Construct an infinite set of functions such that $\|f_i - f_j\|_\infty \geq 1/2$ for all $i\neq j$

Question

Let $C$ be the set of all convex functions on the unit interval bounded by $1$. Identify a subset of $C$ such that $\|f_i - f_j\|_\infty \geq 1/2$ for all $i\neq j$.

I'm having trouble coming up with an example. My attempts so far have been to consider functions like $1/2^{nx}$ which can be far apart near $0$, but I think I need to choose an appropriate subsequence.

Answer 1

Define $f_n: [0, 1] \to \Bbb R$ as $$ f_n(x) = \max(1-2^n x, 0) \, . $$

All functions $f_n$ are convex on $[0, 1]$, and for $i < j$ is $$ \|f_i - f_j\|_\infty \geq f_i\left(\frac {1}{2^{i+1}}\right) - f_j\left(\frac {1}{2^{i+1}}\right) = \frac 12 - 0 = \frac 12 \, . $$

Remark: This construction can be generalized: For any $a > 1$ are the functions $$ f_n(x) = \max(1-a^n x, 0) \, . $$ convex on $[0, 1]$ and satisfy $\|f_i - f_j\|_\infty \geq 1-1/a$ for $i \ne j$, so the mutual distance can be arbitrarily close to one.

Answer 2

Let $\rho> 1$, and $a_n> 0$ be a geometric progression with ratio $\rho$. For $m\ne n$
$$\|x^{a_n}- x^{a_m}\|_{\infty}= \|t^{\frac{a_n}{a_m}}-t\|_{\infty}\ge \|t^{\rho}-t\|_{\infty}$$ so we got our sequence $(x^{a_n})_n$ ( if $\rho$ large enough).