$h_c(x) = \sup \{n^c f(nx):n=1,2,3,\dots\}$ is in $L^1(R)$ if $0<c<1$.

Question

This is exercise 7.19 from Rudin's RCA.

Suppose $f$ is continuous on $R$, $f(x)>0$ if $0<x<1,$ $f(x)=0$ otherwise. Define $$h_c(x) = \sup \{n^c f(nx):n=1,2,3,\dots\}.$$ Prove that $h_c$ is in $L^1(R)$ if $0<c<1$.

I am stuck with this problem. How can I prove the integrability for this function? I would greatly appreciate any help.

Answer 1

We can see that $h_c$ is supported on $[0,1]$. When you look at the definition of $h_c(x)$, the largest $n$ for which $f(nx) \neq 0$ is $1/x$. That means that $h_c(x) \leq \frac{1}{x^c}||f||_{\infty}$.

Now when integrating, we know that $\int_0^1 h_c(x) dx \leq \int_0^1 \frac{1}{x^c}dx ||f||_{\infty} = \frac{1}{1-c}||f||_{\infty}$, this works for $0<c<1$. For $c=1$ it wouldnt work as the integral diverges.

Answer 2

We know that $f$ is Borel measurable since it is continuous on $\mathbb{R}$, hence $n^c f(nx)$ is Borel measurable, hence $h_c(x)$ is Borel measurable. By Rolle's Theorem and the function defintion, a maximal value $M$ must exist when $x \in (0,1)$.

If $h_c$ $x\in (0,1)$ then $0<nx<1 \iff 0<n<1/x $. Hence when $0<c<1$ \begin{equation*} \begin{aligned} \sup_{n\ge 1}\{ n^cf(nx) \} \le (1/x)^c M < M \end{aligned} \end{equation*}

If $x\notin (0,1)$ then $h_c(x)=0$.

Hence \begin{equation*} \begin{aligned} \int_\mathbb{R} |h_c(x)| \,dx = \int_0^1 h_c(x) \,dx &< \int_0^1 M \,dx = M \\ \end{aligned} \end{equation*} Hence $h_c(x) \in L^1(\mathbb{R})$ if $0<c<1$.