Fréchet space
We consider two classes of topological vector spaces, one more general
than the other. Following Rudin [1] we will define a Fréchet space
to be an element of the smaller class, and refer to an instance of the
more general class as an F-space. After giving the
definitions, we will explain why one definition is stronger than the
other.
Definition 1.
An F-space is a complete topological vector space whose topology is
induced by a translation
invariant metric. To be more precise, we say
that U is an F-space if there exists a metric function
d:U×U→ℝ |
such that
d(x,y)=d(x+z,y+z),x,y,z∈U; |
and such that the collection of balls
Bϵ(x)={y∈U:d(x,y)<ϵ},x∈U,ϵ>0 |
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 this
uniform structure. To be more precise, we say that a sequence an∈U,n=1,2,… is Cauchy if for every neighborhood O of the
origin there exists an N∈ℕ such that
an-am∈O for all n,m>N. The completeness condition then takes
the usual form of the hypothesis that all Cauchy sequences
possess a
limit point
.
Note 2.
It is customary to include the hypothesis that U is
Hausdorff in the definition of a topological vector space.
Consequently, a Cauchy sequence in a complete topological space
will 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 of
an 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 scalar
multiplication and vector addition are continuous
with respect to this
metric.
Definition 2.
A Fréchet space is a complete topological vector space (either real
or complex) whose topology is induced by a countable family of
semi-norms. To be more precise, there exist semi-norm functions
∥-∥n:U→ℝ,n∈ℕ, |
such that the collection of all balls
B(n)ϵ(x)={y∈U:∥x-y∥n<ϵ},x∈U,ϵ>0,n∈ℕ, |
is a base for the topology of U.
Proposition 1
Let U be a complete topological vector space. Then, U is a Fréchet 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. The
semi-norm balls are convex; this follows directly from the semi-norm
axioms. Therefore U is locally convex. To obtain the desired
distance function we set
d(x,y)=∞∑n=02-n∥x-y∥n1+∥x-y∥n,x,y∈U. | (1) |
We now show that d satisfies the metric axioms. Let x,y∈U such that x≠y be given. Since U is Hausdorff, there is at least one seminorm such
∥x-y∥n>0. |
Hence d(x,y)>0.
Let a,b,c>0 be three real numbers such that
a≤b+c. |
A straightforward calculation shows that
a1+a≤b1+b+c1+c, | (2) |
as well. The above trick underlies the definition (1) of our metric function. By the seminorm axioms we have that
∥x-z∥n≤∥x-y∥n+∥y-z∥n,x,y,z∈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 prove that it is Fréchet. For every n=1,2,… let Un be an open convex neighborhood of the origin, contained inside a ball of radius 1/n about the origin. Let ∥-∥n be the seminorm with Un as the unit ball. By definition, the unit balls of these seminorms give a neighborhood base for the topology of U. QED.
References
-
1
W.Rudin, Functional Analysis
.
Title | Fréchet space |
---|---|
Canonical name | FrechetSpace |
Date of creation | 2013-03-22 13:06:10 |
Last modified on | 2013-03-22 13:06:10 |
Owner | rmilson (146) |
Last modified by | rmilson (146) |
Numerical id | 51 |
Author | rmilson (146) |
Entry type | Definition |
Classification | msc 57N17 |
Classification | msc 54E50 |
Classification | msc 52A07 |
Related topic | TopologicalVectorSpace |
Related topic | HausdorffSpaceNotCompletelyHausdorff |
Defines | F-space |