# Sequence of functions converges pointwise to identity [closed]

Let

For $$n\in \mathbb{N}$$ and $$k\in \{0, 1, 2, ..., 2^{n}-1 \}$$ is defined

$$I_{k}^{n}=\left[\frac{k}{2^{n}}, \frac{k+1}{2^{n}}\right)$$

and $$f_{n}:[0, 1) \rightarrow \mathbb{R}$$ is defined by

$$f_{n}(x)=\sum_{k=0}^{2^{n}-1} \frac{k}{2^{n}}1_{I_{k}^{n}}(x)$$

Where for $$A\subseteq \mathbb{R}$$, the function $$1_{A}$$ is the characteristic of $$A$$ such that $$1_{A}(x)=1$$ if $$x\in A$$ and $$1 _{A}(x)=0$$ if $$x\notin A$$.

The idea is to prove that $$f_n \rightarrow f$$ in $$[0, 1)$$. So, let's fix $$x_0 \in [0, 1)$$. We need to show by definition that

$$\lim_{{n \to \infty}} f_n(x_0) = f(x_0) = x_0.$$

Given $$\epsilon > 0$$, we must find $$n_0 \in \mathbb{N}$$ such that if $$n \geq n_0$$, then

$$\left| \sum_{{k=0}}^{{2^n-1}} \frac{k}{2^n}1_{I_k^n}(x_0) - x_0 \right| < \epsilon.$$

Now, let's consider this by cases:

If $$x_0 = 0$$, then

$$\lim_{{n \to \infty}} f_n(x_0) = \lim_{{n \to \infty}} \sum_{{k=0}}^{{2^n-1}} \frac{k}{2^n}1_{I_k^n}(x_0) = 0 = x_0.$$

So,

$$\left| \sum_{{k=0}}^{{2^n-1}} \frac{k}{2^n}1_{I_k^n}(x_0) - x_0 \right| = 0 < \epsilon.$$ And for any choice of $$n_{0}$$ it would be enough.

Now, I am unable to determine this limit for $$0 < x < 1$$. The behavior of the sequence of functions in that interval is somewhat unusual. I attempted to find some values in that interval, as described in the following image:

I don't see how to determine the limit of $$f_{n}$$ at those points, any suggestions? I appreciate it!

• This question would better for math.stackexchange.com Sep 27, 2023 at 2:13
• You don't need to think in terms of functions. By definition, for any real $x$ one has $0\le2^nx-\lfloor 2^nx\rfloor<1$ and you just divide by $2^n$ and get a limit. Sep 27, 2023 at 7:10

This is a classical approximation strategy in "Real Analysis". Recall that $$\mathbf{1}_{I_k^n}(x) = \begin{cases} 0,& x\not\in I_k^n\\ 1, & x\in I_k^n\\ \end{cases}$$ in your notations. In your setting, $$[0,1) = \bigsqcup_{i=0}^{i=2^{n}-1} I_k^n$$ so one only need check every segement $$I_k^n$$. but the value of $$f_n$$ on $$I_k^n$$ is its left endpoint. Then for all $$x\in [0,1)$$ one have
$$\lim_{n\to+\infty} \big| f_n (x) - x \big| = \lim_{n\to+\infty} \big|\sum_{i=0}^{i=2^n-1}\frac{k}{2^n}\mathbf{1}_{I_k^n} (x) - x\big| = \lim_{n\to+\infty} |\frac{k_{x,n}}{2^n} - x|$$
where $$x \in I_{k_{x,n}}^n$$ for each $$n=1,2,...$$. Then
$$\lim_{n\to+\infty} \big| f_n (x) - x \big| = \lim_{n\to+\infty} \big|\sum_{i=0}^{i=2^n-1}\frac{k}{2^n}\mathbf{1}_{I_k^n} (x) - x\big| = \lim_{n\to+\infty} |\frac{k_{x,n}}{2^n} - x| \le \lim_{n\to+\infty}\frac{1}{2^n} = 0$$
So for all $$x\in [0,1)$$ $$\lim_{n\to+\infty} f_n (x) = x$$.