Example of a topological vector space which is not locally convex

Question

I'm currently studying Functional Analysis and the professor gave an example for a TVS (which we have defined to be a vector-space $X$ in which addition $X \times X \rightarrow X, (x, y) \mapsto x + y$ and scalar-multiplication $\mathbf{R} \times X \rightarrow X, (\lambda, x) \mapsto \lambda x$ are continuous), which is not locally convex. The example was the following:

Let $L^0([0, 1])$ denote the set of measurable functions $f : [0, 1] \rightarrow \mathbf{R}$ modulo equivalence almost-everywhere for some measure $\mu$. We make this a metric-space by defining:

$$d(f, g) = \int \frac{\vert f - g \vert}{1 + \vert f - g \vert} d \mu$$

and the then claimed that this is a TVS with the topology induced by $d$. I wanted to check this, and addition is not too big an issue, but I got stuck on scalar-multiplication. I'd appreciate some help on this.

He went on explaining that convergence in $d$ of a sequence $(f_n)_{n \in \mathbf{N}}$ is convergence in measure.

The exercise he gave us then (and which would be my question) was: Any non-empty open convex set $A$ in $L^0([0, 1])$ is equal to the whole space.

I have very little idea on how to do this. My idea would have been to pick an element $g \notin A$ and then choosing a sequence $(g_n)_{n \in \mathbf{N}} \subset X - A$ converging to $g$. This may give me some contradiction, but I really do not see how to use the convexity of $A$.

Thanks for any help!

Answer 1

It depends on the measure. For a measure with finite support the space is finite dimensional and hence locally convex.

Let's thus take $\mu$ as the Lebesgue-measure on $[0,1]$ and assume that $0\in A$. Given $g\notin A$ the idea is to write for given $n$ $$ g=\frac 1n \sum_{k=1}^n ng \chi_k $$ where $\chi_k$ is the indicator function of the interval $[(k-1)/n,k/n)$ (to be precise add the right end point to $I_n$). It remains to see that terms of the sum are in $A$ for large $n$ which is easy since $A$ is a neigborhood of $0$ and $g_n\to 0$ if the measure of the support of $g_n$ tends to $0$.

For the continuity of the multiplication it may help to note that (as for any bilinear map) it is enough to check continuity at $(0,0)$.

Answer 2

The proof of this fact is an application of the theorem of Hahn Banach:

Hahn Banach: Let $X$ be a complete topological vector space and $C$ an open convex subset of $X$, suppose that there exists $x\in X$ which is not in $C$, then there exists a closed hyperplane $H$ which contains $x$ and which is disjoint with $C$.

Generally strategy used to solve this question is to show that every continuous linear functional on $X$ is zero. If this fact is true. If you take any open convex subset $C$ of $X$, and $x\neq C$, a closed hyperplane $H$ which contains $X$ and is disjoint with $C$ is defined by a linear bounded function $l$ and is equal to $x+ker(l)$ but $l$ does not exists so $H$ can't exists.

To show that every bounded linear functional $l$ vanishes on $X$, you use the characteristic function: remark that if $E$ is a set such that $\mu(E)\leq \delta$ for every $c>0$, $c\chi(E)\in B(0,\delta)$ this implies that every bounded linear functional defined on $X$ vanishes. To see this recall that there exists $M>0$ such that for every $x\in B(0,\delta), \mid l(x)\mid <M$. Let $E$ such that $\mu(E)<\delta$ for every integer $n>0$, $n\chi(E)\in B(0,\delta)$ implies that $\mid l(n\chi(E)\mid=n\mid l(\chi(E))\mid<M$. This implies that $l(\chi(E))=0$, if $\mu(E)<\delta$ henceforth, $l=0$.

Answer 3

soit E=C(Ω) l espace de fonction continues. montre que le system. de voisinagesvk.ε={ x : IX(:t)II< ε} Ω parte en Rn. pouur tout compact K don Ω definie sur e une stucture espace vevtoriel toppligesue e.v.t