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!
Best Answer
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 $(0,0)$.