Введение
Конструкция в функциональном анализе, полезная для решения дифференциальных уравнений. Спектр линейного оператора, действующего в банаховом пространстве, является фундаментальным понятием функционального анализа. Спектр состоит из всех скаляров, для которых оператор не имеет ограниченного обратного в этом пространстве. Спектр имеет стандартное разложение на три части:
The spectrum of a linear operator that operates on a Banach space is a fundamental concept of functional analysis. The spectrum consists of all scalars such that the operator does not have a bounded inverse on The spectrum has a standard decomposition into three parts:
точечный спектр, состоящий из собственных значений оператора;
непрерывный спектр, состоящий из скаляров, которые не являются собственными значениями, но делают область значений оператора собственным плотным подмножеством пространства;
остаточный спектр, состоящий из всех остальных скаляров в спектре. Это разложение применимо к изучению дифференциальных уравнений и имеет применение во многих областях науки и техники. Хорошо известный пример из квантовой механики – объяснение дискретных спектральных линий и непрерывной полосы в свете, излучаемом возбужденными атомами водорода.
a continuous spectrum, consisting of the scalars that are not eigenvalues but make the range of a proper dense subset of the space;
a residual spectrum, consisting of all other scalars in the spectrum. This decomposition is relevant to the study of differential equations, and has applications to many branches of science and engineering. A well known example from quantum mechanics is the explanation for the discrete spectral lines and the continuous band in the light emitted by excited atoms of hydrogen.
Для операторов без ограничений
Спектр неограниченного оператора можно разделить на три части аналогично случаю с ограниченным оператором, но поскольку оператор не определен всюду, определения области определения, обратного оператора и т.п. становятся более сложными.
Оператор умножения
При наличии σ-конечного пространства с мерой (S, Σ, μ), рассмотрим пространство Банаха Lp(μ). Функция h: S → C называется существенно ограниченной, если h ограничена почти всюду по μ. Существенно ограниченная функция h индуцирует ограниченный оператор умножения Th на Lp(μ):
The operator norm of T is the essential supremum of h. The essential range of h is defined in the following way: a complex number λ is in the essential range of h if for all ε > 0, the preimage of the open ball Bε(λ) under h has strictly positive measure. We will show first that σ(Th) coincides with the essential range of h and then examine its various parts. If λ is not in the essential range of h, take ε > 0 such that h−1(Bε(λ)) has zero measure. The function g(s) = 1/(h(s) − λ) is bounded almost everywhere by 1/ε. The multiplication operator Tg satisfies 1=Tg · (Th − λ) = (Th − λ) · Tg = I. So λ does not lie in spectrum of Th. On the other hand, if λ lies in the essential range of h, consider the sequence of sets 1={Sn =
h^(−1)(B1/n(λ))}. Each Sn has positive measure. Let fn be the characteristic function of Sn. We can compute directly
This shows Th − λ is not bounded below, therefore not invertible. If λ is such that μ( h−1({λ})) > 0, then λ lies in the point spectrum of Th as follows. Let f be the characteristic function of the measurable set h−1(λ), then by considering two cases, we find
so λ is an eigenvalue of Th. Any λ in the essential range of h that does not have a positive measure preimage is in the continuous spectrum of Th. To show this, we must show that Th − λ has dense range. Given f ∈ L^(p)(μ), again we consider the sequence of sets 1={Sn = h^(−1)(B1/n(λ))}. Let gn be the characteristic function of S − Sn. Define
Direct calculation shows that fn ∈ Lp(μ), with Then by the dominated convergence theorem,
in the Lp(μ) norm. Therefore, multiplication operators have no residual spectrum. In particular, by the spectral theorem, normal operators on a Hilbert space have no residual spectrum.
Норма оператора T является существенным супремумом h. Существенный диапазон h определяется следующим образом: комплексное число λ принадлежит существенному диапазону h, если для всех ε > 0, прообраз открытого шара Bε(λ) под h имеет строго положительную меру. Мы сначала покажем, что σ(Th) совпадает с существенным диапазоном h, а затем рассмотрим его различные части. Если λ не принадлежит существенному диапазону h, выберем ε > 0 таким образом, чтобы h⁻¹(Bε(λ)) имела нулевую меру. Функция g(s) = 1/(h(s) − λ) ограничена почти всюду по μ величиной 1/ε. Оператор умножения Tg удовлетворяет 1 = Tg · (Th − λ) = (Th − λ) · Tg = I. Следовательно, λ не лежит в спектре Th. С другой стороны, если λ принадлежит существенному диапазону h, рассмотрим последовательность множеств {Sn = h⁻¹(B1/n(λ))}. Каждое Sn имеет положительную меру. Пусть fn — характеристическая функция множества Sn. Мы можем непосредственно вычислить
The operator norm of T is the essential supremum of h. The essential range of h is defined in the following way: a complex number λ is in the essential range of h if for all ε > 0, the preimage of the open ball Bε(λ) under h has strictly positive measure. We will show first that σ(Th) coincides with the essential range of h and then examine its various parts. If λ is not in the essential range of h, take ε > 0 such that h−1(Bε(λ)) has zero measure. The function g(s) = 1/(h(s) − λ) is bounded almost everywhere by 1/ε. The multiplication operator Tg satisfies 1=Tg · (Th − λ) = (Th − λ) · Tg = I. So λ does not lie in spectrum of Th. On the other hand, if λ lies in the essential range of h, consider the sequence of sets 1={Sn =
h^(−1)(B1/n(λ))}. Each Sn has positive measure. Let fn be the characteristic function of Sn. We can compute directly
This shows Th − λ is not bounded below, therefore not invertible. If λ is such that μ( h−1({λ})) > 0, then λ lies in the point spectrum of Th as follows. Let f be the characteristic function of the measurable set h−1(λ), then by considering two cases, we find
so λ is an eigenvalue of Th. Any λ in the essential range of h that does not have a positive measure preimage is in the continuous spectrum of Th. To show this, we must show that Th − λ has dense range. Given f ∈ L^(p)(μ), again we consider the sequence of sets 1={Sn = h^(−1)(B1/n(λ))}. Let gn be the characteristic function of S − Sn. Define
Direct calculation shows that fn ∈ Lp(μ), with Then by the dominated convergence theorem,
in the Lp(μ) norm. Therefore, multiplication operators have no residual spectrum. In particular, by the spectral theorem, normal operators on a Hilbert space have no residual spectrum.
Это показывает, что Th − λ не ограничено снизу, следовательно, не обратимо. Если λ таков, что μ(h⁻¹({λ})) > 0, то λ принадлежит точечному спектру Th следующим образом. Пусть f — характеристическая функция измеримого множества h⁻¹(λ), тогда, рассматривая два случая, мы находим, что λ является собственным значением Th. Любое λ из существенного диапазона h, которое не имеет прообраза положительной меры, принадлежит непрерывному спектру Th. Чтобы это показать, необходимо доказать, что Th − λ имеет плотный образ. Пусть f ∈ Lp(μ), снова рассмотрим последовательность множеств {Sn = h⁻¹(B1/n(λ))}. Пусть gn — характеристическая функция множества S − Sn. Определим
The operator norm of T is the essential supremum of h. The essential range of h is defined in the following way: a complex number λ is in the essential range of h if for all ε > 0, the preimage of the open ball Bε(λ) under h has strictly positive measure. We will show first that σ(Th) coincides with the essential range of h and then examine its various parts. If λ is not in the essential range of h, take ε > 0 such that h−1(Bε(λ)) has zero measure. The function g(s) = 1/(h(s) − λ) is bounded almost everywhere by 1/ε. The multiplication operator Tg satisfies 1=Tg · (Th − λ) = (Th − λ) · Tg = I. So λ does not lie in spectrum of Th. On the other hand, if λ lies in the essential range of h, consider the sequence of sets 1={Sn =
h^(−1)(B1/n(λ))}. Each Sn has positive measure. Let fn be the characteristic function of Sn. We can compute directly
This shows Th − λ is not bounded below, therefore not invertible. If λ is such that μ( h−1({λ})) > 0, then λ lies in the point spectrum of Th as follows. Let f be the characteristic function of the measurable set h−1(λ), then by considering two cases, we find
so λ is an eigenvalue of Th. Any λ in the essential range of h that does not have a positive measure preimage is in the continuous spectrum of Th. To show this, we must show that Th − λ has dense range. Given f ∈ L^(p)(μ), again we consider the sequence of sets 1={Sn = h^(−1)(B1/n(λ))}. Let gn be the characteristic function of S − Sn. Define
Direct calculation shows that fn ∈ Lp(μ), with Then by the dominated convergence theorem,
in the Lp(μ) norm. Therefore, multiplication operators have no residual spectrum. In particular, by the spectral theorem, normal operators on a Hilbert space have no residual spectrum.
Прямой расчет показывает, что fn ∈ Lp(μ), причём по доминирующей теореме сходимости,
The operator norm of T is the essential supremum of h. The essential range of h is defined in the following way: a complex number λ is in the essential range of h if for all ε > 0, the preimage of the open ball Bε(λ) under h has strictly positive measure. We will show first that σ(Th) coincides with the essential range of h and then examine its various parts. If λ is not in the essential range of h, take ε > 0 such that h−1(Bε(λ)) has zero measure. The function g(s) = 1/(h(s) − λ) is bounded almost everywhere by 1/ε. The multiplication operator Tg satisfies 1=Tg · (Th − λ) = (Th − λ) · Tg = I. So λ does not lie in spectrum of Th. On the other hand, if λ lies in the essential range of h, consider the sequence of sets 1={Sn =
h^(−1)(B1/n(λ))}. Each Sn has positive measure. Let fn be the characteristic function of Sn. We can compute directly
This shows Th − λ is not bounded below, therefore not invertible. If λ is such that μ( h−1({λ})) > 0, then λ lies in the point spectrum of Th as follows. Let f be the characteristic function of the measurable set h−1(λ), then by considering two cases, we find
so λ is an eigenvalue of Th. Any λ in the essential range of h that does not have a positive measure preimage is in the continuous spectrum of Th. To show this, we must show that Th − λ has dense range. Given f ∈ L^(p)(μ), again we consider the sequence of sets 1={Sn = h^(−1)(B1/n(λ))}. Let gn be the characteristic function of S − Sn. Define
Direct calculation shows that fn ∈ Lp(μ), with Then by the dominated convergence theorem,
in the Lp(μ) norm. Therefore, multiplication operators have no residual spectrum. In particular, by the spectral theorem, normal operators on a Hilbert space have no residual spectrum.
в норме Lp(μ). Следовательно, операторы умножения не имеют остаточного спектра. В частности, по спектральной теореме, нормальные операторы на гильбертовом пространстве не имеют остаточного спектра.
The operator norm of T is the essential supremum of h. The essential range of h is defined in the following way: a complex number λ is in the essential range of h if for all ε > 0, the preimage of the open ball Bε(λ) under h has strictly positive measure. We will show first that σ(Th) coincides with the essential range of h and then examine its various parts. If λ is not in the essential range of h, take ε > 0 such that h−1(Bε(λ)) has zero measure. The function g(s) = 1/(h(s) − λ) is bounded almost everywhere by 1/ε. The multiplication operator Tg satisfies 1=Tg · (Th − λ) = (Th − λ) · Tg = I. So λ does not lie in spectrum of Th. On the other hand, if λ lies in the essential range of h, consider the sequence of sets 1={Sn =
h^(−1)(B1/n(λ))}. Each Sn has positive measure. Let fn be the characteristic function of Sn. We can compute directly
This shows Th − λ is not bounded below, therefore not invertible. If λ is such that μ( h−1({λ})) > 0, then λ lies in the point spectrum of Th as follows. Let f be the characteristic function of the measurable set h−1(λ), then by considering two cases, we find
so λ is an eigenvalue of Th. Any λ in the essential range of h that does not have a positive measure preimage is in the continuous spectrum of Th. To show this, we must show that Th − λ has dense range. Given f ∈ L^(p)(μ), again we consider the sequence of sets 1={Sn = h^(−1)(B1/n(λ))}. Let gn be the characteristic function of S − Sn. Define
Direct calculation shows that fn ∈ Lp(μ), with Then by the dominated convergence theorem,
in the Lp(μ) norm. Therefore, multiplication operators have no residual spectrum. In particular, by the spectral theorem, normal operators on a Hilbert space have no residual spectrum.
Самодобавленные операторы в гильбертовом пространстве
Пространства Гильберта являются пространствами Банаха, поэтому вышеизложенное обсуждение применимо и к ограниченным операторам на пространствах Гильберта. Тонкий момент касается спектра T*. Для пространства Банаха T* обозначает транспонированный оператор и σ(T*) = σ(T). Для гильбертова пространства T* обычно обозначает сопряженный оператор T ∈ B(H), а не транспонированный, и σ(T*) – это не σ(T), а его образ при комплексном сопряжении. Для самосопряженного оператора T ∈ B(H) борелевское функциональное исчисление предоставляет дополнительные способы естественного разбиения спектра.