Let f : [0, 1]→R be a differentiable function such that f ′ (0) < 2 < f ′ (1) .

9.2.8. Let f : [0, 1]→R be a differentiable function such that f ′ (0) < 2 < f ′ (1) .

Prove that there exists x 0 ∈ (0,1) such that f ′ (x 0 )=2 .

Solution. Consider the function g (x) = f (x) − 2x. We have g ′ (0) < 0 < g ′ (1), so g (x) < g(0) for x sufficiently close to 0 and g(x) < g(1) for x sufficiently close to 1. So, the minimum value of g in [0, 1] is achieved for some x 0 ∈ (0,1). Thus, by Fermat’s theorem, g ′ (x 0 ) = 0. It follows that f ′ (x 0 ) = 2. ⊓ ⊔

9.2.9. Prove that the following limit exists and is finite:

Solution. Let

It is enough to show that this function is increasing and bounded from below. Indeed, for any x

4 ≥x 4 +t ,t ≥ 0 we have (x + t) 4 . Hence " 1 dx

" 1 +t du

I (t) ≥ + lnt =

+ lnt = ln(1 + t) ≥ 0, ∀t ≥ 0. +t

Next, we prove that I ′ (t) ≥ 0. We first observe that " t

dx

" 1 dx

4 0 + t [(x/t) + 1] 1 /4 t t [(x/t) 4 + 1] 1 /4 + lnt. By the substitution y = x/t we obtain

I (t) =

" 1 dy

" 1 /t

dy

I (t) =

+ lnt.

0 (y 4 + 1) 1 /4

1 (y 4 + 1) 1 /4

9.2 Elementary Examples 333 Therefore

t 2 (1/t 4 1 /4

9.2.10. Let f : [0, a]→[0,∞)

be a continuous function such that f (0) = 0 and f has a right derivative and f d ′ (0) = 0 . Moreover, we assume that

∀t ∈ [0,a].

Prove that f ≡0 . Solution. Let

Then F is well defined because the function f (s)/s can be extended by continuity in s = 0, since lim s → 0 f (s)/s = f ′ (0) = 0. Furthermore, we have

≤ 0, ∀t ∈ [0,a].

Thus, the mapping F (t)/t is decreasing. Since F(0) = 0, we find that F(t) ≤ 0. But F (t) ≥ 0. Hence F ≡ 0, that is, f ≡ 0. ⊓ ⊔

9.2.11. Let f : [0, 1]→R

be an integrable function such that

f (x)dx =

xf (x)dx = 1.

Prove that # 1

0 f 2 (x)dx ≥ 4 . Romanian Mathematical Olympiad, 2004

Solution. A direct application of the Cauchy–Schwarz inequality implies " 1 " 1 " 1 2

f 2 (x)dx ·

x 2 dx ≥

xf (x)dx

so # 1

0 f 2 (x)dx ≥ 3. In order to find a stronger bound we start from # 1 2

0 ( f (x) + ax + b) dx ≥ 0, for any real numbers a and b. Therefore

f (x)dx ≥ − 2

−b − 2a − 2b − ab. We are interested in finding a and b such that E (a, b) = −a 2 /3−b 2 −2a−2b−ab

is maximum. For this purpose we first fix the real parameter b and consider the

334 9 Riemann Integrability

mapping ϕ (a) = −a 2 /3 − (b + 2)a − b 2 − 2b, which achieves its maximum for

a M = −3(b + 2)/2. Then ϕ (a M ) = 4 − (b − 2) 2 /4, which is maximum for b = 2. The above argument implies that the maximum of E (a, b) is 4 and it is attained for

a = −6 and b = 2, so the minimum is achieved for f (x) = 6x − 2. This concludes the proof. ⊓ ⊔

9.2.12. Fix real numbers a , b , c , d such that a and let g :

[a, b]→R continuous function satisfying b a # g (x)dx = 0 . Find a function f such that b 2

be a

a f (x)dx is a minimum subject to the conditions b a f (x)dx = c and b a f (x)g(x)dx = d .

J.R. Hatcher, AMM E 1104, 1954 Solution. For any real numbers λ and µ we have

[ f (x) − λ g (x) − µ ] 2 dx ≥ 0.

Set A # b =

a g 2 (x)dx. If A > 0 then the above inequality can be rewritten as " b d 2 c 2 (A λ − d) 2 [(b − a) µ − c] 2

f 2 (x)dx ≥

−a

b −a This implies that the required minimum is d 2 A −1 +c 2 (b − a) −1 , and it is achieved

if and only if f (x) = dA −1 g (x) + c(b − a) −1 .

If A = 0 (such a case is possible only if d = 0) then g = 0. The Cauchy–Schwarz inequality implies that the minimum of # b f 2 (x)dx is c 2 (b − a) −1

a , and it is attained if and only if f (x) = c(b − a) −1 . ⊓ ⊔