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A locally convex topological vector space is a topological vector space in which the origin has a local base of absolutely convex *absorbent* sets.

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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,

  1. 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?
  2. 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"?

Thanks and regards!

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up vote 3 down vote accepted

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?

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.

But I am not sure why the first definition requires "absolutely convex" and "absorbent" while the second just "convex"?

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

  • Francois Treves: "Topological Vector Spaces, Distributions and Kernels"

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?

Yes :-)

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+1 Thanks! In "there is a basis of neighborhoods consisting of barrels (absorbing, blanced, convex, closed)", do you mean a basis of the TVS, or a local basis of the origin? –  Tim Feb 26 '12 at 19:36
    
A basis of neighborhoods of zero, I just corrected my answer accordingly. –  Tim van Beek Feb 28 '12 at 7:45
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