Proof of Hahn Banach Theorem corollary: the dual of a LCS sperates points I am having troubles in understanding the proof of some corollaries to the Hahnn Banach Theorem, from Treves' "Topological Vector Space, Distribution and Kernels"

Now summing up my doubts:


*

*In corollary 1, where is the hypothesis of local convexity exploited? Why  $E/M_0$ Hausdorff automatically implies $f'$ continuous?

*In corollary 2 where is the hypothesys of Hausdorfness used ? The
only reason I can come up with, adapting the proof of Corollary 1, as
the author suggests, is that if $E$ is Hausdorff, then
$$E/Cl\{0\}=E/\{0\}=E$$, and hence the proof is simplified, we do not
have to mess with the projection $\phi$. But why is it essential?

*Finally, Schechter in his Handbook of Analysis and its foundations,
claims the Corollary 2 is actually equivalent, and hence not
properly a corollary,  to the Hahn Banach theorem.  Any direct way to
see this ?  


For reference, I am using the following two versions of Hahn Banach (those of Treves).  
[(HB1): Geometric Hahn Banach Theorem] Let E be a TVS, M a linear manifold in E and A a nonempty convex open subset of E such that $M\cap A=\emptyset$. Then there exists an hyperplane containing M and not intersecting A. 
[(HB2): Analytical Hahn Banach Theorem] 
Let E be a vector space, p  a seminorm on E and M a subspace of E. If f is a linear form on M such that $|f(x)|\leq p(x)\ \forall x\in M$ then there exist a linear extension to E such that $|f_1(x)|\leq p(x)\ \forall x\in E$
 A: I'll take real spaces for simplicity.

As for why local convexity is needed in the corollaries: The proof of Corollary 1 states "In virtue of the Hahn-Banach theorem[...]". But notice that geometric HB deals with an open convex set, and analytical HB deals with a seminorm, none of which we have during the proof! So what's the deal? Some details are being ommited. When we expand them, the necessity for local convexity becomes clear:


*

*Let us try to use geometric HB: By local convexity, and since $x_0$ does not belong to the closure $\overline{M_0}$, there exists a convex open set containing $x_0$ and which does not intersect $M_0$. Then we can use geometric HB directly, separating $\overline{M_0}$ and $x_0$ by a closed hyperplane $N$. So the quotient $M/N$ is a Hausdorff one-dimensional TVS, hence topologically isomorphic to $\mathbb{R}$ (Treves, Theorem 9.1(a)). Let $g\colon M/N\to\mathbb{R}$ be a linear topological isomorphism taking $x_0+N$ to $1$. Then the composition of $g$ with the canonical quotient map $M\to M/N$ is nonzero (because it takes $x_0$ to $1$) but vanishes on $M_0$.

*We can also use analytical HB: Since $M$ is locally convex, $M/\overline{M_0}$ is also locally convex. Denote $\phi\colon M\to M/\overline{M_0}$ the quotient map. Since $\phi(x_0)\neq 0$, $M/\overline{M_0}$ is Hausdorff and locally convex, there exists a continuous seminorm $p$ on $M/\overline{M_0}$ such that $p(\phi(x_0))\neq 0$. Then proceed in the same manner as Treves' original proof.
As for why the Hausdorff property implies that $f$ is continuous, there are two options:


*

*You have probably seen that every finite-dimensional vector space admits a unique Hausdorff topological vector space topology. So the topology of $\mathbb{R}\phi(x_0)$ is the one coming from the linear isomorphism $f$, which automatically makes it continuous.

*Alternatively, since we're dealing with locally convex spaces, we can take a continuous seminorm $p$ on $M/\overline{M_0}$ for which $p(\phi(x_0))=1$. Thus $p(\lambda\phi(x_0))=|\lambda|$, from which continuity of $f'$ follows.

To see why we need the Hausdorff property in Corollary $2$, we should try to mimic the proof of Corollary 1: If $x_0\neq 0$, define $f'\colon\mathbb{R}x_0\to\mathbb{R}$ as $\lambda x_0\mapsto x_0$, and extend it by HB to all of $M$. But the problem is that the map $f'$ is the application suppose that $V$ is a non-Hausdorff topological vector space. This means that $\left\{0\right\}$ is not a closed set. Let $y_0$ be any nonzero vector in $\overline{\left\{0\right\}}$. Since $\overline{\left\{0\right\}}$ is the closure of a subspace, the it is a subspace as well, so the line $\mathbb{R}y_0$ passing through $y_0$ is contained in it. Moreover, the zero vector $0$ is dense in $\mathbb{R}y_0$, from which you can conclude that the subspace topology of $\mathbb{R}y_0$ is the undiscrete one: Only $\varnothing$ and $\mathbb{R}y_0$ are open. Therefore the map $\lambda y_0\mapsto y_0$ is not continuous on $\mathbb{R}y_0$.

As for the last question, I do not know any direct way to see why the separation of points property implies the usual Hahn-Banach theorems. I tried using Minkowski functionals, but to no avail.
However, I will assume all proof of Schechter's book are correct, although I did not check them. He does indeed provide the equivalence of 26 forms of Hahn-Banach (all of which are, thus, equivalent weaker forms of the Axiom of Choice). They are numbered $(HB1)-(HB26)$. The analytic HB is (HB2); geometric HB is (HB18); Corollary 2 is (HB22)
I will not write down the statements, but the way their equivalence is proven is:


*

*p. 322-323:
$$(HB1)\to(HB2)\to(HB3)\to(HB1)$$

*p. 322:
$$(HB2)\to(HB4)\to(HB5)\to(HB1)$$

*p. 618-619:
$$(HB2)\to(HB7)\to(HB8)\to(HB9)\to(HB11)\to(HB12)\to(HB1)$$

*p.618-619:
$$(HB8)\to(HB10)\to(BH11)$$

*p. 620-621:
$$(HB12)\to(HB13)\to(HB14)\to(HB12)$$

*p. 717:
$$(HB15)\leftrightarrow(HB2)\leftrightarrow(HB16)$$

*p. 754:
$$(HB4)\to(HB17)\to(HB18)\to(HB19)\to(HB20)\to(HB21)\to(HB22)\to(HB9)$$

*p. 754:
$$(HB20)\to(HB23)\to(HB11)$$

*p. 760:
$$(HB20)\to(HB24)\to(HB21)$$

*p. 805-806:
$$(HB2)\to(HB25)\to(HB26)\to(HB12)$$
So the shortest path from (HB22) to either (HB2) or (HB18) is
$$(HB22)\to(HB9)\to(HB11)\to(HB12)\to(HB1)\to(HB2)$$
The first implications are basically trivial, however (HB12) deals with "Luxembourg's measure" and (HB1) with Banach limits, so there is no easy way to adapt his proof.
