# Sequences of Simple functions

Let $$f(x)=x^2$$ on $$E=[0,1]$$. Provide an explicit increasing sequence of nonnegative simple functions $$\varphi_n(x)$$ which converges pointwise to $$f(x)$$.

I'm having a hard time finding this sequence of simple functions, any help would be greatly appreciated.

The series of functions I would use would be defined on $$n\in\mathbb N$$ as $$\varphi_n(x) = \bigg(\frac{\lfloor x\cdot2^n\rfloor}{2^n}\bigg)^2$$It's obvious by properties of the floor function that each $$\varphi_n$$ is simple.

Moreover, if $$\varphi_n(\alpha) = y$$, then $$\bigg(\frac{\lfloor \alpha\cdot2^n\rfloor}{2^n}\bigg)^2=y$$Assume for the sake of contradiction that $$\varphi_{n+1}(\alpha) = zThis means that $$\bigg(\frac{\lfloor \alpha\cdot2^n\rfloor}{2^n}\bigg)^2<\bigg(\frac{\lfloor \alpha\cdot2^{n+1}\rfloor}{2^{n+1}}\bigg)^2$$$$\frac{\lfloor \alpha\cdot2^n\rfloor}{2^n}<\frac{\lfloor \alpha\cdot2^{n+1}\rfloor}{2^{n+1}}$$$$2\lfloor \alpha\cdot2^n\rfloor<\lfloor \alpha\cdot2^{n+1}\rfloor$$If we set $$\alpha\cdot 2^n = \lambda$$ $$2\lfloor\lambda\rfloor<\lfloor2\lambda\rfloor$$which contradicts the properties of the floor function. This proves that $$\varphi_n$$ is strictly increasing.

You can prove that this sequence of functions converges to $$x^2$$ pretty easily by yourself.

Consider finer and finer partitions of $$[0,1]$$ in subintervals and assign every interval the square of some of its elements.

E.g.

$$[0,1]=\bigcup_{k=0}^{n-1}\left[\frac kn,\frac{k+1}n\right)\cup \{1\},$$

$$\left[\frac kn,\frac{k+1}n\right)\to\frac{k^2}{n^2}.$$

In other words,

$$\phi_n(x)=\left(\frac{\lfloor nx\rfloor}{n}\right)^2.$$

• This is incorrect. Suppose I take $x=\frac13$. $$\varphi_3(x) = \frac19$$$$\varphi_4(x)=\frac1{16}$$This fails the strictly increasing requirement. – Don Thousand Oct 9 '18 at 22:35
• @RushabhMehta: yep, I missed the strictly increasing requirement. – Yves Daoust Oct 10 '18 at 7:20

You can also use an explicit general result, that will work for $$any$$ non-negative function $$f$$, by considering inverse images of slices of the range of $$f$$:

$$E_{nj}=\left \{ x\in \mathbb R:\frac{j-1}{2^{n}}\le f(x)\le \frac{j}{2^{n}} \right \};\ F_n=\left \{x\in \mathbb R: f(x)>n \right \}$$

Now , if we define

$$\phi_n(x)=\sum_{j=1}^{n2^{n}}\frac{j-1}{2^{n}}\chi_{E_{nj}}(x)+n\chi_{F_n}(x)$$

where $$\chi_A$$ is the characteristic function on $$A$$, it's easy to see (draw a picture!) that the $$\phi_n$$ increase to $$f$$.