Is the unit ball on the set of continuous functions of a space $X$ strictly convex?

Question

I have been trying to show that $C(X)$ is not strictly convex but I have been having a tough time, any help would be appreciated.

Answer 1

I assume that your $X$ is compact (so that $\|f\|_\infty$ is defined for all continous $f\colon X\to\Bbb R$). If we additionally assume that $X$ allow the existence of non-constant continuous functions (e.g., $X$ is not endowed with the indiscrete topology), then the following works:

Let $f\colon X\to\Bbb R$ be non-constant continouus. Then $\|f\|\ne 0$ and $g:=\frac1{\|f\|}f$ has norm $1$. Then there exists $x_0\in X$ with $g(x_0)=\pm 1$. Define $h(x)=g(x_0)$. Then all convex combinations of $g$ and $h$ have norm $1$.

Answer 2

For $X = \mathbb{R}$, take a function $f$ which is $0$ before $-2$, strictly increase until $-1$ where $f(-1)=1$, which is constant until $0$, and is even. Take $g(x) = f(x-1)$. Then $f \neq g$, $\sup |f| = \sup|g| = \sup \dfrac{|f+g|}{2}=1$, thus the unit ball in $C(\mathbb{R})$ is not strictly convex.

If you can construct such similar functions on $X$, then the answer will be the same