Step functions are not dense in $L_\infty$

Question

I have seen here here that step functions are not dense in $L_\infty$.

I can't seem to understand why necessarily $||f-s||_\infty \ge\frac{1}{2}$.

Any help would be appreciated.

Answer 1

Any step function $s$ is constant, say $=c,$ on $(0,\epsilon)$ for some $\epsilon>0.$ Hence $\|f-s\|_\infty\ge\max(|0-c|,|1-c|)\ge\frac12.$