RADU GOLOGAN, CEZAR LUPU. Abstract. We consider a generalization of an
olympiad problem which can be regarded as a result for a Volterra operator. 1.
AN OLYMPIAD PROBLEM: ZEROES OF FUNCTIONS IN THE IMAGE OF A VOLTERRA OPERATOR RADU GOLOGAN, CEZAR LUPU Abstract. We consider a generalization of an olympiad problem which can be regarded as a result for a Volterra operator.
1. Introduction & Main result One of the problems of the Romanian National Olympiad in 2006 was the following Let f : [0, 1] → R be a continuous function with Z 1 f (x)dx = 0. 0
Show that there exists c ∈ (0, 1) such that Z c xf (x)dx = 0. 0
In what follows we give two proofs to this problem. In the second proof we shall use a mean value theorem due to Flett. For more details we recommend [2] and [3]. Z t First proof. We assume by contradiction that xf (x)dx 6= 0, ∀t ∈ (0, 1). With0 Z t Z t out loss of generality, let xf (x)dx > 0, ∀t ∈ (0, 1) and let F (t) = f (x)dx. 0
Integrating by part, we obtain Z t Z t 0< xf (x)dx = tF (t) − F (x)dx, ∀t ∈ (0, 1). 0
0
0
Now, by passing to the limit when t → 1, and taking into account that F (1) = 0, we deduce that Z 1 F (x)dx ≤ 0.(∗) 0
Key words and phrases. mean value theorem, integrals. 2000 Mathematics Subject Classification. 26A24, 26A33. 1
2
RADU GOLOGAN, CEZAR LUPU
Now, we consider the differentiable function, H : [0, 1] → R defined by Z t F (x)dx 0 H(t) = , if t 6= 0 . t 0, if t = 0 Z t F (x)dx tF (t) − 0 0 It is easy to see H (t) = > 0, so µ is increasing on the interval t2 (0, 1), so it is increasing on the interval [0, 1] (by continuity argument). Because H(0) = 0, it follows that Z 1 F (x)dx > 0, 0
which is in contradiction with (∗). So, there exists c ∈ (0, 1) such that Z c xf (x)dx = 0. 0
Second proof. We consider the following differentiable function H : [0, 1] → R, defined by Z t Z t H(t) = t f (x)dx − xf (x)dx 0 0
Z
with H (t) =
0
t 0
Z
0
f (x)dx. It is clear that H (0) = H (1) = 0
1
f (x)dx = 0. Applying 0
Flett’s mean value theorem (see [1]), there exists c ∈ (0, 1) such that H(c) − H(0) c Z c Z c Z c f (x)dx − xf (x)dx f (x)dx = c c H0 (c) =
or
0
0
0
which is equivalent to Z
c
xf (x)dx = 0. 0
An extension of theorem 1.1 was given in [4], namely Theorem 1.1. Let f, g : [0, 1] → R be two continuous functions. There exists c ∈ (0, 1) such that Z 1 Z c Z 1 Z c f (x)dx xg(x)dx = g(x)dx xf (x)dx. 0
0
0
0
AN OLYMPIAD PROBLEM: ZEROES OF FUNCTIONS IN THE IMAGE OF A VOLTERRA OPERATOR 3
The proof is almost the same with the second proof of theorem 1.1, only this time ˜ : [0, 1] → R defined by we shall consider the function H ˜ H(t) =
Z
1
Z t Z Z t f (x)dx t g(x)dx − xg(x)dx −
0
0
0
1
Z t Z t g(x)dx t f (x)dx − xf (x)dx .
0
0
0
The proof of the main result involves some non-elementary facts. The following lemma will be used. Lemma 1.2. Let h : [0, 1] → R be a continuous function and φ : [0, 1] → R is nondecreasing, continuous in 0 and φ(0) = 0. Then Z t h(x)φ(x)dx 0 lim = 0. t→0+ φ(t) Z t h(x)φ(x)dx 6= 0. Thus, there Proof. We assume by contradiction that lim+ 0 t→0 φ(t) Z tn h(x)φ(x)dx 0 exists a sequence tn > 0 such that ≥ c > 0, which is equivalent to φ(tn ) Z tn h(x)φ(x)dx ≥ cφ(tn ) > 0. 0
On the other hand, using the continuity and the fact that φ is nondecreasing we obtain Z tn h(x)φ(x)dx ≤ tn φ(tn ) 0< 0
and by letting tn → 0 we have a contradiction. We are now able to state the general form of our intermediate value result. Theorem 1.3. Let f, g, φ : [0, 1] → R such that f, g are continuous functions and φ is nondecreasing, continuous in 0 and φ(0) = 0. Then there exists c ∈ (0, 1) such that Z Z Z Z 1
c
f (x)dx 0
˜ = Proof. Let H(t)
1
g(x)φ(x)dx = 0
Z
c
g(x)dx 0
f (x)φ(x)dx. 0
t
h(x)φ(x)dx, where h : [0, 1] → R is a continuous function. 0
˜ H(t) = 0. Integrating by parts in the t→0 φ(t) Riemann-Stieltjes integral setting, we have Z 1 Z 1 Z 1 ˜ h(x)φ(x) H(x) 1 1 ˜ h(x)dx = dx = | − H(x)d = φ(x) φ(x) φ By the preeceding lemma we have lim+
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RADU GOLOGAN, CEZAR LUPU
Z 1 ˜ ˜ H(1) H() 1 ˜ = − − H(x)d . φ(1) φ() φ Z 1 h(x)dx = 0, we get Now, by letting → 0, if we assume that 0
Z 1 ˜ H(1) 1 ˜ = H(x)d . φ(1) φ 0 ˜ This implies that the function H(x) cannot be of constant sign on (0, 1). Thus ˜ there is c ∈ (0, 1) such that H(c) = 0. In the particular case when Z 1 Z 1 h(t) = f (t) g(x)φ(x)dx − g(t) f (x)φ(x)dx, 0
Z
0
1
we clearly have h(x)dx = 0, so by the considerations above there exists c ∈ (0, 1) 0 Z c such that h(x)φ(x)dx = 0 which is equivalent to 0 1
Z
Z
Z
0
1
Z
0
c
f (x)φ(x)dx.
g(x)dx
g(x)φ(x)dx =
f (x)dx 0
c
0
To formulate a consequence, denote by C([0, 1]) the Banach space of continuous functions on [0, 1] and by Cnull the subspace of functions having zero integral. Theorem 1.4. Let φ : [0, 1] → R a nondecreasing function continuous at 0 and such that φ(0) = Z0, and consider the Volterra operator Vφ : C([0, 1]) → C([0, 1]) given by x
φ(t)f (t)dt. Then, all functions in Vφ (Cnull ) have at least one zero in
Vφ (f )(x) = 0
(0, 1). References [1] T.M. Flett, A mean value problem, Mathematical Gazette 42(1958), 38–39. [2] T.L. R˘ adulescu, V.D. R˘ adulescu, T. Andreescu, Problems in Real Analysis: asvanced calculus on the real axis, Springer Verlag, 2009. [3] T. Lupu, Probleme de Analiz˘ a Matematic˘ a: Calcul Integral, GIL Publishing House, 1996. [4] C. Lupu, T. Lupu Problem 11290, American Mathematical Monthly, no. 4/2007. Polytechnic University of Bucharest, Department of Mathematics II, Str. Splaiul Independentei 313, RO–060042 and Insitute of Mathematics ”Simon Stoilow” of the Romanian Academy, Str. Calea Grivitei nr. 21, RO–014700, Romania E–mail address:
[email protected] University of Bucharest, Faculty of Mathematics, Str. Academiei 14, RO–70109, Bucharest and University of Craiova, Faculty of Mathematics, Str. A.I. Cuza 10, RO–200585, Craiova, Romania E–mail address:
[email protected],
[email protected]
