DF space

* **Functional Analysis**
## Overview diagrams
* topological vector spaces
* locally convex topological vector spaces
## Basic concepts
* topological vector space
* locally convex topological vector space
* Banach Spaces
* reflexive
* Hilbert Spaces, Fréchet Spaces, Sobolev spaces, Lebesgue Spaces
* Bornological Vector Spaces
* Barrelled Vector Spaces
* linear operator
* bounded, unbounded, self-adjoint, compact, Fredholm
* spectrum of an operator
* operator algebras
* functional calculus
## Theorems
* Stone-Weierstrass theorem
* spectral theory
* spectral theorem
* Gelfand duality
* functional calculus
* Riesz representation theorem
* measure theory
## Topics in Functional Analysis
* Bases
* Algebraic Theories in Functional Analysis
* An Elementary Treatment of Hilbert Spaces
* When are two Banach spaces isomorphic?

A **DF space** is a type of locally convex topological vector space. The basic idea is that a DF space is *morally* the strong dual of a Fréchet space. That is, it has all the nice structure that such a dual would have but without the bother of actually having to be a dual space.

A locally convex topological vector space is a **DF space** if it possesses a fundamental sequence? of bounded sets and if every strongly bounded countable union of equicontinuous subsets of the dual is again equicontinuous.

- The strong dual of a metrisable locally convex topological vector space is a DF space.
- Every normable space is a DF space.
- Every infrabarrelled space that has a fundamental sequence? of bounded sets is a DF space.
- In particular, every LB space? is a DF space, since any bounded set in an LB space is contained in one of the factors.

A locally convex topological vector space that is metrisable and is a DF space is normable.

Let $E$ be a metrisable DF space.

To prove the result, we shall use a proposition recorded in Schaefer (IV.6.7). That says that a convex, circled subset $V$ of $E$ is a neighbourhood of $0$ if (and only if) for every convex, circled bounded subset $B \subseteq E$, $B \cap V$ is a $0$-neighbourhood in $B$.

As $E$ is metrisable, it has a countable $0$-neighbourhood base, say $(V_n)$. As $E$ is a DF space, it has a countable fundamental family of bounded sets, say $(B_n)$. Let us assume, without loss of generality, that this family is increasing.

For each $k \in \mathbb{N}$, as $B_k$ is bounded, there is some $\lambda_k \gt 0$ such that $B_k \subseteq \lambda_k V_k$. Since, by assumption, the $B_k$ are an increasing family, we have $B_k \subseteq \lambda_l V_l$ for all $l \ge k$. Let $B \coloneqq \bigcap \lambda_k V_k$. This is a bounded set since it is contained in $\lambda_k V_k$ for each $k$. Let $l \in \mathbb{N}$ and consider $B_l \cap B$. We can write this as

$B_l \cap \big(\bigcap_{k=1}^{l-1} \lambda_k V_k\big) \cap B_l \cap \big( \bigcap_{k \ge l} \lambda_k V_k \big).$

Since $B_l \subseteq \lambda_k V_k$ for $k \ge l$, the last part is unnecessary and so we see that

$B_l \cap B = B_l \cap \big(\bigcap_{k=1}^{l-1} \lambda_k V_k \big).$

This is then a finite intersection of open sets in $B_l$ and so is open in $B_l$.

As this holds for any of the $B_l$s, it holds for any bounded set, whence the proposition from Schaefer applies to show that $B$ is a $0$-neighbourhood. Thus $E$ possesses a bounded $0$-neighbourhood, whence is normable.

Revised on May 21, 2010 10:53:25
by Andrew Stacey
(129.241.15.70)