Let $(C,d)$ be the metric space of all continuous complex-valued functions on $[0,1]$ where $$d(f,g)=\int_0^1\frac{|f(x)-g(x)|}{1+|f(x)-g(x)|}dx$$ Show that $(C,d)$ contains no non-empty proper open convex set.
My attempt:
Let $V\neq \emptyset$ be open and convex in $(C,d)$. We need to show $V=C.$ Let $f \in C$. Since $V \neq \emptyset,$ there exists $h \in V.$ Without loss of generality, $h=0.$ Now $V$ is open, therefore there exists $r>0$ such that $B(0,r) \subseteq V$, where $$B(0,r)=\{f \in C:d(f,0)<r\}$$
Now to show $f \in V$, I need to write $f$ as a convex combination of some elements in $V.$ So I need to find $t_i,f_i$ such that $\sum t_i=1$ and $f=\sum t_ig_i$ where $f_i \in V$. It will be sufficient to have $f_i \in B(0,r)$ ie $$\int_0^1\frac{|f_i(x)|}{1+|f_i(x)|}dx<r$$
I'm having trouble finding suitable $f_i.$
If the support of $f_i$ is contained within an interval of length $r$, that guarantees $d(f_i, 0)<r$. So, it suffices to write $f$ as the sum of functions with small support. The key term is partition of unity: have some continuous functions $\phi_i$ with small support, such that $\sum \phi_i\equiv 1$; then $$f = \frac{nf \phi_1+\dots +nf\phi_n}{n} $$ is a desired convex combination.
A typical continuous partition of unity on $[0,1]$ is given by functions $$ \phi_i(x) = \max(0, 1 - n|x-i/n|), \quad i=0,1,\dots,n $$ which are piecewise linear functions that are equal to $1$ at one of the points $i/n$, and $0$ at all other such points.