Question

Assume that \(T\) is a bounded linear operator from a Hilbert space \(H\) into \(H\) such that \((T(x), x) \geq(x, x)\) for every \(x \in H .\) Show that \(T\) is an invertible operator on \(H\).

Short answer

The operator \( T \) is invertible because it is injective and surjective by the given condition and the Banach Open Mapping Theorem.

Step-by-step solution

- Understand the given condition

The exercise states that for every element \( x \) in the Hilbert space \( H \), the inner product \((T(x), x) \geq (x, x)\) holds.

- Use the condition to show that \( T \) is injective

Assume \( T(x) = 0 \) for some \( x \in H\). Then, the condition gives us \((T(x), x) = 0 \geq (x, x)\). Since \( (x, x) \geq 0 \) and equality holds if and only if \( x = 0 \), it follows that \( x = 0 \). Thus, \( T \) is injective.

- Show that \( T \) is surjective using the Banach Open Mapping Theorem

Since \( T \) is a bounded, linear, injective operator on a Hilbert space, by the Banach Open Mapping Theorem, \( T \) maps onto \( H \) (surjective). Therefore, \( T \) is bijective.

- Conclude that \( T \) is invertible

Since \( T \) is both injective and surjective, it is bijective. Hence, there exists an inverse operator \( T^{-1} \) on \( H \).

Key concepts

bounded linear operator

A bounded linear operator is a fundamental concept in functional analysis and operator theory. An operator, denoted by \( T \), is called linear if it satisfies two properties for all vectors \( x \) and \( y \) in a Hilbert space \( H \) and all scalars \( \alpha \):

• Additivity: \( T(x + y) = T(x) + T(y) \)
• Scalar multiplication: \( T( \alpha x ) = \alpha T(x) \)

Moreover, the operator is bounded if there exists a constant \( M \) such that \( \|T(x)\| \leq M \|x\| \) for all \( x \in H \). Bounded linear operators are continuous, meaning small changes in the input result in small changes in the output. These operators ensure a controlled behavior, which is crucial in many applications and proofs, such as the one we explore in the exercise.

inner product

An inner product is a generalization of the dot product in vector spaces, providing a way to define angles and lengths. For a Hilbert space \( H \), the inner product \( ( \cdot , \cdot ) \) associates each pair of vectors \( x \) and \( y \) in \( H \) with a complex number (or real number if \( H \) is a real Hilbert space). Its main properties are:

• Linearity in the first argument: \( (\alpha x + \beta y, z) = \alpha (x, z) + \beta (y, z) \)
• Conjugate symmetry: \( (x, y) = \overline{(y, x)} \)
• Positive definiteness: \( (x, x) \geq 0 \) with equality if and only if \( x = 0 \)

In our exercise, the condition \( (T(x), x) \geq (x, x) \) uses the inner product to establish a relationship between the operator \( T \) and the elements of the Hilbert space.

Banach Open Mapping Theorem

The Banach Open Mapping Theorem is a pivotal result in functional analysis. It states that if \( T \) is a bounded and surjective linear operator between Banach spaces, then \( T \) maps open sets to open sets. In the context of our Hilbert space exercise, let's break it down:

• Hilbert spaces are also Banach spaces since they are complete under their norm.
• If \( T \) is injective (one-to-one), the theorem guarantees that \( T \) is also surjective (onto).

Applying this theorem helps show that \( T \) is bijective (both injective and surjective), which is necessary to conclude that \( T \) is invertible. This ensures every vector in the Hilbert space has a unique preimage under \( T \), establishing the existence of \( T^{-1} \).