Show that $\phi_n(x) \leq \phi_{n+1}(x)$ (Folland Theorem 2.10)

Question

I am reading the proof of Theorem 2.10 in Folland's Real Analysis.

I'm stuck to show the sentence:

It is easily checked that $$\phi_n \leq \phi_{n+1}$$ for all $n$.

I first noticed that $E_{n}^{k}=E_{n+1}^{2k} \mathop{\dot{\cup}} ~ E_{n+1}^{2k+1}$.

Indeed,

\begin{eqnarray*} x \in E_{n}^{k} &\Leftrightarrow& k2^{-n} <f(x) \leq (k+1)2^{-n}\\ &\Leftrightarrow& 2k2^{-(n+1)} <f(x) \leq (2k+2)2^{-(n+1)}\\ &\Leftrightarrow& f(x) \in (2k2^{-(n+1)}, (2k+1)2^{-(n+1)}] \mathop{\dot{\cup}} ((2k+1)2^{-(n+1)},(2k+1+1)2^{-(n+1)}]\\ &\Leftrightarrow& x \in E_{n+1}^{2k} \mathop{\dot{\cup}} ~ E_{n+1}^{2k+1}. \end{eqnarray*}

I would like to write $$\phi_n(x)=\sum_{k=0}^{2^{2n}-1}k2^{-n} \chi_{E_{n}^{k}}+\chi_{F_n} \hbox{ and } \phi_{n+1}(x)=\sum_{k=0}^{2^{2(n+1)}-1}k2^{-(n+1)} \chi_{E_{n+1}^{k}}+\chi_{F_{n+1}}$$ and compare the values of $\phi_n(x)$ and $\phi_{n+1}(x)$.

But I don't know how to proceed.

Answer 1

Notice that if $0\leq f(x)<2^{2n}$, $\phi_n(x)$ is defined as $2^{-n}k$ where $k$ is the integer such that $k\leq 2^nf(x)<k+1$, that is $k=\lfloor 2^n f(x)\rfloor$, where $\lfloor \cdot\rfloor$ is the maximum integer function: $\lfloor t\rfloor=k\in\mathbb{Z}$ iff $k\leq t<k+1$. This all means that $$\phi_n=\min\big(2^{-n}\lfloor 2^n f\rfloor,2^n\big)$$

Now

• If $0\leq f(x)<2^{2n}$ and $k=\lfloor 2^nf(x)\rfloor$ then $2^{-n}k\leq f(x)<2^{-n}(k+1)$. Then, either
  $$2^{-n-1}2k=2^{-n}k\leq f(x)<\big(2^{-n}(k+1)+2^{-n}k\big)/2 = 2^{-n-1}(2k+1)$$ or $$2^{-n-1}(2k+1)\leq f(x)<2^{-n}(k+1)=2^{-n-1}(2k+2)$$ That is $2k=\lfloor 2^{n+1}f(x)\rfloor$, or $2k+1=\lfloor 2^{n+1}f(x)\rfloor$. This in turn means that either $$\phi_{n+1}(x)=2^{-n-1}\lfloor 2^{n+1}f(x)\rfloor=2^{-n}k=\phi_n(x)$$ or $$\phi_{n+1}(x)=2^{-n-1}\lfloor 2^{n+1}f(x)\rfloor=2^{-n-1}(2k+1)=2^{-n}k+\frac12> 2^{-n}k=\phi_n(x)$$
• If $f(x)\geq 2^{2n}$ then $2^{n+1}f(x)\geq 2^{3n+1}$ and so, $\lfloor 2^{n+1}f(x)\rfloor \geq 2^{3n+1}$. Consequently $$\phi_{n+1}(x)=\min\big(2^{-n-1}\lfloor 2^{n+1}f(x)\rfloor, 2^{n+1}\big)\geq 2^{2n}=\phi_n(x)$$