From Wiki
A locally convex topological vector space is a topological vector
space in which the origin has a local base of absolutely convex
absorbent sets.
Also from Wiki
Locally convex topological vector spaces: here each point has a local
base consisting of convex sets.
From PlanetMath
Let $V$ be a topological vector space over a subfield of the complex
numbers (usually taken to be $\mathbb{R}$ or $\mathbb{C}$ ). If the
topology of $V$ has a basis where each member is a convex set, then
$V$ is a locally convex topological vector space.
I was wondering if the three definitions are equivalent? Specifically,
-
The last two definitions seem to agree with each other, because the union of local bases, each for each point, is a base of the topology, and the restriction of a base to a point is a local base of that point?
-
Between the first two definitions:
- Because any translation of any open subset is still open, so I think it doesn't matter to specified for a local base of the
origin or a local base of every point? - But I am not sure why the first definition requires "absolutely convex" and "absorbent" while the second just "convex"?
- Because any translation of any open subset is still open, so I think it doesn't matter to specified for a local base of the
Thanks and regards!
Best Answer
Yes, the filter of neighborhoods $F(x)$ of any point x is the family of sets $V + x$ where $V$ runs over all elements of the filter of neighborhoods $F(0)$ of the origin, because addition is a continuous with continuous inverse. So saying that the origin has a local base of convex sets is the same as saying that every point has as local base of convex sets.
It should be just convex, the rest is a theorem: For a filter of neighborhoods of the origin $F(0)$ of a topological vector space it is true that
a) every $U \in F(0)$ is absorbing,
b) every $U \in F(0)$ contains a $V \in F(0)$ that is balanced.
For a locally convex topological vector space (the origin has a local base consisting of convex sets) it is true that there is a basis of neighborhoods of zero consisting of barrels (absorbing, blanced, convex, closed). This is proposition 7.2 in
Yes :-)