Algebraic facts about the topology of uniform convergence #
THIS FILE IS SYNCHRONIZED WITH MATHLIB4. Any changes to this file require a corresponding PR to mathlib4.
This file contains algebraic compatibility results about the uniform structure of uniform
convergence / 𝔖
-convergence. They will mostly be useful for defining strong topologies on the
space of continuous linear maps between two topological vector spaces.
Main statements #
uniform_fun.uniform_group
: ifG
is a uniform group, thenα →ᵤ G
a uniform groupuniform_on_fun.uniform_group
: ifG
is a uniform group, then for any𝔖 : set (set α)
,α →ᵤ[𝔖] G
a uniform group.uniform_on_fun.has_continuous_smul_of_image_bounded
: letE
be a TVS,𝔖 : set (set α)
andH
a submodule ofα →ᵤ[𝔖] E
. If the image of anyS ∈ 𝔖
by anyu ∈ H
is bounded (in the sense ofbornology.is_vonN_bounded
), thenH
, equipped with the topology induced fromα →ᵤ[𝔖] E
, is a TVS.
Implementation notes #
Like in topology/uniform_space/uniform_convergence_topology
, we use the type aliases
uniform_fun
(denoted α →ᵤ β
) and uniform_on_fun
(denoted α →ᵤ[𝔖] β
) for functions from α
to β
endowed with the structures of uniform convergence and 𝔖
-convergence.
TODO #
uniform_on_fun.has_continuous_smul_of_image_bounded
unnecessarily asks for𝔖
to be nonempty and directed. This will be easy to solve once we know that replacing𝔖
by its noncovering bornology (i.e not whatbornology
currently refers to in mathlib) doesn't change the topology.
References #
Tags #
uniform convergence, strong dual
Equations
Equations
Equations
Equations
Equations
Equations
Equations
Equations
Equations
Equations
Equations
Equations
Equations
Equations
Equations
Equations
- uniform_fun.module = pi.module α (λ (ᾰ : α), β) R
Equations
- uniform_on_fun.module = pi.module α (λ (ᾰ : α), β) R
If G
is a uniform group, then α →ᵤ G
is a uniform group as well.
If G
is a uniform additive group, then α →ᵤ G
is a uniform additive group
as well.
Let 𝔖 : set (set α)
. If G
is a uniform additive group, then α →ᵤ[𝔖] G
is a
uniform additive group as well.
Let 𝔖 : set (set α)
. If G
is a uniform group, then α →ᵤ[𝔖] G
is a uniform group as
well.
Let E
be a TVS, 𝔖 : set (set α)
and H
a submodule of α →ᵤ[𝔖] E
. If the image of any
S ∈ 𝔖
by any u ∈ H
is bounded (in the sense of bornology.is_vonN_bounded
), then H
,
equipped with the topology of 𝔖
-convergence, is a TVS.
For convenience, we don't literally ask for H : submodule (α →ᵤ[𝔖] E)
. Instead, we prove the
result for any vector space H
equipped with a linear inducing to α →ᵤ[𝔖] E
, which is often
easier to use. We also state the submodule
version as
uniform_on_fun.has_continuous_smul_submodule_of_image_bounded
.
Let E
be a TVS, 𝔖 : set (set α)
and H
a submodule of α →ᵤ[𝔖] E
. If the image of any
S ∈ 𝔖
by any u ∈ H
is bounded (in the sense of bornology.is_vonN_bounded
), then H
,
equipped with the topology of 𝔖
-convergence, is a TVS.
If you have a hard time using this lemma, try the one above instead.