Question

Let \(T \in \mathcal{B}(X, Y)\). Prove the following: (i) \(\underline{\operatorname{Ker}(T)}=T^{*}\left(Y^{*}\right)_{\perp}\) and \(\underline{\operatorname{Ker}\left(T^{*}\right)}=T(X)^{\perp}\). (ii) \(\overline{T(X)}=\operatorname{Ker}\left(T^{*}\right)_{\perp}\) and \(\overline{T^{*}\left(Y^{*}\right)} \subset \operatorname{Ker}(T)^{\perp}\) Hint: (i): Assume \(x \in T^{*}\left(Y^{*}\right)_{\perp} .\) Then for any \(g \in Y^{*}\) we have \(g(T(x))=\) \(T^{*}(g)(\underline{x})=0\), and hence \(T(x)=0\). Thus \(x \in \operatorname{Ker}(T)\). (ii): \(\overline{T(X)}=\overline{\operatorname{span}}(T(X))=\left(T(X)^{\perp}\right)_{\perp}=\operatorname{Ker}\left(T^{*}\right)_{\perp}\)

Short answer

(i) \(\underline{\operatorname{Ker}(T)} = T^*(Y^*)_{\perp}\) and \(\underline{\operatorname{Ker}(T^*)} = T(X)^{\perp}\). (ii) \(\overline{T(X)} = \operatorname{Ker}(T^*)_{\perp}\) and \(\overline{T^*(Y^*)} \subset \operatorname{Ker}(T)^{\perp}.

Step-by-step solution

Definitions and Preliminary Information

Review the definitions and properties of the kernel and range of linear operators and their adjoints. Recall that for a linear operator \(T \in \mathcal{B}(X, Y)\) and its adjoint \(T^* \in \mathcal{B}(Y^*, X^*)\), we have properties involving orthogonal complements and kernels in Banach spaces.

Prove (i) \(\underline{\operatorname{Ker}(T)} = T^*\left(Y^*\right)_{\perp}\)

Assume \(x \in T^*(Y^*)_{\perp}\). By definition, for any \(g \in Y^*\), we have \(T^*(g)(x) = \langle g, T(x) \rangle = 0\). This implies \(T(x) = 0\), hence \(x \in \operatorname{Ker}(T)\). Therefore, \(T^*(Y^*)_{\perp} \subseteq \operatorname{Ker}(T)\). For the reverse inclusion, if \(x \in \operatorname{Ker}(T)\), then for any \(g \in Y^*\), \(T(x) = 0\) implies \(\langle g, T(x) \rangle = 0\), so \(x \in T^*(Y^*)_{\perp}\). Thus, \(\operatorname{Ker}(T) = T^*(Y^*)_{\perp}\).

Prove (i) \(\operatorname{Ker}(T^*) = T(X)^{\perp}\)

Assume \(f \in \operatorname{Ker}(T^*)\). By definition, for any \(x \in X\), we have \(T^*(f)(x) = \langle f, T(x) \rangle = 0\), implying \(f \in T(X)^{\perp}\). Hence, \(\operatorname{Ker}(T^*) \subseteq T(X)^{\perp}\). For the reverse inclusion, if \(f \in T(X)^{\perp}\), then \(\langle f, T(x) \rangle = 0\) for any \(x \in X\). Hence, \(T^*(f) = 0\), so \(f \in \operatorname{Ker}(T^*)\). Thus, \(T(X)^{\perp} = \operatorname{Ker}(T^*)\).

Prove (ii) \(\overline{T(X)} = \operatorname{Ker}(T^*)_{\perp}\)

Using the Hahn-Banach theorem and the properties of orthogonal complements, it's known that \(\overline{T(X)} = (T(X)^{\perp})_{\perp}\). From part (i), we know \(T(X)^{\perp} = \operatorname{Ker}(T^*)\). Hence, \(\overline{T(X)} = (\operatorname{Ker}(T^*))_{\perp}\).

Prove (ii) \(\overline{T^*(Y^*)} \subseteq \operatorname{Ker}(T)^{\perp}\)

Using the fact that \(\overline{T(X)} = \operatorname{Ker}(T^*)_{\perp}\) as shown in step 4, consider the set \(\operatorname{Ker}(T)\). By definition, \(y^* \in T^*(Y^*)\) means there exists some \(y \in Y^*\) such that \(T^*(y) = y^*\). Hence, for any \(x \in X\), \(\langle y^*, x \rangle = \langle y, T(x) \rangle\), and if \(y^* \in \operatorname{Ker}(T)^{\perp}\), then these equalities hold in the closed subspace. Therefore, \(\overline{T^*(Y^*)} \subseteq \operatorname{Ker}(T)^{\perp}\).

Key concepts

Banach Space

A Banach space is a specific type of vector space that is fundamental in the study of functional analysis. It is a complete normed vector space, which means that it is a vector space equipped with a norm and is complete in the sense that every Cauchy sequence converges within the space.

Key properties of Banach spaces include:

• Norm: A function assigning a non-negative length or size to each vector in the space.
• Completeness: Every Cauchy sequence (a sequence where the elements get arbitrarily close to each other) eventually converges in the space.
• Linearity: The operations of vector addition and scalar multiplication behave in the standard linear fashion.

Banach spaces are crucial in analysis because they allow the application of the Banach Fixed-Point Theorem and various other results that require completeness. They serve as the setting for many problems in functional analysis, including the ones involving linear operators and their properties.

Kernel of a Linear Operator

The kernel (or null space) of a linear operator is a fundamental concept in linear algebra and functional analysis. Given a linear operator \(T\) from one vector space \(X\) to another vector space \(Y\):

• The kernel of \(T\), denoted as \(\operatorname{Ker}(T)\), consists of all vectors \(x\) in \(X\) such that \(T(x) = 0\).
• Formally, it is defined as: \(\operatorname{Ker}(T) = \{ x \in X \mid T(x) = 0 \}\).
• The kernel is always a subspace of \(X\).

This means it satisfies the properties of a subspace: it contains the zero vector, is closed under vector addition, and is closed under scalar multiplication.

In the context of adjoint operators, the kernel plays a vital role in determining orthogonal complements. For example, one important result is that the kernel of the adjoint operator \(T^*\) is the orthogonal complement of the range of \(T\). This relationship is tightly linked with the following result:

• \(\operatorname{Ker}(T^*) = T(X)^{\perp}\), where \(T(X)\) denotes the range of \(T\).

Adjoint Operators

Adjoint operators are crucial when studying linear operators in functional analysis. Given a linear operator \(T \in \mathcal{B}(X, Y)\) between Banach spaces, its adjoint \(T^*\) operates on the dual spaces \(Y^*\) and \(X^*\), which consist of all continuous linear functionals on \(Y\) and \(X\) respectively:

• The adjoint \(T^*\) is defined to satisfy the relation \(\langle T(x), y^* \rangle = \langle x, T^*(y^*) \rangle\) for all \(x \in X\) and \(y^* \in Y^*\).
• This means that for any linear functional \(y^* \in Y^*\), the composition \(T^*(y^*)\) is a linear functional in \(X^*\).

Adjoint operators help transfer properties of \(T\) between spaces and are used in various theorems involving orthogonal complements and kernels. For instance, we showed that:

• \(\operatorname{Ker}(T) = (T^*(Y^*))_{\perp}\), meaning the kernel of \(T\) is the orthogonal complement of the image of the adjoint \(T^*\).
• \(\operatorname{Ker}(T^*) = T(X)^{\perp}\), which tells us that the kernel of the adjoint \(T^*\) is the orthogonal complement of the range of \(T\).

Orthogonal Complements

An orthogonal complement is a key concept in understanding the structure of subspaces within a vector space. For a subspace \(M\) of an inner product space \(H\), the orthogonal complement \(M^{\perp}\) consists of all vectors that are orthogonal to every vector in \(M\):

• Mathematically, \(M^{\perp} = \{ x \in H \mid \langle x, m \rangle = 0, \forall m \in M \}\).
• The orthogonal complement \( M^{\perp} \) is always a closed subspace.
• If \(M\) is closed, then \(H\) is the direct sum of \(M\) and \(M^{\perp}\).

Orthogonal complements are extremely useful in functional analysis, especially in the context of adjoint operators and Banach spaces. For example, important results include:

• The orthogonal complement of the range of \(T\) is the kernel of \(T^*\): \(\operatorname{Ker}(T^*) = T(X)^{\perp}\).
• The closed span of the range of \(T\) is the orthogonal complement of the kernel of \(T^*\): \( \overline{T(X)} = \operatorname{Ker}(T^*)_{\perp} \).

These relationships help explicate how linear operators and their adjoints interact with subspace structures, offering deep insights into the solutions and properties of functional equations.