Fréchet space

We consider two classes of topological vector spaces, one more generalthan the other. Following Rudin [1] we will define a Fréchet spaceto be an element of the smaller class, and refer to an instance of themore general class as an F-space. After giving thedefinitions, we will explain why one definition is stronger than theother.

Definition 1.

An F-space is a complete topological vector space whose topology isinduced by a translation invariant metric. To be more precise, we saythat $U$ is an F-space if there exists a metric function

$$d:U\times U\to ℝ$$

such that

$$d(x,y)=d(x+z,y+z),x,y,z\in U;$$

and such that the collection of balls

is a base for the topology of $U$.

Note 1.

Recall that a topological vector space is a uniform space.The hypothesis that $U$ is complete is formulated in reference to thisuniform structure. To be more precise, we say that a sequence $a_n\in U,n=1,2,\mathrm{\dots }$ is Cauchy if for every neighborhood $O$ of theorigin there exists an $N\in \mathbb{N}$ such that$a_n-a_m\in O$ for all $n,m>N$. The completeness condition then takesthe usual form of the hypothesis that all Cauchy sequences possess alimit point.

Note 2.

It is customary to include the hypothesis that $U$ isHausdorff in the definition of a topological vector space.Consequently, a Cauchy sequence in a complete topological spacewill have a unique limit.

Note 3.

Since $U$ is assumed to be complete, the pair$(U,d)$ is a complete metric space. Thus, an equivalent definition ofan F-space is that of a vector space equipped with a complete, translation-invariant (but not necessarily homogeneous (http://planetmath.org/NormedVectorSpace)) metric, such that the operations of scalarmultiplication and vector addition are continuous with respect to thismetric.

Definition 2.

A Fréchet space is a complete topological vector space (either realor complex) whose topology is induced by a countable family ofsemi-norms. To be more precise, there exist semi-norm functions

$$\parallel -{\parallel }_n:U\to ℝ,n\in \mathbb{N},$$

such that the collection of all balls

is a base for the topology of $U$.

Proposition 1

Let $U$ be a complete topological vector space. Then, $U$ is aFréchet space if and only if it is a locally convex F-space.

Proof.First, let us show that a Fréchet space is a locally convex F-space,and then prove the converse. Suppose then that $U$ is Fréchet. Thesemi-norm balls are convex; this follows directly from the semi-normaxioms. Therefore $U$ is locally convex. To obtain the desireddistance function we set

$$d(x,y)=\sum _{n=0}^{\mathrm{\infty }}{2}^{-n}\frac{{\parallel x-y\parallel }_n}{1+{\parallel x-y\parallel }_n},x,y\in U.$$(1)

We now show that $d$ satisfies the metric axioms. Let $x,y\in U$such that $x\ne y$ be given. Since $U$ is Hausdorff, there is atleast one seminorm such

$${\parallel x-y\parallel }_n>0.$$

Hence $d(x,y)>0$.

Let $a,b,c>0$ be three real numbers such that

$$a\le b+c.$$

A straightforward calculation showsthat

$$\frac{a}{1+a}\le \frac{b}{1+b}+\frac{c}{1+c},$$(2)

as well. The above trick underlies the definition (1) ofour metric function. By the seminorm axioms we have that

$${\parallel x-z\parallel }_n\le {\parallel x-y\parallel }_n+{\parallel y-z\parallel }_n,x,y,z\in U$$

for all $n$. Combining this with (1) and(2) yields the triangle inequality for $d$.

Next let us suppose that $U$ is a locally convex F-space, and provethat it is Fréchet. For every $n=1,2,\mathrm{\dots }$ let $U_n$ be an openconvex neighborhood of the origin, contained inside a ball of radius$1/n$ about the origin. Let $\parallel -{\parallel }_n$ be the seminorm with$U_n$ as the unit ball. By definition, the unit balls of theseseminorms give a neighborhood base for the topology of $U$. QED.