) there exists lim x → ∞ f (x)/x .

Solution. (a) Let (x n ) be a sequence of real numbers such that lim n → ∞ x n = +∞ and lim n → ∞ f (x n )/x n = ℓ. For y > 0 arbitrarily chosen, there exist k n ∈ N and

3.3 Qualitative Results 127 z ∈ [0,y) such that x n =k n y + z. On the other hand, we obtain by induction that

f (nx) ≥ n f (x), for all n ∈ N and x ≥ 0. Hence

f (x n ) ≥ f (k n y ) + f (z) ≥ k n f (y) + f (z), that is,

Passing to the limit as n →∞ we obtain ℓ ≥ f (y)/y, for all y > 0. (b) Assume, by contradiction, that there are two sequences (x n ) and (y n ) such that

lim n → ∞ x n = lim n → ∞ y n = ∞, lim n → ∞ f (x n )/x n =ℓ 1 , and lim n → ∞ f (y n )/y n = ℓ 2 , with ℓ 1 2 . Then, by (a), ℓ 1 > f (y)/y, for all y > 0. It follows that in the neighborhood (ℓ 1 +ℓ 2 /2, ℓ 1 +ℓ 2 ) of ℓ 2 there exists no element of the form

f (y n )/y n , contradiction. This shows that ℓ 1 =ℓ 2 , so lim x → ∞ f (x)/x exists. ⊓ ⊔

3.3.5. (a) Construct a function f :R →[0,∞) such that any point x ∈Q is a local

strict minimum point of f .

(b) Construct a function f :Q →[0,∞) such that any point is a local strict minimum point and f is unbounded on any set of the form I ∩Q , where I is a nondegen- erate interval. (c) Let f :R →[0,∞)

be a function that is unbounded on any set of the form I ∩Q , where I is a nondegenerate interval. Prove that f does not have the property stated in (a).

Romanian Mathematical Olympiad, 2004 Solution. (a) Define f by f (x) = 1 if x ∈ R \ Q and f (x) = 1 − 1/p, provided

that x = n/p, (n, p) = 1, p > 0. We observe that x = n/p is a local strict minimum point because there exists a neighborhood of x not containing rational numbers of the form a /b with (a, b) = 1, b > 0 and b ∈ {1,2,..., p}.

(b) Consider the function f (n/p) = p. (c) We argue by contradiction and assume that such a function exists. We con-

struct a sequence of nondegenerate intervals as follows. Let I = [a 0 ,b 0 ] be a nondegenerate interval, and for given n ≥ 1, consider the intervals I k = [a k ,b k ], k = 0, 1, . . . , n − 1, such that I 0 ⊃I 1 ⊃··· ⊃I n −1 and, for any 0 ≤ k ≤ n − 1,

f (I k ) ⊂ [k,∞). Fix x n ∈Q∩I n −1 with f (x n ) ≥ n and consider a neighborhood

I n = [a n ,b n ]⊂I n −1 of x n such that for all x ∈I n ,f (x) ≥ f (x n ) ≥ n. Since (a n ) n ≥0 is nondecreasing, (b n ) n ≥0 is nonincreasing, and a n <b n , there exists c ∈∩ n ≥0 I n . Thus, for all positive integers n, f (c) ≥ n. This contradiction concludes our proof. ⊓ ⊔

3.3.6. Let f : (A, ∞)→R be a function satisfying lim x → ∞ [ f (x + 1) − f (x)] = +∞ and such which f is bounded on every bounded interval contained in (A, ∞) . Prove that lim x → ∞ f (x)/x = +∞ .

128 3 Limits of Functions Solution. Arguing by contradiction, there exists a sequence (x n ) n ≥1 and a real

number M (which can be supposed to be positive) such that

x n →∞ as n→∞ and f (x n ) ≤ Mx n , for all integers n ≥ 1.

On the other hand, since lim x → ∞ [ f (x + 1) − f (x)] = +∞, there exists a > A such that f (x + 1) − f (x) ≥ 2M, for all x ≥ a. For all n ≥ 1, write x n =a+k n +r n , where k n = [x n − a]. Thus,r n ∈ [0,1), and by our hypothesis, k n →∞ as n→∞. So, for all n ≥ 1,

Mx n − f (a) ≥ f (x n ) − f (a)

= ∑ [ f (x n − j + 1) − f (x n − j)] + f (a + r n ) − f (a)

j =0

≥ 2Mk n + f (a + r n ) − f (a) > 2Mx n − 2M(a + 1) + inf f

a (x) − f (a). ⊓ ≤x<a+1 ⊔ Taking n sufficiently large, the above inequality yields a contradiction.