Construction of a sequence of continuous functions that are pairwise separated by 1 in the uniform norm

Question

I need to construct a sequence of continuous functions $f_n\in C[a,b]$ such that for each $n\neq m \in \mathbb{N}$ we have $\left \| f_n-f_{m} \right \|_\infty = 1$

I can totally understand how to draw such series, but I lack the knowledge on how to construct it.

Heres my drawing:

Hope it's understandable. The upper bound is 1.

Answer 1

WLOG suppose we are working on $[0,1]$. All other cases are similar.

Split $[0,1]$ in $E_n:= [\frac{1}{2n+2}, \frac{1}{2n}]$ and $F_n := [\frac{1}{2n+1.5}, \frac{1}{2n + 0.5}]$. Then $F_n \subseteq E_n$ for each $n$. Then for each $n$, define $f_n: [0,1] \to [0,1]$ by $f_n = 0$ on $E_n^C$, $f_n = 1$ on $F_n$ and $f_n$ is linear in between (i.e. draw a line from the left endpoint of $E_n$ to the left endpoint of $F_n$ and a line from the right endpoint of $F_n$ to the right endpoint $E_n$). Then the $f_n$ are continuous functions that are 0 outside of $E_n$ and $1$ somewhere in $E_n$. But note that $m\neq n$ implies $F_n \cap E_m = \emptyset$ so $\|f_n -f_m\|_\infty \geq 1$ (since for $x \in F_n$, $f_n(x) =1$ but $f_m(x) = 0$). But also since $f_j$ has codomain $[0,1]$, $\|f_n -f_m\|_\infty \leq 1$. So $\|f_n - f_m\|_\infty = 1$.

Edit:

Explicit formula for $f_n$.

$$f_n = \begin{cases} 0 & 0 \leq x \leq \frac{1}{2n+2}\\ \left(\frac{1}{2n+1.5} - \frac{1}{2n+2}\right)^{-1}(x-\frac{1}{2n+2}) & \frac{1}{2n+2} \leq x \leq \frac{1}{2n+1.5} \\ 1 & \frac{1}{2n+1.5} \leq x \leq \frac{1}{2n+0.5} \\ \left(\frac{1}{2n+0.5} - \frac{1}{2n}\right)^{-1}(x-\frac{1}{2n+0.5}) +1 & \frac{1}{2n+0.5} \leq x \leq \frac{1}{2n} \\ 0 & \frac{1}{2n} \leq x \leq 1 \end{cases} $$

Answer 2

Let $$F(x) = \begin{cases} 0 & x\le a \\ \frac{x-a}{b-a} & a \le x \le b \\ 1 & x \ge b \end{cases}$$ And define $$F_n(x)=F[n(x - a) + a]$$ Not exactly like the ones you drew but very close and the idea is the same.