Rolle s Theorem for Polynomials of Degree Four in a Hilbert Space 1

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1 Journal of Mathematical Analysis and Applications 265, (2002) doi:0.006/jmaa , available online at on Rolle s Theorem for Polynomials of Degree Four in a Hilbert Space Jesús Ferrer Departamento de Análisis Matemático, Universidad de Valencia, Dr. Moliner 50, 4600 Burjasot, Valencia, Spain jesus.ferrer@uv.es Submitted by John Horváth Received November 7, 2000 In an infinite-dimensional real Hilbert space, we introduce a class of fourthdegree polynomials which do not satisfy Rolle s Theorem in the unit ball. Extending what happens in the finite-dimensional case, we show that every fourth-degree polynomial defined by a compact operator satisfies Rolle s Theorem Elsevier Science Key Words: Rolle s Theorem; Hilbert spaces; polynomials.. INTRODUCTION As is well known, Rolle s Theorem states that, if U is a bounded open subset of n and f U is a continuous function such that it is differentiable in every point of U having constant value on the boundary U, then there is a point x 0 U in which the Fréchet derivative f x 0 is zero. Since the proof of this theorem clearly relies on the compactness of the closure U, it is plain that a similar argument cannot be used in real Banach spaces of infinite dimension. As a matter of fact, a wide class of Banach spaces for which this result fails is introduced in [2, 3, 0]. In this paper we are interested in the study of Rolle s Theorem applied to continuous polynomials that vanish in the unit sphere of a real Hilbert space. Answering a question posed in [9], we gave a counterexample to this result in [6], although the function described was not a polynomial. We have also studied this problem for usual real Banach spaces such as c 0 l,orl, finding that the only continuous polynomial that vanishes in The author has been partially supported by Programa Sectorial, Dirección General de Estudios Superiores pro X/02 $ Elsevier Science All rights reserved. 322

2 rolle s theorem for polynomials of degree four 323 the unit sphere is the zero polynomial (see [7]). Concerning continuous polynomials in l,orl, it was already shown in [5, Corollary.5] that this was so; nevertheless we even see in [7] that the same can be said for any algebraic polynomial in c 0 or l. In what follows X will be a real Hilbert space, and B U, and S will stand for the closed unit ball, its interior and boundary, respectively. By a continuous polynomial in X we mean a real-valued function of the form n f x =u 0 + u j x j where u 0, u j denotes a symmetric continuous j-linear functional, and the symbol x j stands for x x x in X j j n. As we did in [8] one can easily prove that a polynomial vanishing in S can be factored as f x = x 2 g x where g is another polynomial which we called the associated factor of f. Also in [8], we proved that such a polynomial satisfies Rolle s Theorem when its associated factor is weakly continuous; hence every polynomial vanishing in S whose degree is less than four does satisfy Rolle s Theorem. S. A. Shkarin gave in [0] an example of a polynomial of degree four in L 2 0 for which the result failed, so that we know that Rolle s Theorem does not hold for continuous polynomials in general. Due to the interest recently shown in this counterexample (see [4]), in the next section we shall introduce a class of fourth-degree polynomials, containing Shkarin s counterexample, for which Rolle s Theorem does not hold. In the third section we give some sufficient conditions for polynomials of degree four that fulfill this theorem. j= 2. POLYNOMIALS NOT SATISFYING ROLLE S THEOREM The counterexample used by S. A. Shkarin in [0] was the following. In L 2 0, he considered the fourth-degree polynomial P x = x 2 Q x where Q x = Ax x +2 ϕ x and Ax t =tx t ϕ t =t t x L 2 0 and t 0 The class of polynomials that we introduce in the coming definition generalizes Shkarin s example.

3 324 jesús ferrer Definition. In a real Hilbert space X, by a Shkarin polynomial we mean a real-valued function of the form P x = x 2 Q x, with Q x = Ax x +2 ϕ x +k, where the following conditions hold:. A is a strictly positive operator in X, i.e., x 0 implies Ax x > For such an operator A, general notions of spectral theory tell us that, for λ, I λa is always an isomorphism and thus there is a vector x λ such that I λa x λ =λϕ. In this condition we assume that there is 0 <ρ< such that x λ ρ, λ<. 3. is not an eigenvalue, i.e., the operator I A is one to one. 4. ϕ is not in the range of I λa, λ>. Notice that this implies that the spectrum of A contains the interval 0, and so X has to be of infinite dimension. We assume in all of the following that ϕ 0; otherwise P 0 =0. Lemma. Let P x = x 2 Q x be a polynomial of degree four satisfying condition. Then, the real-valued function ϕ x λ is strictly increasing, while x λ decreases in 0 and increases strictly in 0. Proof. Making use of the implicit function theorem, we have that x λ defines a continuous and differentiable function from into X. d Hence, ϕ x λ is differentiable, and we show next that ϕ x λ > 0, dλ λ<. Taking the derivative in I A x λ =ϕ, λ<, λ 0, we obtain λ λ 2 x λ + ( λ I A )x λ =0 Hence, λ I A x λ = /λ 2 x λ, and, since I A is self-adjoint, we λ have d dλ ϕ x λ = ϕ x λ ( ) = λ I A x λ x λ ( ) = x λ λ I A x λ = x λ 2 λ 2 > 0 noticing that x λ =0 implies that λϕ = 0, which is not so. For the second part of our statement, it suffices to show the desired monotonicity for x λ 2.

4 rolle s theorem for polynomials of degree four 325 If λ<0, then d dλ x λ 2 = 2 x λ x λ = 2λ 2 λ I A x λ x λ = 2λ 2 x λ 2 /λ Ax λ x λ < 0, since x λ =0would imply x λ =0. If λ 0, then we can obtain the vectors x λ, x λ by means of the power series expansions and so x λ = λ n+ A n ϕ x λ = n + λ n A n ϕ n=0 d dλ x λ 2 = 2 which is clearly positive. m n=0 n=0 n + λ m+n+ A m+n ϕ ϕ Lemma 2. Let P x be a polynomial as in Lemma satisfying conditions and 2. Then ϕ belongs to both the range of A and the range of I A. Proof. Since, after condition 2, the set x λ λ< is bounded, we may find sequences λ j j= and λ 2j j= such that lim j λ j =, lim j λ 2j =, and the vector sequences x λ j j=, x λ 2j j= converge weakly in X to, say, ψ and φ, respectively. By taking weak limits in the equations ( ) I A x λ λ j =ϕ I λ 2j A x λ 2j =λ 2j ϕ j j we obtain Aψ = ϕ and the result now follows. ( I A ) φ = ϕ Notice that, since x λ ρ, λ<, then ψ ρ, φ ρ, too. Also, it is convenient to recall that, after Lemma, sup x λ 2 λ<0 = lim λ x λ 2 { r = sup x λ 2 0 <λ< } Notice also that φ 2 r. = lim x λ 2 λ Proposition. A sufficient condition for a Shkarin polynomial not to satisfy Rolle s Theorem is that Aψ ψ <k< φ 2 2r + Aφ φ

5 326 jesús ferrer Proof. Let P x = x 2 Q x be a Shkarin polynomial for which the above inequality holds. Since Q x =Q ψ +x ψ x ψ, x X, we have Q x Q ψ =k Aψ ψ > 0, x X. Assuming P x satisfied Rolle s Theorem, there would exist λ 0 x 0 U such that I λ 0 A x 0 = λ 0 ϕ λ 0 = x 0 2 Q x 0 Hence, after condition 4, it follows that λ 0 0, If we now consider the real-valued function h λ = x λ 2 + λq x λ. = 2 x λ 2 + λ ϕ x λ + k λ < we have that, after Lemma, h λ is strictly increasing in 0. Hence, doing some computations, ϕ φ +k sup h λ = lim h λ =2r + < 0<λ< λ and we take into account two possibilities: One. λ 0 0. Then, I λ 0A is an isomorphism, and so x 0 = x λ 0. Consequently, h λ 0 =, which contradicts the inequality obtained before. Two. λ 0 =. In this case, after condition 3, it follows that x 0 = φ, and we have also a contradiction. = φ 2 + Q φ = ϕ φ +k 2 φ 2 + 2r + ϕ φ +k Theorem. If P x = x 2 Q x is a Shkarin polynomial such that φ 2 = r, then the condition given before is also necessary. That is, P x does not satisfy Rolle s Theorem if and only if Aψ ψ <k< 3 φ 2 + Aφ φ

6 rolle s theorem for polynomials of degree four 327 Proof. Obviously, we need only show its necessity. Assuming that P x does not satisfy the theorem, then k Aψ ψ ; otherwise, 0 = Q ψ = P ψ and ψ ρ<. Now, if k< Aψ ψ, we consider as before the function h λ = x λ 2 + λq x λ = 2 x λ 2 + λ ϕ x λ + k λ < Hence, since after Lemma, ϕ x λ is increasing in lim h λ lim λ ϕ x λ + k λ λ = ϕ ψ +k = k Aψ ψ =+, we have Thus, since h 0 =0, continuity ensures that there is λ 0 < 0 such that h λ 0 =. This means that P x λ 0 = 0 and x λ 0 ρ<, a contradiction. Therefore, k> Aψ ψ. We show next the other inequality. The hypothesis φ 2 = r implies that, by defining x =φ h λ can be extended continuously to. Hence, h < ; otherwise there would be λ 0 0 such that h λ 0 =. This implies that P x λ 0 = 0 and x λ 0 ρ<, a contradiction. Now, ( ) >h = 2 φ 2 ϕ φ +k + = 3 φ 2 k Aφ φ + and the result follows. As a consequence of the previous results we find an easy way to generate counterexamples to Rolle s Theorem, clearly containing Shkarin s original one. Corollary. In L 2 0, let Ax t = tx t, ϕ t = t t, and Q x = Ax x +2 ϕ x +k. Then, the polynomial P x = x 2 Q x does not satisfy Rolle s Theorem if and only if 2 <k< 4. Proof. It is plain that A is a strictly positive operator with =. For λ<, we have that λt t x λ = x λ 2 /3 < λt It is also clear that I A is one to one, and, for λ>, we have that ϕ is not in the range of I λa; otherwise the function t t would be in λt L 2 0, which is not so. Hence, P x is a Shkarin polynomial. Now, the vectors ψ and φ obtained in Lemma 2 can easily be computed to be ψ t =t φ t =t

7 328 jesús ferrer Hence, making use of the Lebesgue Convergence Theorem, we have that, in L 2 0, and so lim x λ =φ λ r = lim λ x λ 2 = φ 2 Therefore, since Aψ ψ = 2, and 3 φ 2 + Aφ φ =/4, the result follows after Theorem. 3. POLYNOMIALS SATISFYING ROLLE S THEOREM The aim of this section is to give some sufficient conditions to obtain some classes of fourth-degree polynomials for which Rolle s Theorem holds. To start with, let us recall that, in general, a continuous polynomial of degree four vanishing in the unit sphere must have the form P x = x 2 Q x Q x = Ax x +2 ϕ x +k where k, ϕ X, and A is a non-zero bounded self-adjoint operator in X. Again, we assume ϕ 0. Calculating the derivative P x, one easily observes that P x will satisfy Rolle s Theorem if we find a solution λ x X for the system x = λ Ax + ϕ x 2 + λq x = x < () For λ <, we have that the vector x λ = n=0 λ n+ A n ϕ is a solution of the first equation, such that the real-valued functions ϕ x λ x λ 2 h λ = x λ 2 + λq x λ are all differentiable in the interval the first section, = 2 x λ 2 + λ ϕ x λ + k. Moreover, as we saw in d x λ 2 ϕ x λ = > 0 0 < λ < dλ λ 2 which implies that λ ϕ x λ > 0, 0 < λ <. Proposition 2. With the terminology set above, if k >, then P x satisfies Rolle s Theorem.

8 rolle s theorem for polynomials of degree four 329 Proof. If 0 < λ <, making use of the mean value theorem, we can find a value µ such that 0 < µ < λ and h λ =2 x λ 2 + λ2 µ 2 x µ 2 + kλ > kλ Hence, since k <, we have that h >. Thus, since h 0 =0, there k is a value λ 0, with the same sign as k, such that 0 < λ 0 < k h λ 0 = then, x λ 0 2 = 2 λ 0 ϕ x λ 0 kλ 0 < /2 < and λ 0 x λ 0 is a solution of system (). Proposition 3. If either or is an eigenvalue of the operator A, then the polynomial P x satisfies Rolle s Theorem. Proof. We assume that is an eigenvalue; the other part of the result can be shown in an analogous manner. We also assume that k ; otherwise the result follows after Proposition 2. Now, let z be a non-zero element of Ker I A. We consider two possibilities: One. There is λ 0 0 such that h λ 0 =. Then, x λ 0 2 = 2 λ 0 ϕ x λ 0 kλ 0 < 2 2 λ 0 ϕ x λ 0 < Hence, λ 0 x λ 0 is a solution of system () and we are through. Two. h λ <, 0 <λ<. Then, the set x λ 0 <λ< is bounded, and so we can find a sequence λ j j= in 0 and a vector φ X such that lim j λ j = weakly to φ in X. Thus, I A φ = and the sequence x λ j j= converges ϕ and φ 2 + Q φ = ϕ φ +k 2 φ 2 + We now consider the function h t = φ + tz 2 + Q φ + tz sup h λ j j t For t, since A φ + tz = φ + tz ϕ, and noticing that ( ϕ z = I A ) ( φ z = φ I A ) z = 0

9 330 jesús ferrer we have Q φ + tz =φ + tz φ+ tz +2 ϕ φ + tz +k = φ + tz 2 + ϕ φ +k Therefore, h t =2 φ + tz 2 + ϕ φ +k t Now, h t being continuous, h 0 =2 φ 2 + ϕ φ +k and lim t h t =+ imply the existence of t 0 0 such that h t 0 =. Clearly, since ϕ x λ is strictly increasing in 0, ϕ φ =lim j ϕ x λ j > 0, and we have φ + t 0 z 2 = ( ϕ φ 2 k ) ( ) ϕ φ 2 < 2 so that φ+ t 0z is a solution of system (). After recalling that a compact self-adjoint operator A always has or as an eigenvalue, we obtain the following result, which clearly extends the classical result for finite dimension. Corollary 2. Theorem. If A is a compact operator, then P x satisfies Rolle s Before finishing, this last result can also be shown, and even extended, by noticing that when A is compact the polynomial Q x = Ax x + 2 ϕ x +k is weakly continuous (see []), and we showed in [8] that every function of the form f x = x 2 g x, with g x weakly continuous, satisfies Rolle s Theorem. Hence, since every compact polynomial is weakly continuous [], it follows that Rolle s Theorem applies for polynomials of the type P x = x 2 Q x, provided Q x is compact. REFERENCES. R. M. Aron, C. Hervés, and M. Valdivia, Weakly continuous mappings on Banach spaces, J. Funct. Anal. 52, No. 2 (983), D. Azagra, J. Gómez, and J. A. Jaramillo, Rolle s theorem and the negligibility of points in infinite-dimensional Banach spaces, J. Math. Anal. Appl. 23 (997), D. Azagra and M. Jiménez-Sevilla, The failure of Rolle s theorem in infinite-dimensional Banach spaces, J. Funct. Anal. 82 (200), D. Azé and J.-B. Hiriart-Urruty, Sur an air de Rolle et Rolle, Res. Math. Ens. Sup. 3 4 (2000),

10 rolle s theorem for polynomials of degree four R. Deville, G. Godefroy, and V. Zizler, Smoothness and Renormings in Banach Spaces, Pitman Monographs and Surveys in Pure and Applied Mathematics, Vol. 64, Longman, Harlow, J. Ferrer, Rolle s theorem fails in l 2, Amer. Math. Monthly 03 (996), J. Ferrer, Real normed spaces whose unit sphere determines polynomials, J. Math Comp. Sci. (Math. Ser.) 9, No. (996), J. Ferrer, On Rolle s theorem in spaces of infinite dimension, Indian J. Math. 42, No. (2000), M. Furi and M. Martelli, A multidimensional version of Rolle s theorem, Amer. Math. Monthly 02 (995), S. A. Shkarin, On Rolle s theorem in infinite-dimensional Banach spaces, Mat. Zametki 5, No. 3 (992),