Show That T T T Is Bounded If And Only If V V V Is Closed.

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Introduction

In functional analysis, a closed linear operator is a linear operator that maps a subspace of a Banach space to another Banach space, and its graph is closed. In this discussion, we will explore the relationship between a closed linear operator and the closedness of its domain. Specifically, we will show that a closed linear operator $T$ is bounded if and only if its domain $V$ is closed.

Preliminaries

Before we dive into the main result, let's recall some definitions and properties.

• A Banach space is a complete normed vector space.
• A linear operator $T$ between two Banach spaces $X$ and $Y$ is said to be closed if its graph is closed, i.e., if for any sequence $(x_n)$ in $X$ such that $x_n \to x$ and $Tx_n \to y$, we have $x \in V$ and $Tx = y$.
• A linear subspace $V$ of a Banach space $X$ is said to be closed if it contains all its limit points, i.e., if for any sequence $(x_n)$ in $V$ such that $x_n \to x$, we have $x \in V$.

The Main Result

We will show that $T$ is bounded if and only if $V$ is closed.

The Forward Direction

Suppose $T$ is bounded. We need to show that $V$ is closed. Let $(x_n)$ be a sequence in $V$ such that $x_n \to x$. Since $T$ is bounded, it is continuous, and therefore, $Tx_n \to Tx$. Since $T$ is closed, its graph is closed, and therefore, $x \in V$ and $Tx = Tx$. This shows that $V$ is closed.

The Reverse Direction

Suppose $V$ is closed. We need to show that $T$ is bounded. Let $x \in V$ and let $(x_n)$ be a sequence in $V$ such that $x_n \to x$. Since $V$ is closed, we have $x \in V$. Since $T$ is closed, its graph is closed, and therefore, $Tx_n \to Tx$. Since $T$ is linear, we have $Tx_n = Tx$ for all $n$. This shows that $T$ is bounded.

Conclusion

In this discussion, we have shown that a closed linear operator $T$ is bounded if and only if its domain $V$ is closed. This result highlights the importance of the closedness of the domain of a linear operator in functional analysis.

Proof of the Main Result

The Forward Direction

Suppose $T$ is bounded. We need to show that $V$ is closed. Let $(x_n)$ be a sequence in $V$ such that $x_n \to x$. Since $T$ is bounded, it is continuous, and therefore, $Tx_n \to Tx$. Since $T$ is closed, its graph is closed, and therefore, $x \in V$ and $Tx = Tx$. This shows that $V$ is closed.

The Reverse Direction

Suppose $V$ is closed. We need to show that $T$ is bounded. Let $x \in V$ and let $(x_n)$ be a sequence in $V$ such that $x_n \to x$. Since $V$ is closed, we have $x \in V$. Since $T$ is closed, its graph is closed, and therefore, $Tx_n \to Tx$. Since $T$ is linear, we have $Tx_n = Tx$ for all $n$. This shows that $T$ is bounded.

Corollary

As a corollary of the main result, we have the following:

• If $T$ is a bounded linear operator, then its domain $V$ is closed.
• If $V$ is a closed linear subspace of a Banach space $X$, then any closed linear operator $T$ from $V$ to a Banach space $Y$ is bounded.

Example

Let $X$ and $Y$ be Banach spaces and let $V \subset X$ be a linear subspace. Let $T : V \to Y$ be a closed linear operator. Suppose $T$ is bounded. Then, by the main result, we have that $V$ is closed.

Conversely, suppose $V$ is closed. Then, by the main result, we have that $T$ is bounded.

Counterexample

Let $X$ and $Y$ be Banach spaces and let $V \subset X$ be a linear subspace. Let $T : V \to Y$ be a closed linear operator. Suppose $V$ is not closed. Then, by the main result, we have that $T$ is not bounded.

Open Problem

Let $X$ and $Y$ be Banach spaces and let $V \subset X$ be a linear subspace. Let $T : V \to Y$ be a closed linear operator. Suppose $T$ is not bounded. Is it true that $V$ is not closed?

References

• [1] Rudin, W. (1991). Functional Analysis. McGraw-Hill.
• [2] Yosida, K. (1980). Functional Analysis. Springer-Verlag.

Acknowledgments

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Introduction

In our previous discussion, we showed that a closed linear operator $T$ is bounded if and only if its domain $V$ is closed. In this Q&A article, we will address some common questions and concerns related to this result.

Q: What is the significance of the closedness of the domain of a linear operator?

A: The closedness of the domain of a linear operator is crucial in functional analysis. It ensures that the operator is well-defined and that its graph is closed. This, in turn, implies that the operator is bounded.

Q: Can you provide an example of a closed linear operator that is not bounded?

A: Yes, consider the operator $T : l^2 \to l^2$ defined by $T(x_1, x_2, \ldots) = (x_1, 0, 0, \ldots)$. This operator is closed, but it is not bounded.

Q: How does the closedness of the domain of a linear operator relate to the closedness of its range?

A: The closedness of the domain of a linear operator does not necessarily imply the closedness of its range. However, if the operator is bounded, then its range is closed.

Q: Can you provide a counterexample to show that the closedness of the domain of a linear operator does not imply the closedness of its range?

A: Yes, consider the operator $T : l^2 \to l^2$ defined by $T(x_1, x_2, \ldots) = (x_1, x_2, \ldots)$. This operator is closed, but its range is not closed.

Q: How does the main result relate to the Hahn-Banach theorem?

A: The main result is a consequence of the Hahn-Banach theorem. The Hahn-Banach theorem states that if a linear functional on a subspace of a Banach space is bounded, then it can be extended to a bounded linear functional on the entire Banach space. The main result can be viewed as a generalization of this theorem to linear operators.

Q: Can you provide a proof of the main result using the Hahn-Banach theorem?

A: Yes, here is a proof of the main result using the Hahn-Banach theorem:

Suppose $T$ is bounded. Then, by the Hahn-Banach theorem, we can extend $T$ to a bounded linear operator on the entire Banach space $X$. This implies that $V$ is closed.

Conversely, suppose $V$ is closed. Then, by the Hahn-Banach theorem, we can extend $T$ to a bounded linear operator on the entire Banach space $X$. This implies that $T$ is bounded.

Q: What are some applications of the main result?

A: The main result has several applications in functional analysis and operator theory. For example, it can be used to study the properties of linear operators on Banach spaces, such as their boundedness and closedness.

Q: Can you provide some open problems related to the main result?

A: Yes, here are some open problems related to the main result:

• Is it true that if a linear operator on a Banach space is closed, then its domain is closed?
• Can you provide a characterization of the closed linear operators on a Banach space in terms of their domains?
• How does the main result relate to other results in functional analysis, such as the Hahn-Banach theorem and the open mapping theorem?

References

• [1] Rudin, W. (1991). Functional Analysis. McGraw-Hill.
• [2] Yosida, K. (1980). Functional Analysis. Springer-Verlag.

Acknowledgments

The author would like to thank [Name] for helpful discussions and suggestions.