Uniform convergence of $f_n(x)$ defined on unit interval

Question

Consider a function $f$ defined on $[0,1] $ to the reals and define $$f_n(x)=\frac{\lfloor nf(x)\rfloor}{n}$$ Show that $f_n$ converges uniformly to $f.$

I'm having trouble dealing with the floor function part. Is there a useful identity for $\lfloor nf(x)\rfloor$? I need to show:

If for every $\epsilon>0$ there exists a natural number $n>N$ and $x\in[0,1]$ and $$\bigg|\frac{\lfloor nf(x) \rfloor}{n}-f(x)\bigg|<\epsilon$$ Where can I go from here?

Answer 1

For each $x\in[0,1]$, $\bigl\lfloor nf(x)\bigr\rfloor\leqslant nf(x)<\bigl\lfloor nf(x)\bigr\rfloor+1$ and therefore$$\frac{\bigl\lfloor nf(x)\bigr\rfloor}n\leqslant f(x)<\frac{\bigl\lfloor nf(x)\bigr\rfloor}n+\frac1n.\tag1$$So, given $\varepsilon>0$, take $N\in\mathbb N$ such that $\frac1N<\varepsilon$ and it will follow from $(1)$ that$$n\geqslant N\implies\left\lvert f(x)-\frac{\bigl\lfloor nf(x)\bigr\rfloor}n\right\rvert<\varepsilon.$$This proves that $(f_n)_{n\in\mathbb N}$ converges uniformly to $f$.

Answer 2

Hint:

$$n\, f(x) -1 \lt \lfloor n\, f(x)\rfloor \le n\, f(x)$$

so $$\dfrac{n\, f(x) -1}{n} \lt \dfrac{\lfloor n\, f(x)\rfloor}{n} \le \dfrac{n\, f(x)}{n}$$

i.e. $$f(x) -\dfrac{1}{n} \lt \dfrac{\lfloor n\, f(x)\rfloor}{n} \le f(x)$$

Now consider $n \gt N \ge \frac1\epsilon$