Existence of Non-Trivial, Convex, Open Set in $C_{\mathbb{C}}[0,1]$ Under $L^{0}$ Metric

Question

I've been struggling with this problem for the last four hours. The problem is to show that the space of $\mathbb{C}$-valued continuous functions on $[0,1]$ under the metric $d(f,g)=\int_{0}^{1}\frac{\vert f-g\vert}{1+\vert f-g\vert}dx$ has no non-trivial convex open sets. I refer to d as the $L^{0}$ metric since such a metric is often used on the space of random variables on a probability space.

I've taken two thus far fruitless approaches. In the first approach, I assumed there existed such a non-trivial convex set U then tried to show that any $f\in C_{\mathbb{C}}[0,1]$ can be expressed as convex combination of elements of U.

Second thought was to use the fact that $C_{\mathbb{C}}[0,1]$ is a dense subspace of $L^{p}$, with $0<p<1$, under the $L^{0}$ metric. I wanted to show that since such a set does not exist in $L^{p}$ under this metric then it cannot exist in $C_{\mathbb{C}}[0,1]$. I was able to get as far as showing that these $L^{p}$-spaces have no non-trivial, convex, open sets under this metric.

Answer 1

Suppose $U$ is a nonempty convex open set. By translation, we can assume that $0\in U$. Thus there is a $\delta > 0$ such that $B_\delta(0) = \{ f : d(f,0) < \delta\} \subset U$. Fix an integer $n > 2/\delta$, and for $0 \leqslant k \leqslant n$, let

$$\varphi_k(x) = \begin{cases} n\left(x - \frac{k-1}{n}\right) &, \frac{k-1}{n}\leqslant x \leqslant \frac{k}{n}\\ n\left(\frac{k+1}{n}-x\right) &, \frac{k}{n} \leqslant x \leqslant \frac{k+1}{n}\\ \qquad 0 & x < \frac{k-1}{n} \lor x > \frac{k+1}{n}\end{cases}$$

where, for $k = 0$ resp $k = n$ the part outside $[0,1]$ is of course ignored. Then we have

$$\sum_{k=0}^n \varphi_k(x) = 1$$

for all $x\in [0,1]$, and for any continuous $f\colon [0,1]\to\mathbb{C}$,

$$f = \sum_{k=0}^n \frac{1}{n+1}\cdot \underbrace{\left((n+1)\cdot \varphi_k\cdot f\right)}_{\psi_k}.\tag{1}$$

Now,

$$d(\psi_k,0) = \int_0^1 \frac{\lvert \psi_k(x)\rvert}{1+\lvert\psi_k(x)\rvert}\,dx = \int_{(k-1)/n}^{(k+1)/n} \frac{\lvert \psi_k(x)\rvert}{1+\lvert\psi_k(x)\rvert}\,dx < \int_{(k-1)/n}^{(k+1)/n}\,dx = \frac{2}{n} < \delta$$

for $1 \leqslant k < n$, and adjusting the integral limits to $0$ resp. $1$ for $k = 0$ resp. $k = n$ shows that also $d(\psi_0,0) < \delta$ and $d(\psi_n,0) < \delta$.

Thus $(1)$ shows that $f$ is a convex combination of functions in $B_\delta(0) \subset U$, and since $U$ is convex, it follows that $f\in U$. Since $f$ was completely arbitrary, this shows that $U = C([0,1],\mathbb{C})$. Thus there are only two convex open sets - $\varnothing$ and $C([0,1],\mathbb{C})$ - in the topology defined by $d$.