co.combinatorics – Largest number N for which injective mapping \$f: 2^N to 2^8 times 2^8 times 2^8\$ which is Lipschitz-1 CT with \$Kleq 3\$ exists

I have a function on $$h: (0,1) to (0,1)$$ whose output is smooth (polynomial of low degree), and I need to discretize it but I need to save it with three 8 bit numbers. These three 8 bit numbers need to be ok to interpolate, so basically the definition of Lipschitz continuity.

That is I want to find a way to store my $$xin (0,1)$$ numbers as $$f(x)in 2^8 times 2^8 times 2^8$$ and I would like:

$$|f(x_1)-f(x_2)| leq K |x_1 – x_2|$$

With K not too large, and store as many $$x$$s as possible (I am interested in inverting this map later).

Now, I know how to do it with $$K=3$$ and for $$2^8$$: Just divide $$(0,1)$$ into $${ frac{1,2,…,256}{256} }$$ and have $$f(x) mapsto (256x,256x,256x)$$.

The question is how much better can I do? Can I save $$2^N$$ $$x$$s with a small $$K$$ and large $$N$$? I know I could save $$2^{24}$$ if $$K$$ was huge, but I would like to find a middle ground.

As I understand the folding solution from another question will have a big $$K$$ in my case if we consider neighboring elements in the image of $$f$$ which cut across the folds.