Let 0 ≤a≤1 . Find all continuous functions f : [0, 1]→[0,∞) such that
9.6.5. Let 0 ≤a≤1 . Find all continuous functions f : [0, 1]→[0,∞) such that
f (x)dx = 1,
xf (x)dx = a,
x 2 f (x)dx = a 2 .
Solution. Let f be such a function. Applying the Cauchy–Schwarz inequality, we have
a = xf
x (x)dx ≤ 2 f (x)dx
f (x)dx ≤ a.
So, all the above inequalities are, in fact, equalities. Taking into account the condi- tions that ensure equality in the Cauchy–Schwarz inequality, we deduce that there exists a nonnegative constant C such that
x f (x) = C f (x), ∀x ∈ [0,1]. Hence # 1 f (x) ≡ 0, which contradicts
0 f (x)dx = 1. In conclusion, there is no func- tion satisfying our hypotheses. ⊓ ⊔ Alternative proof. Multiplying the three equalities in our statement by α 2 , −2 α ,
resp. 1, we obtain " 1
f (x)( α − x) 2 dx = 0, ∀ α ∈ R.
This implies f # 1 ≡ 0, which contradicts
0 f (x)dx = 1. Thus, there does not exist such
a function.
(0) = 1 and such that lim
9.6.6. Let f : [0, ∞)→[0,∞)
be a continuous function satisfying f
t → ∞ x f (s)ds ≤ f (x) , for all x ≥0 . Find the maximum value of f (x) , for x >0 .
354 9 Riemann Integrability Solution. The maximum value of f
(x) is 1 if 0 ≤ x ≤ 1 and e if x > 1. Let F(x) = lim
1 −x
t → ∞ x f (s)ds. By hypothesis we deduce that F(0) ≤ 1 and
F (x) ≤ −F ′ (x), for all x. Hence
(ln F(x)) ′
(x)
F (x) ≤ −1, ∀x ∈ R. By integration from 0 to x we obtain ln (F(x)/F(0)) ≤ −x. Since F(0) ≤ 1, it
follows that F (x) ≤ e −x , for all x > 0. Using now the monotony of f , we obtain
f (x) ≤ 1, for any x ≥ 0. We also have " x
f (x) ≤
f (t)dt ≤ F(x − 1) ≤ e 1 −x , ∀x > 1.
x −1
To prove that this bound is achieved, fix x ≥ 0. For any x ∈ [0,1], define
1 if 0 ≤ t ≤ 1,
(t) =
0 if t > 1. The function f ∗ satisfies our assumptions and f ∗ (x) = 1. For any x > 1, define
⎪ −t ⎨ e if 0 ≤ t < x − 1,
f ∗ (t) = e 1 −x if x − 1 ≤ t ≤ x,
⎩ ⎪ 0 if t >x. Then f ∗ satisfies our hypotheses and f ∗ (x) = e 1 −x . ⊓ ⊔
be a continuous function such that the limits (finite or infinite) f (−∞) = lim t → −∞ f (t) and f (+ ∞) = lim t → +∞ f (t) exist, and f (+ ∞) < f (t) < f (−∞) , for all t ∈R . Prove that there exists a function u satisfying u ′′ + f (u) − g(x) = 0 in (0, 1) and u ′ (0) = u ′ (1) = 0 if and only if
9.6.7. Assume that g : (0, 1)→R is a continuous function and let f :R →R
f (+ ∞) <
g (x)dx < f (−∞).
Ph. Korman, SIAM Review, Problem 00-001 Solution. (Michael Renardy) The necessity of the condition follows by integrat-
ing the differential equation on (0, 1). We obtain that if such a function u exists, then
f (+ ∞) <
f (u(x))dx =
g (x)dx < f (−∞).
To prove the sufficiency, consider the initial value problem
u ′′ + f (u) − g(x) = 0, u(0) = a, u ′ (0) = 0.
9.6 Qualitative Results 355 We obtain that if a →∞,
" 1 " 1 " 1 u ′ (1) = u ′′ (x)dx =
[g(x) − f (u(x))]dx −→
g (x)dx − f (+∞).
Similarly, as a → − ∞,
u ′ (1) −→
g (x)dx − f (−∞). ⊓ ⊔
Using the hypothesis in combination with the intermediate value property, we obtain
a value of a ∈ R for which u ′ (1) = 0.