Proof of limiting function

Question

I need a help with a proof.

Let $\displaystyle\phi_n:\mathbb{R}_0^+ \rightarrow \mathbb{R}_0^+:x\mapsto \begin{cases} \frac{\displaystyle\lfloor x\cdot 2^n \rfloor}{\displaystyle2^n},&x<2^n,\\ 2^n, &\text{otherwise}.\end{cases}$

Prove that $\displaystyle \lim_{n\rightarrow \infty} \phi_n = x$.

Could anyone please help me with that? Thank you a lot!

Answer 1

Hint: Write $\lfloor x\cdot 2^n\rfloor=x\cdot2^n-\epsilon_{n,x}$, where $\epsilon_{n,x}\in[0,1)$ for all $x\in\mathbb{R}_0^{+}$ and $n\in\mathbb{N}$.

Answer 2

Hint: How far apart are $x$ and $[2^n x]/2^n$? How far apart are $2^n x$ and $[2^n x]$?

Answer 3

you can use the beautiful result $$[kx]=\sum_{m=0}^{k-1}[x+\frac{m}{k}]$$ here putting $k=2^n$ you get the limit $$\lim_{n\to \infty} \sum_{m=0}^{2^n-1}[x+\frac{m}{2^n}]$$ you observe the $\frac{m}{2^n}$ in every terms tends to $0$.also there are $2^n$ terms in total each equal to $x$. therefore you get $$\frac{2^nx}{2^n}$$ which is $x$.