As stated in the title, I am trying to make sense of the equivalent definitions of locally convex spaces. Especially, I am confused about the part, where I try to prove, that the topologies coincide.
Let me first give the definition I am working with:
A topological vector space $(X,\tau)$ is called locally convex, if there is a family of seminorms $(p_i)_{i\in I}$, such that
$\tau$ is the initial topology with respect to the quotient mappings $q_i: X \rightarrow X/\ker(p_i)$, $i\in I$.
The seminorms are separating, i.e. if $x\ne 0$, there is an $i\in I$ with $p_i(x)\ne 0$.
Let me mention, that for any $i\in I$, $\epsilon >0$ and $x\in X$, the sets
$$B^i_{\epsilon}(x):=\{y\in X: p_i(y-x)<\epsilon\}$$
form a subbase of the initial topology of the mappings $q_i: X \rightarrow X/\ker(p_i)$, $i\in I$.
Note: There is a norm $\lVert\cdot\rVert:X/\ker(p_i)\rightarrow \mathbb{R}$, such that $p_i=\lVert q_i\rVert$, especially $p_i(y-x)=\lVert q_i(y)-q_i(x)\rVert$. Then, $B^i_{\epsilon}(x)=q_i^{-1}[ B_\epsilon^{X/\ker(p_i)} (q_i(x))] $, i.e. can be written as preimage of $q_i$ of an $\epsilon$-ball in $X/\ker(p_i)$.
I want to prove the following statement:
A topological vector space $(X,\tau)$ is a locally convex space, if it contains a neighborhood basis of $0$ of convex sets.
Proof of the statement: Every convex neighborhood of $0$ contains an open convex balanced neighborhood of $0$. Therefore one can find a neighborhood basis of $0$ of open convex balanced sets, which will be denoted as $\mathcal{B}^{disc}$. Then one can prove, that the family of Minkowski functionals
$$\{p_U: U\in\mathcal{B}^{disc}\}$$
is a separating family of seminorms, which are continuous with respect to $\tau$. Moreover, for any $U\in \mathcal{B}^{disc}$, it holds $U=\{x\in X: p_U(x)<1\}.$
Now, it is only left to show, that the initial topology with respect to the quotient mappings $q_U: X \rightarrow X/\ker(p_U)$, $U\in \mathcal{B}^{disc}$ coinicides with $\tau$.
For this, I am not sure, if my proof for $\tau\subset \tau^{init}$ is right. Proof:
$\tau^{init}\subset \tau:$ For any $U\in \mathcal{B}^{disc}$, the seminorm $p_U$ is continuous with respect to $\tau$. Then, for any $U\in \mathcal{B}^{disc}$, $\epsilon >0$, $x\in X$, one finds $B^U_\epsilon(x)=(\epsilon U)+x\in\tau$, since translation and scalar multiplication are homeomorphisms, i.e. all subbase sets of $\tau^{init}$ are contained in $\tau$, such that $\tau^{init}\subset \tau$.
$\tau\subset \tau^{init}:$ Pick any $V\in \tau$. For any $x\in V$, the set $V-x$ is an $\tau-$open neighborhood of $0$, since translations are homeomorphisms. This means, for any $x\in V$, one can find a $\tau-$open $U^x\in\mathcal{B}^{disc}$, such that $U^x\subseteq V-x$, leading to $U^x+x\subseteq V$ and $\cup_{x\in V}(U^x+x) = V$. However, any $U^x+x\in \tau^{init}$, since
$$U^x+x = \{y+x\in X: p_{U^x}(y)<1\}=\{z\in X: p_{U^x}(z-x)<1\}=B_1^{U^x}(x)\in \tau^{init}.$$
This yields
$$V = \cup_{x\in V}(U^x+x) = \cup_{x\in V}B_1^{U^x}(x)\in \tau^{init}.$$
Here is what is confusing for me overall: I have shown, that $\tau^{init}\subset \tau$ and that any set in $\tau$ can be written as a union of sets in of the subbase of $\tau^{init}$. However, wouldn't this imply, that any set of the initial topology could be written as union of sets of the subbase? This shouldn't be true, since a subbase is not necessarily a base, right? Is there anything wrong with my proof, or am I just running in circles for another reason?
Further thoughts on this: Please correct me, if I am wrong with this. If $\mathcal{B}^{disc}$ is a open neighborhood basis of $0$, then $\{U+x:U\in \mathcal{B}^{disc}, x\in E\}$ should be a base of $\tau$. (Again using, that translation with $x$ is a homeomorphism.) Once again, you can rewrite $$U+x = \{y+x\in X: p_U(y)<1\}=\{z\in X: p_U(z-x)<1\}=B^U_1(x),$$ such that the mentioned subbase contains a base for $\tau$. Since $\tau^{init}\subset \tau$, it is a base for $\tau^{init}$. But then it seems, that the definition I stated above is slightly more general? Is there some definition for something like a convex neighborhood subbasis ?
Any hint or help is appreciated! Thank you in advance!
Best Answer
The proof of $\tau=\tau^{init}$ looks correct.
In the proof you are not writing $V$ as a union of elements of $\mathcal B$, where $$\mathcal B:=\{U+x\,|\, U\in \mathcal B^{disc}, x\in X\}.$$ The family $\mathcal B$ is actually a full basis, so there is no contradiction.
Edit. Maybe I know where your confusion arises. The subbase you mentioned at the beginning is in general not a base (e.g., take the weak topology on a Banach space $X$, where you can choose the $p_i$ to be elements of $X^*$), but sometimes it can be a base. In fact (I am fairly sure), for a given TVS $X$ with associated seminorms $\{p_j\}$, you can always choose a new family of seminorms $\{r_j\}$ such that the resulting subbase is in fact a base of the topology. In the proof of the above statement, the seminorms $\{p_j\}$ that are being generated with that method give birth to a full base of $(X,\tau)$, not simply a subbase (this fact is clear from the proof).