On a problem of nieminen

William Donoghue

doi:10.1007/BF02684290

Donoghue, William

2007-08-04 00:00:00

By WILLIAM F. DONOGHUE, Jr. In a recent publication [3] T. Nieminen raises the following question: If T is a bounded operator on a Hilbert space, the spectrum of which is a subset of [z[ = I and whose resolvent satisfies the inequality I]lZ,[I--<]]z]--I ]-~ on the resolvent set, does it follow that T is unitary? We show that it does. It is convenient to recall a definition: if T is a bounded operator on a Hilbert space, the set W(T) of complex numbers of the form (Tx, x) where ][x[] =I is called the numerical range of T and is a convex set contained in a circle of radius I IT I[ [4, i]. Our Theorem is then the following. Theorem. -- The following three classes of bounded operators on the Hilbert space d(f are identical: (I) The unitary operators. (II) The operators T for which (a) T -1 exists and is everywhere defined and []Txl]>l[x][ for all x in W and (b) W(T) is a subset of the unit disc. (III) The operators T for which (a') o is in the resolvent set and ][R 011<I and (b') for an unbounded sequence of numbers tn>i, llR.[l~(t,--I) -1 if iz[ =tn. Proof. -- It is easy to see that the unitary operators are in class III. We shall show first that operators in the class III are in class II, and then that the operators in II are unitary. It is obvious that conditions (a) and (a') are equivalent. If, now, T satisfies (a') and (b') and X = the ~~ we will have for anyy in Jt ~ [lYl] ~ (t~--1) -1 [[ (T--x)yl[ whence (t.-- I )21lyll2< ll Tyli2 + t~ llyil~--t,,[e'~ (Ty, y) -+-e-'~ (Ty, y) ]. Now if [ly[l=I and (Yy, y)=pe ~ we obtain 2t.[o cos (0--m)--i]< llYyll~- i. Since the sequence t, is unbounded and 0 arbitrary we infer that p< I, i.e. (b). It T satisfies (a), then, by a known theorem [2], T=VA where V is unitary and A is positive and also satisfies (a). It is clear that the spectrum of the self-adjoint A lies in the interval I<X<[]A[], hence we may write A=I+H where I is the identity operator and H is positive. Accordingly, T = V + VH and it is obvious that any operator of this form, where V is unitary and H is positive and bounded, satisfies (a). 127 WILLIAM F. DONOGHUE Let x be a normalized element of aft for which ] (Vx, x) ] = i --~ for some small which may be o. We will have (Tx, x) = (Vx, x) + (VHx, x) = (Vx, x) + (Hx, V-ix). Choose a normalized element y such that (x,y) -- o and V-Ix =- =x q- ~y ; evidently 0c = (Vx, x) and l el 2-t- I~ 15= I. Thus, from (b) I(Tx, x) l= [(Vx, x)[i -~- (Hx, x)] nt-~ (Hx, y)l<,. Let h = (Hx, x) ; from the Schwarz inequality we obtain [ (Hx, y)12< (Ux, x) (gy, y)<ll H [I h and we know that I ~1--<\/~-~ ; hence (I--~) (I q- h) < ~ + \/2 [1HII ~h. It follows that h We infer that vanishes if ~ does and that the ratio h/~ is bounded as e approaches o. there exists an M such that (Hx, x)<M[I--](Vx, x)1] for all normalized x. Let V= (te2"iXdE be the spectral representation of V; for an integer N let J0 x Pk= E~,,z~--EIk-1),N, k = i, 2, ..., N. Ifx is a normalized element in the range of Pk, then the number (Vx, x)= J0 (le2"~X d(E ~ x x , x) is evidently in the convex hull of the complex numbers on an arc of length 2rc/N of the unit circle, whence i-- t (Vx, x)t<~2/N ~. It will follow that liHxl[~ = (H2x, x)<ilHl[ Mn2/N 2 and therefore [IH+II <CIIxlI/N for all x in the range of Pk, the constant C being independent of N. If, now, Z is an arbitrary vector in the space and y = Hz, then, for such x, I(Y, x)[= I(Hz, x) l= [(z, Hx)[<l[zl[-liUxll<[lzl[ .[Ixl[c/N and for x----= PkY [l P yl[ 2= (y, Pky) ilzll. IIPky II C/N or II P YlI IIzlI C/N. Accordingly Ilyl[ 5 Ih PkYlI2 N Ilzll2C /Nz= Ilzll2C2/N- Since N is arbitrary, y = o, and since z is arbitrary, H = o and T =- V, whence T is unitary. If we recall that a contraction on a Hilbert space is any linear transformation 128 ON A PROBLEM OF NIEMINEN of bound at most I, we can state the essential part of our result in the following convenient form: Corollary. -- If S is a contraction on a Hilbert space which has a bounded inverse for which W(S -1) is a subset of the unit disc, then S is unitary. One might be led to the conjecture that any bounded operator T for which the resolvent satisfies the inequality [[ R~[I< I/d(z) where d(z) is the distance from z to the spectrum of T must be normal. That this is not the case is shown by the following simple counterexample. Let A be an operator of bound < I which is not normal; W(A) will be a subset of the unit disc and, by a known theorem, for ]z[>I the resolvent of A will be bounded by the reciprocal of the distance from z to W(A), a fortiori by ([ z [--I)-1. Let B be an operator which is normal and whose spectrum is exactly the unit disc. If o~ff is the direct sum of two Hilbert spaces Yt~x and ~r176 and T is defined as the operator A on ~~ 1 and B on o~ff~, then it is easy to see that the spectrum ofT is the unit disc, that T is not normal, and that IIR~ll<(lzI--I)-I on the resolvent set, where R, is the resolvent of T. We remark that the conjecture would be correct if the Hilbert space were finite dimensional. REFERENCES [I] DONOGHUE (W. F.), On the numerical range of a bounded operator, Michigan Mathematical aTournal, vol. 4 (1957), pp. 26x-263. [2] NAOY (B. v. Sz.), Spektraldarstellung linearer transformationen des Hilbertschen Raumes, Berlin, 1942. [3] NmMINEN (T.), A condition for the self-adjointness of an operator, Annales Academiae Fennicae, I962. [4] STONE (M. H.), Linear Transformations in Hilbert Space, Amer. Math. Soc. Colloquium Publications, 15 (x932). Refu le 15 janvier 1963.

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$By WILLIAM F. DONOGHUE, Jr. In a recent publication [3] T. Nieminen raises the following question: If T is a bounded operator on a Hilbert space, the spectrum of which is a subset of [z[ = I and whose resolvent satisfies the inequality I]lZ,[I--<]]z]--I ]-~ on the resolvent set, does it follow that T is unitary? We show that it does. It is convenient to recall a definition: if T is a bounded operator on a Hilbert space, the set W(T) of complex numbers of the form (Tx, x) where ][x[] =I is called the numerical range of T and is a convex set contained in a circle of radius I IT I[ [4, i]. Our Theorem is then the following. Theorem. -- The following three classes of bounded operators on the Hilbert space d(f are identical: (I) The unitary operators. (II) The operators T for which (a) T -1 exists and is everywhere defined and []Txl]>l[x][ for all x in W and (b) W(T) is a subset of the unit disc. (III) The operators T for which (a') o is in the resolvent set and ][R 011<I and (b') for an unbounded sequence of numbers tn>i, llR.[l~(t,--I) -1 if iz[ =tn. Proof. -- It is easy to see that the unitary operators are in class III. We shall show first that operators in the class III are in class II, and then that the operators in II are unitary. It is obvious that conditions (a) and (a') are equivalent. If, now, T satisfies (a') and (b') and X = the ~~ we will have for anyy in Jt ~ [lYl] ~ (t~--1) -1 [[ (T--x)yl[ whence (t.-- I )21lyll2< ll Tyli2 + t~ llyil~--t,,[e'~ (Ty, y) -+-e-'~ (Ty, y) ]. Now if [ly[l=I and (Yy, y)=pe ~ we obtain 2t.[o cos (0--m)--i]< llYyll~- i. Since the sequence t, is unbounded and 0 arbitrary we infer that p< I, i.e. (b). It T satisfies (a), then, by a known theorem [2], T=VA where V is unitary and A is positive and also satisfies (a). It is clear that the spectrum of the self-adjoint A lies in the interval I<X<[]A[], hence we may write A=I+H where I is the identity operator and H is positive. Accordingly, T = V + VH and it is obvious that any operator of this form, where V is unitary and H is positive and bounded, satisfies (a). 127 WILLIAM F. DONOGHUE Let x be a normalized element of aft for which ] (Vx, x) ] = i --~ for some small which may be o. We will have (Tx, x) = (Vx, x) + (VHx, x) = (Vx, x) + (Hx, V-ix). Choose a normalized element y such that (x,y) -- o and V-Ix =- =x q- ~y ; evidently 0c = (Vx, x) and l el 2-t- I~ 15= I. Thus, from (b) I(Tx, x) l= [(Vx, x)[i -~- (Hx, x)] nt-~ (Hx, y)l<,. Let h = (Hx, x) ; from the Schwarz inequality we obtain [ (Hx, y)12< (Ux, x) (gy, y)<ll H [I h and we know that I ~1--<\/~-~ ; hence (I--~) (I q- h) < ~ + \/2 [1HII ~h. It follows that h We infer that vanishes if ~ does and that the ratio h/~ is bounded as e approaches o. there exists an M such that (Hx, x)<M[I--](Vx, x)1] for all normalized x. Let V= (te2"iXdE be the spectral representation of V; for an integer N let J0 x Pk= E~,,z~--EIk-1),N, k = i, 2, ..., N. Ifx is a normalized element in the range of Pk, then the number (Vx, x)= J0 (le2"~X d(E ~ x x , x) is evidently in the convex hull of the complex numbers on an arc of length 2rc/N of the unit circle, whence i-- t (Vx, x)t<~2/N ~. It will follow that liHxl[~ = (H2x, x)<ilHl[ Mn2/N 2 and therefore [IH+II <CIIxlI/N for all x in the range of Pk, the constant C being independent of N. If, now, Z is an arbitrary vector in the space and y = Hz, then, for such x, I(Y, x)[= I(Hz, x) l= [(z, Hx)[<l[zl[-liUxll<[lzl[ .[Ixl[c/N and for x----= PkY [l P yl[ 2= (y, Pky) ilzll. IIPky II C/N or II P YlI IIzlI C/N. Accordingly Ilyl[ 5 Ih PkYlI2 N Ilzll2C /Nz= Ilzll2C2/N- Since N is arbitrary, y = o, and since z is arbitrary, H = o and T =- V, whence T is unitary. If we recall that a contraction on a Hilbert space is any linear transformation 128 ON A PROBLEM OF NIEMINEN of bound at most I, we can state the essential part of our result in the following convenient form: Corollary. -- If S is a contraction on a Hilbert space which has a bounded inverse for which W(S -1) is a subset of the unit disc, then S is unitary. One might be led to the conjecture that any bounded operator T for which the resolvent satisfies the inequality [[ R~[I< I/d(z) where d(z) is the distance from z to the spectrum of T must be normal. That this is not the case is shown by the following simple counterexample. Let A be an operator of bound < I which is not normal; W(A) will be a subset of the unit disc and, by a known theorem, for ]z[>I the resolvent of A will be bounded by the reciprocal of the distance from z to W(A), a fortiori by ([ z [--I)-1. Let B be an operator which is normal and whose spectrum is exactly the unit disc. If o~ff is the direct sum of two Hilbert spaces Yt~x and ~r176 and T is defined as the operator A on ~~ 1 and B on o~ff~, then it is easy to see that the spectrum ofT is the unit disc, that T is not normal, and that IIR~ll<(lzI--I)-I on the resolvent set, where R, is the resolvent of T. We remark that the conjecture would be correct if the Hilbert space were finite dimensional. REFERENCES [I] DONOGHUE (W. F.), On the numerical range of a bounded operator, Michigan Mathematical aTournal, vol. 4 (1957), pp. 26x-263. [2] NAOY (B. v. Sz.), Spektraldarstellung linearer transformationen des Hilbertschen Raumes, Berlin, 1942. [3] NmMINEN (T.), A condition for the self-adjointness of an operator, Annales Academiae Fennicae, I962. [4] STONE (M. H.), Linear Transformations in Hilbert Space, Amer. Math. Soc. Colloquium Publications, 15 (x932). Refu le 15 janvier 1963.$

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On a problem of nieminen

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