Convex interior topology I have found a fascinating example of topology on a vector space $V$, but I cannot prove its interesting properties to myself. Let $\mathcal{B}$ be the family of all convex symmetric (i.e. $\forall U\in\mathcal{B}\quad x\in U\Rightarrow-x\in U$) sets $U\subset V$ coinciding with its interior, defined as $\{x\in U:\quad\forall y\in V\quad\exists \varepsilon(y)=\varepsilon>0:\forall t\in\mathbb{R}(|t|<\varepsilon\Rightarrow x+ty\in U)\}$.
I read that


*

*
$\mathcal{B}$ is a local basis for $0$ for a locally convex topology $\tau$ satisfying axiom $T_1$;

*$\tau$ is the strongest among locally convex topologies where the linear operations on $V$ are continuous;

*Any linear functional $V\to K$ (where I think $K$ can be both $\mathbb{C}$ or $\mathbb{R}$) is continuous with respect to $\tau$.


I can find no on line resource on the issue and my book only has half a page on locally convex topological vector spaces. Can anyone help me in understanding these properties with links or proofs?
I heartily thank you all!
 A: Sorry, it's gotten a bit long, but here you go:
Let's begin with the definitions. A topological $\mathbb{R}$-vector space is a real vector space $E$ endowed with a topology $\mathscr{T}$ such that the maps
$$\begin{gather}
+ \colon E\times E \to E\\
\cdot \colon \mathbb{R}\times E \to E
\end{gather}$$
are continuous (where $\mathbb{R}$ carries the standard topology, and the domains are endowed with the product topologies).
An easy consequence is that for every $x\in E$ the translation $\tau_x \colon y \mapsto y+x$ is continuous - and since its inverse is the translation $\tau_{-x}$, which is continuous, it is a homeomorphism.
Similarly, for every $r\in\mathbb{R}$ the multiplication $\mu_r\colon x \mapsto r\cdot x$ is continuous, and for $r\neq 0$, it is a homeomorphism, and for every $x\in E$, the map $\varepsilon_x \colon r \mapsto r\cdot x$ is continuous [but rarely a homeomorphism].
Since the translations are homeomorphisms, the topology is completely determined by the filter $\mathscr{N}$ of neighbourhoods of $0$. For any $x\in E$, the system of neighbourhoods of $x$ is
$$\mathscr{V}(x) = \{\tau_x(U) : U\in \mathscr{N}\} = \{ U + x : U\in\mathscr{N}\}.$$
In order to be able to obtain a vector space topology from a family of sets we want to be a neighbourhood basis of $0$, we must identify some conditions that a neighbourhood basis of $0$ in a topological vector space must have, and then show that these conditions are also sufficient.
The continuity of the addition in $(0,0)$ means that for every neighbourhood $U$ of $0$, there are neighbourhoods $V,W$ of $0$ with $+(V,W) \subset U$. Since the intersection of two neighbourhoods of $0$ is a neighbourhood of $0$, we can assume that $V = W$, so "for every $U\in\mathscr{N}$ there is a $V\in\mathscr{N}$ with $V+V\subset U$".
The continuity of $\mu_r$ in $0\in E$ means that for every $U\in\mathscr{N}$ there is a $V\in \mathscr{N}$ such that $r\cdot V\subset U$. For $r\neq 0$, that can also be written as $V \subset r^{-1}\cdot U$, and hence $r^{-1}\cdot U$ is a neighbourhood of $0$ if and only if $U$ is one.
The continuity of $\varepsilon_x$ in $0$ means that for every $U\in\mathscr{N}$ there is a $\delta > 0$ such that $(-\delta,\delta)\cdot x \subset U$. We give that property a name, we call subsets $A$ of $E$ with the property that for all $x\in E$ there is a $\delta(x) > 0$ such that $(-\delta(x),\delta(x))\cdot x \subset A$ absorbing. So "every neighbourhood of $0$ in a topological vector space is absorbing".
The continuity of scalar multiplication in $(0,0) \in \mathbb{R}\times E$ means that for every $U\in \mathscr{N}$ there is a $\delta > 0$ and a $V\in \mathscr{N}$ with $(-\delta,\delta)\cdot V \subset U$. The set
$$W = (-\delta,\delta)\cdot V = \{ r\cdot x : \lvert r\rvert < \delta,\, x\in V\}$$
has the property that for all $y\in W$ and all $t\in\mathbb{R}$ with $\lvert t\rvert \leqslant 1$ we have $t\cdot y \in W$ (if $y = r\cdot x$ with $\lvert r\rvert < \delta$ and $x\in V$, then $t\cdot y = (tr)\cdot x$, and $\lvert tr\rvert \leqslant \lvert r\rvert < \delta$), or $[-1,1]\cdot W = W$. We call such sets balanced. So, every neighbourhood of $0$ contains a balanced neighbourhood of $0$.
Together with the properties that the neighbourhood filter of a point in any topological space must have, we obtain the list of required properties:


*

*$\bigl(\forall U\in \mathscr{N}\bigr) (0\in U)$,

*$\bigl(\forall U\in\mathscr{N}\bigr)(U\subset V \implies V\in\mathscr{N})$,

*$U_1,\dotsc, U_n\in \mathscr{N} \implies U_1 \cap \dotsc \cap U_n \in \mathscr{N}$,

*$\bigl(\forall U\in\mathscr{N}\bigr)\bigl(\exists V\in\mathscr{N}\bigr)(V+V\subset U)$,

*$\bigl(\forall U\in\mathscr{N}\bigr)\bigl(\forall r\in\mathbb{R}\setminus\{0\}\bigr)(r\cdot U\in\mathscr{N})$,

*all $U\in\mathscr{N}$ are absorbing,

*$\bigl(\forall U\in\mathscr{N}\bigr)\bigl(\exists \delta > 0\bigr)\bigl(\exists V\in\mathscr{N}\bigr)\bigl((-\delta,\delta)\cdot V\subset U\bigr)$.


We remark that the inclusion of 1. in the list is redundant, that follows from 6., as well as from 7.
Next, we show that if $\mathscr{N}$ is a system of subsets of an $\mathbb{R}$-vector space $E$ with the properties 1-7, then there is a unique vector space topology $\mathscr{T}_{\mathscr{N}}$ on $E$ such that $\mathscr{N}$ is the family of neighbourhoods of $0$ in $\mathscr{T}_{\mathscr{N}}$. The uniqueness is clear, since by a remark above, if such a topology exists, then we have $\mathscr{V}(x) = \tau_x(\mathscr{N})$ for all $x\in E$, and since a topology is determined by the family of neighbourhood filters of all points, we have no choice but to define the neighbourhood filters such.
Next we have to verify that the family $\{ \mathscr{V}(x) : x\in E\}$ of filters satisfies the requirements for neighbourhood systems in a topological space.


*

*$\bigl(\forall x\in E\bigr)\bigl(\forall U\in \mathscr{V}(x)\bigr)(x\in U)$ follows from 1. and $U = \tau_x(V) = x+V$ for some $V\in\mathscr{N}$.

*$\bigl(\forall x\in E\bigr)\bigl(\forall U\in \mathscr{V}(x)\bigr)\bigl(U\subset A\implies A\in\mathscr{V}(x)\bigr)$ follows from 2.

*$\bigl(\forall x\in E\bigr)\bigl(U_1,\dotsc,U_n\in\mathscr{V}(x)\implies U_1 \cap \dotsc \cap U_n\in \mathscr{V}(x)\bigr)$ follows from 3.

*$\bigl(\forall x\in E\bigr)\bigl(\forall U\in\mathscr{V}(x)\bigr)\bigl(\exists V\in\mathscr{V}(x)\bigr)\bigl(y\in V\implies U\in\mathscr{V}(y)\bigr)$ follows from 4. If $U = x+W$ for a $W\in\mathscr{N}$ and $Y\in\mathscr{N}$ with $Y+Y\subset W$, then $V = x+Y \in\mathscr{V}(x)$, and for every $y \in V$ we have $y+Y\subset x+Y+Y \subset x+W = U$, so $U\in\mathscr{V}(y)$ since $y+Y\in\mathscr{V}(y)$.


Thus there is a unique topology $\mathscr{T}_\mathscr{N}$ on $E$ such that for all $x\in E$ the filter $\mathscr{V}(x) = \tau_x(\mathscr{N})$ is the neighbourhood filter of $x$ in $\mathscr{T}_\mathscr{N}$.
It remains to be seen that the topology $\mathscr{T}_\mathscr{N}$ is a vector space topology on $E$.
The continuity of addition is very easy: Given $x,y\in E$ and $U = \tau_{x+y}(V) \in \mathscr{V}(x+y)$, we choose a $W\in\mathscr{N}$ with $W+W\subset V$ by property 4, and see that $$\tau_x(W) + \tau_y(W) = (x+W) + (y+W) = (x+y)+(W+W)\subset U,$$ so the addition is continuous in $(x,y)$.
For the continuity of scalar multiplication we need a little more work. Given $(r,x)\in\mathbb{R}\times E$ and a neighbourhood $U = rx + V$ of $rx$, we write
$$sy - rx = (s-r)x + r(y-x) + (s-r)(y-x)$$
and choose, by repeated application of 4, a $W\in\mathscr{N}$ with $W+W+W\subset V$. By property 6, there is a $\delta_1 > 0$ such that $(s-r)x\in W$ for all $s$ with $\lvert s-r\rvert < \delta_1$. By property 5, if $r\neq 0$, there is an $Y\in\mathscr{N}$ such that $r\cdot Y\subset W$. If $r = 0$, the existence of such an $Y$ is trivial (we can then take $Y = E$). By property 7, there are $\delta_2 > 0$ and $Z\in\mathscr{N}$ such that $(-\delta_2,\delta_2)\cdot Z\subset W$. Thus, when $\lvert s-r\rvert < \min \{\delta_1,\delta_2\}$ and $y-x\in Y\cap Z$, we have $sy-rx \in W+W+W\subset V$, and hence scalar multiplication is continuous in $(r,x)$.
Now we can look at the properties a family $\mathscr{B}$ of sets must have in order to be a neighbourhood basis of $0$ in a vector space topology. We can skip the redundant property 1, and property 2 need not hold for a neighbourhood basis. We can reduce property 3 to the case of two sets (we could have also done that for the full neighbourhood system), but since the intersection of two sets from a neighbourhood basis need not belong to the neighbourhood basis itself, it becomes $\bigl(\forall U_1,U_2\in \mathscr{B}\bigr)\bigl(\exists V\in\mathscr{B}\bigr)(V\subset U_1\cap U_2)$. Properties 4,6, and 7 must be taken quasi verbatim (replacing $\mathscr{N}$ with $\mathscr{B}$), and property 5 becomes $\bigl(\forall U\in\mathscr{B}\bigr)\bigl(\forall r > 0\bigr)\bigl(\exists V\in\mathscr{B}\bigr)(V\subset r\cdot U)$ [property 7 allows to restrict our attention to positive $r$, we could also have done that above].
Since every neighbourhood of $0$ in a topological vector space contains a balanced neighbourhood of $0$, one often is interested in families of balanced sets that should be a neighbourhood basis of $0$ in a topological vector space. Any family of balanced sets automatically has property 7, so then we only need to check the modified properties 3-6 to see whether the family of balanced sets is a neighbourhood basis of $0$ in a vector space topology.
Things become nicer if we are interested in locally convex topologies. (A vector space is locally convex, if $0$ has a neighbourhood basis consisting of convex sets.) Since the convex hull of a balanced set is convex and balanced, in a locally convex space, $0$ has neighbourhood bases consisting of balanced convex sets. For families of balanced convex sets, the list of conditions to check becomes nicely short:
Proposition: Let $E$ be an $\mathbb{R}$-vector space, and $\mathscr{B}$ be a family of balanced convex subsets of $E$. Then $\mathscr{B}$ is a neighbourhood basis of $0$ in a locally convex vector space topology on $E$ if and only if


*

*$\bigl(\forall U_1,U_2\in\mathscr{B}\bigr)\bigl(\exists V\in\mathscr{B}\bigr)(V\subset U_1\cap U_2)$,

*$\bigl(\forall U\in\mathscr{B}\bigr)\bigl(\forall r > 0\bigr)\bigl(\exists V\in\mathscr{B}\bigr)(V\subset r\cdot U)$,

*all $U\in\mathscr{B}$ are absorbing.


The $V+V\subset U$ condition follows from the homothety invariance and convexity, since $\frac{1}{2} U + \frac{1}{2}U \subset U$ for convex $U$.
An easy consequence is the
Proposition: The family $\mathscr{B}_a$ of all absorbing balanced convex sets in an $\mathbb{R}$-vector space $E$ is a neighbourhood basis for the finest locally convex vector space topology on $E$.
Proof sketch: If $U$ is convex, balanced and absorbing, so is $r\cdot U$ for all $r > 0$ [elementary verification]. If $U_1$ and $U_2$ are convex, balanced and absorbing, so is $U_1\cap U_2$ [also elementary verification]. So $\mathscr{B}_a$ is a neighbourhood basis for a locally convex vector space topology $\mathscr{T}$ on $E$. It remains to see that for every locally convex vector space topology $\mathscr{T}'$, we have $\mathscr{T}' \subset \mathscr{T}$. But $\mathscr{T}'$ has a neighbourhood basis $\mathscr{B}'$ of $0$ consisting of convex, balanced and absorbing sets, so $\mathscr{B}'\subset \mathscr{B}_a$, and that implies $\mathscr{V}_{\mathscr{T}'}(0) \subset \mathscr{V}_\mathscr{T}(0)$, which in turn implies $\mathscr{T}'\subset \mathscr{T}$.

Now we prove that in the finest locally convex topology $\mathscr{T}$ every linear functional is continuous - in fact, every linear map $\lambda \colon E\to F$ where $F$ is a locally convex $\mathbb{R}$-vector space is continuous, and that $\mathscr{T}$ is a $T_1$ topology.
We start with the continuity of linear maps. Let $\Lambda \colon E\to F$ be linear, and $F$ a locally convex space. Let $U\subset F$ be any neighbourhood of $0$, and $V\subset U$ be a convex balanced neighbourhood of $0$. By linearity of $\Lambda$, $W = \Lambda^{-1}(V)$ is convex and balanced. Since $V$ is a neighbourhood of $0$, $W$ is also absorbing: Let $x\in E$. Then there is a $\delta > 0$ such that $\delta\Lambda x = \Lambda(\delta x) \in V$, and by balancedness $r\cdot x \in W$ for all $r$ with $\lvert r\rvert < \delta$ follows. So $W = \Lambda^{-1}(V)$ is a convex, balanced and absorbing set, hence a neighbourhood of $0$ in $\mathscr{T}$. Thus $\Lambda$ is continuous in $0$, hence everywhere.
From the continuity of all linear forms ($\mathbb{R}$ is locally convex) follows the $T_1$ property: Let $x\neq 0$. Then there is a linear form $\varphi \colon E\to \mathbb{R}$ with $\varphi(x) = 1$. The set $U = \varphi^{-1}((-1/2,1/2))$ is a neighbourhood of $0$ in $\mathscr{T}$ that does not contain $x$. (In fact, we even have $U \cap (x+U) = \varnothing$.) Hence $\mathscr{T}$ is a $T_1$ topology.

Now, finally, we show that your family $\mathcal{B}$ is a neighbourhood basis of $0$ in $\mathscr{T}$.
For convex sets in a real vector space, symmetry ($U = -U$) is equivalent to balancedness [easy verification], so all sets in $\mathcal{B}$ are convex and balanced. Furthermore, they are absorbing - that the sets coincide with their interior means that for all $U\in\mathcal{B}$ and $x\in U$ the set $U-x$ is absorbing; taking $x = 0$ shows in particular that $U$ is absorbing.
So every $U\in\mathcal{B}$ is a (convex and balanced) neighbourhood of $0$ in $\mathscr{T}$.
Every convex, balanced and open (in $\mathscr{T}$) set belongs to $\mathcal{B}$, since for an open $U$ and $x\in U$, the set $U-x$ is a neighbourhood of $0$, hence absorbing, and that means that $x$ is in the interior of $U$ as defined in the question [the terminology is a bit unfortunate since interior has a (different) meaning in topology; I know the property as all points of $U$ being internal points]. So every convex balanced open set belongs to $\mathcal{B}$, and since these form a neighbourhood basis of $0$ in $\mathscr{T}$, so does $\mathcal{B}$.
