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A FUNCTIONAL CALCULUS BASED ON THE NUMERICAL RANGE. APPLICATIONS MICHEL CROUZEIX ABSTRACT. We develop a functional calculus for both bounded and unbounded operators in Hilbert spaces based on a simple inequality related to polynomial functions of a square matrix and involving the numerical range. We present some applications in different areas of mathematics. 1. INTRODUCTION The starting point of this work is the following inequality sup u∗p(A)v ; u, v Cd, u∗u = v∗v = 1 12 sup p(v∗Av) ; v Cd, v∗v = 1 , {| | ∈ }≤ {| | ∈ } which holds for all square matrices A Cd,d, all polynomials p : C C, and all values of d. This inequality may be equivalently written∈ as → p(A) 12 sup p(z) , (1) k k≤ z∈W (A) | | where W (A)= v∗Av ; v Cd, v∗v = 1 is the numerical range of A. We refer to [5] for a proof of (1). { ∈ } Recall that the spectrum σ(A) of a matrix A is contained in its numerical range. Thus if a matrix A is normal we have p(A) = sup p(λ) ; λ σ(A) , and in this particular case inequality (1) is valid with 1 in placek of 12.k The great{| interest| ∈ of (1)} is to be valid for any square matrix. The author conjectures that, in this inequality, 12 may be replaced by 2. An outline of this paper is the following. In Section 2 we show how this inequality allows us to develop a holomorphic calculus for bounded and unbounded operators in a Hilbert space H. In Sections 3 and 4 we describe some applications in different areas of mathematics: second order differential equations, probability, numerical linear algebra, operator theory ... 2. FROM THE INEQUALITY TOWARDS A FUNCTIONAL CALCULUS We assume that the inequality (1) is valid for all square matrices A Cd,d, all polynomials ∈ p : C C, and all values of d. → 2.1. First step : extension to bounded operators in a Hilbert space. From the independence of the dimension we deduce : the inequality p(A) 12 sup p(z) k k≤ z∈W (A) | | holds for any bounded linear operator A (H) on any complex Hilbert space H and for any ∈ L polynomial p. 2000 Mathematics Subject Classification. 47A11,47A25. Key words and phrases. functional calculus, numerical range. 1 2 MICHEL CROUZEIX Proof. Indeed this is clearly true in the case of finite dimensional H. Otherwise let an operator A (H) and a polynomial p of degree n be given ; we consider a vector u H and define the Krylov∈ L subspace ∈ := Span (u, Au, . , Anu). K We set A˜ = Π A, where Π denotes the orthogonal projection from H to . Then we have K K K p(A)u = p(A˜)u and, on the finite dimensional space , the linear operator A˜ satisfies W (A˜) K ⊂ W (A). Thus p(A)u = p(A˜)u 12 u sup p(z) 12 u sup p(z) . k k k k≤ k k z∈W (A˜) | |≤ k k z∈W (A) | | The result follows, since u is arbitrary. 2.2. Second step : extension to rational functions. We now claim that the inequality r(A) 12 sup r(z) k k≤ z∈W (A) | | is valid for any bounded linear operator A (H) on a Hilbert space H and for any rational ∈ L function r, bounded on the numerical range W (A). Proof. Indeed such a rational function r is a uniform limit of polynomials on the bounded set W (A) (as is easily seen by writing the rational function in simple partial fraction form). 2.3. Third step : extension to unbounded operators. We now consider a closed, densely de- fined unbounded operator A : D(A) H H on the Hilbert space H. We assume that its ⊂ → spectrum is contained in the closure of the numerical range: σ(A) W (A). Then we claim that the inequality ⊂ r(A) 12 sup r(z) k k≤ z∈W (A) | | is valid for any rational function r, bounded on the numerical range W (A). Proof. From its boundedness it follows that the rational function r has its poles off W (A) ⊃ σ(A). Thus r(A) is well defined. Note that the numerical range is convex and unbounded for an unbounded operator A. Thus only 3 situations can occur : W (A)= C ; then the inequality is trivially satisfied, since the only bounded rational functions on•C are the constant ones. W (A) is a strip. • There exist θ R and β C such that (0, + ) W (eiθA + β) z C ; Re z 0 . • ∈ ∈ ∞ ⊂ ⊂ { ∈ ≥ } We only look at this last case. Without loss of generality, we can assume that θ = β = 0, and thus (0, + ) W (A) z C ; Re z 0 and A is maximal accretive. Then with ε > 0 we ∞ ⊂ −1 ⊂ { ∈ ≥ } set Aε := A(1+εA) . The operator Aε is bounded, and it is easily verified that W (Aε) W (A). We deduce from the previous step that ⊂ r(Aε) 12 sup r(z) 12 sup r(z) . k k≤ z∈W (Aε) | |≤ z∈W (A) | | Writing the rational function in simple partial fraction form we see that r(Aε) strongly converges to r(A) as ε 0, which shows the inequality. → A similar proof applies in the strip case, see [5]. A FUNCTIONAL CALCULUS BASED ON THE NUMERICAL RANGE. APPLICATIONS 3 2.4. Last step : extension to holomorphic functions. We now consider the algebra Cb(z) of rational functions, bounded on W (A), and the algebra (W (A)) of continuous and bounded Hb functions in W (A) which are holomorphic in the interior of W (A). We equip these two algebras with the norm f ∞,A := sup f(z) ; z W (A) . The following result shows the existence of a functional calculusk k [12] based{| on the| numerical∈ range.} Theorem 1. For any closed linear unbounded operator A such that σ(A) W (A), the homomor- ⊂ phism r r(A) from the algebra C (z), with norm . , into the algebra (H), is bounded 7→ b k k∞,A L with constant 12. It admits a unique bounded extension from the algebra (W (A)) into (H). Hb L This extension is bounded with constant 12 as well as completely bounded with constant 12. Sketch of proof. As in the previous step we only look at the case of maximal accretive A. First we consider a function f (W (A)) with lim f(z) = 0. Such a function f is a uniform limit ∈Hb z→∞ of rational functions rn on W (A), which allows us to define f(A) = lim rn(A). For a general function f (W (A)), we associate f and ε > 0 with f (z) := (1+εz)−1f(z) (note that ∈ Hb ε f ( ) = 0) and define f(A) = (1+εA)f (A). Then we have, if u D(A), ε ∞ ε ∈ f(A)u = (1+εA)f (A)u = f (A)(1+εA)u k k k ε k k ε k 12 f ( u + ε Au ) 12 f ( u + ε Au ). ≤ k εk∞,A k k k k ≤ k k∞,A k k k k Passing to the limit as ε 0 and using the density of D(A) in H we deduce f(A) (H) and → ∈L f(A) 12 sup f(z) . k k≤ z∈W (A) | | From the fact that the inequality (1) is still valid in tensorial form we conclude that the map is completely bounded (see [5] for more details). 3. AN APPLICATION TO SECOND ORDER DIFFERENTIAL EQUATIONS Let us consider two complex Hilbert spaces V H with V continuously and densely embedded ⊂ in H. We also consider a continuous sesquilinear form a( , ) on V V and assume that there · · × exist α> 0, λ, M, N R such that ∈ α v 2 Re a(v, v)+ λ v 2 M v 2 , k kV ≤ k kH ≤ k kV v V. (2) and Im a(v, v) N v v , ∀ ∈ ≤ k kV k kH To this form we associate a closed operator A defined by D(A) := u V ; there exists K such that a(u, v) K v , v V , { ∈ u | |≤ uk kH ∀ ∈ } and Au, v = a(u, v). h iH It is well known that D(A) is dense in H and σ(A) W (A). The assumption (2) clearly implies ⊂ α W (A) := x+iy ; x + λ y2 . ⊂ P { ≥ N 2 } Note that, if z belongs to the parabolic set , then Im z1/2 ω, with ω := max(λ, N/(2√α)). P | |≤ Thus the function cos(t√z) satisfies cos(t√z) eω|t| for all z and all t in R. Also this function is holomorphic in C, since the| cosine function| ≤ is even. Theorem∈ P 1 allows us to define C(t) := cos(t√A) and we have the following properties : • C(t) (H) and C(t) 12 eω|t| for all t R. ∈L k k≤ ∈ • C(0) = I and C(t+s) 2C(t)C(s)+ C(t s) = 0 for all t,s R. − − ∈ 4 MICHEL CROUZEIX • For all u H, u(t) := C(t) u satisfies u C(R; H), (strong continuity on H). 0 ∈ 0 ∈ • For all u D(A), u(t) := C(t) u satisfies u C(R; D(A)). 0 ∈ 0 ∈ Proof. Since the map f f(A) is an algebraic homomorphism, the two equalities are direct 7→ consequences of the well-known formulas cos 0 = 1 and cos(a+b) + cos(a b) = 2 cos a cos b. − Since we have C(t) 12 eω|t| and D(A) is dense in H, it suffices to show u(.) C(R; H) k k ≤ ∈ for all u D(A) in order to establish the strong continuity in H.

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