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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.

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  • $\begingroup$ please add your own thoughts $\endgroup$ – qbert Oct 9 '18 at 21:40
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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 $\varphi_{n+1}(\alpha) = z<y$This 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.

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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,

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

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

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