# teorema de Fuglede

Teorema de Fuglede em matemática, O teorema de Fuglede é um resultado da teoria dos operadores, named after Bent Fuglede.

Conteúdo 1 The result 2 Putnam's generalization 3 C*-algebras 4 References The result Theorem (Fuglede) Let T and N be bounded operators on a complex Hilbert space with N being normal. If TN = NT, then TN* = N*T, where N* denotes the adjoint of N.

Normality of N is necessary, as is seen by taking T=N. When T is self-adjoint, the claim is trivial regardless of whether N is normal: {displaystyle TN^{*}=(NT)^{*}=(TN)^{*}=N^{*}T.} Tentative Proof: If the underlying Hilbert space is finite-dimensional, the spectral theorem says that N is of the form {displaystyle N=sum _{eu}lambda _{eu}P_{eu}} where Pi are pairwise orthogonal projections. One expects that TN = NT if and only if TPi = PiT. Indeed it can be proved to be true by elementary arguments (por exemplo. it can be shown that all Pi are representable as polynomials of N and for this reason, if T commutes with N, it has to commute with Pi...). Therefore T must also commute with {estilo de exibição N^{*}=soma _{eu}{{bar {lambda }}_{eu}}P_{eu}.} No geral, when the Hilbert space is not finite-dimensional, the normal operator N gives rise to a projection-valued measure P on its spectrum, p(N), which assigns a projection PΩ to each Borel subset of σ(N). N can be expressed as {displaystyle N=int _{sigma (N)}lambda dP(lambda ).} Differently from the finite dimensional case, it is by no means obvious that TN = NT implies TPΩ = PΩT. Desta forma, it is not so obvious that T also commutes with any simple function of the form {displaystyle rho =sum _{eu}{bar {lambda }}P_{Ômega _{eu}}.} De fato, following the construction of the spectral decomposition for a bounded, normal, not self-adjoint, operator T, one sees that to verify that T commutes with {estilo de exibição P_{Ômega _{eu}}} , the most straightforward way is to assume that T commutes with both N and N*, giving rise to a vicious circle!

That is the relevance of Fuglede's theorem: The latter hypothesis is not really necessary.

Putnam's generalization The following contains Fuglede's result as a special case. The proof by Rosenblum pictured below is just that presented by Fuglede for his theorem when assuming N=M.

Teorema (Calvin Richard Putnam)[1] Let T, M, N be linear operators on a complex Hilbert space, and suppose that M and N are normal, T is bounded and MT = TN. Then M*T = TN*.

Primeira prova (Marvin Rosenblum): Por indução, the hypothesis implies that MkT = TNk for all k. Thus for any λ in {estilo de exibição mathbb {C} } , {estilo de exibição e^{{bar {lambda }}M}T=Te^{{bar {lambda }}N}.} Consider the function {estilo de exibição F(lambda )=e^{lambda M^{*}}Te^{-lambda N^{*}}.} This is equal to {estilo de exibição e^{lambda M^{*}}deixei[e^{-{bar {lambda }}M}Te^{{bar {lambda }}N}certo]e^{-lambda N^{*}}=U(lambda )TV(lambda )^{-1},} Onde {estilo de exibição U(lambda )=e^{lambda M^{*}-{bar {lambda }}M}} Porque {estilo de exibição M} is normal, e da mesma forma {estilo de exibição V(lambda )=e^{lambda N^{*}-{bar {lambda }}N}} . However we have {estilo de exibição U(lambda )^{*}=e^{{bar {lambda }}M-lambda M^{*}}=U(lambda )^{-1}} so U is unitary, and hence has norm 1 for all λ; the same is true for V(λ), assim {estilo de exibição |F(lambda )|leq |T| forall lambda .} So F is a bounded analytic vector-valued function, and is thus constant, and equal to F(0) = T. Considering the first-order terms in the expansion for small λ, we must have M*T = TN*.

The original paper of Fuglede appeared in 1950; it was extended to the form given above by Putnam in 1951.[1] The short proof given above was first published by Rosenblum in 1958; it is very elegant, but is less general than the original proof which also considered the case of unbounded operators. Another simple proof of Putnam's theorem is as follows: Segunda prova: Consider the matrices {displaystyle T'={começar{bmatriz}0&0\T&0end{bmatriz}}quadrilátero {texto{e}}quad N'={começar{bmatriz}N&0\0&Mend{bmatriz}}.} The operator N' is normal and, por suposição, T' N' = N' T' . By Fuglede's theorem, um tem {displaystyle T'(N')^{*}=(N')^{*}T'.} Comparing entries then gives the desired result.

From Putnam's generalization, one can deduce the following: Corollary If two normal operators M and N are similar, então eles são unitariamente equivalentes.

Prova: Suppose MS = SN where S is a bounded invertible operator. Putnam's result implies M*S = SN*, ou seja.

{estilo de exibição S^{-1}M^{*}S=N^{*}.} Take the adjoint of the above equation and we have {estilo de exibição S^{*}M(S^{-1})^{*}=N.} Então {estilo de exibição S^{*}M(S^{-1})^{*}=S^{-1}MSquad Rightarrow quad SS^{*}M(SS^{*})^{-1}=M.} Let S*=VR, with V a unitary (since S is invertible) and R the positive square root of SS*. As R is a limit of polynomials on SS*, the above implies that R commutes with M. It is also invertible. Então {displaystyle N=S^{*}M(S^{*})^{-1}=VRMR^{-1}V^{*}=VMV^{*}.} Corollary If M and N are normal operators, and MN = NM, then MN is also normal.

Prova: The argument invokes only Fuglede's theorem. One can directly compute {estilo de exibição (MN)(MN)^{*}=MN(NM)^{*}=MNM^{*}N^{*}.} By Fuglede, the above becomes {displaystyle =MM^{*}NN^{*}=M^{*}MN^{*}N.} But M and N are normal, assim {displaystyle =M^{*}N^{*}MN=(MN)^{*}MN.} C*-algebras The theorem can be rephrased as a statement about elements of C*-algebras.

Teorema (Fuglede-Putnam-Rosenblum) Let x, y be two normal elements of a C*-algebra A and z such that xz = zy. Then it follows that x* z = z y*.