TITLE: Definition of locally convex topological vector space
QUESTION [6 upvotes]: 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"?


Thanks and regards!

REPLY [9 votes]: 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 :-)