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In page 39 of the book Fourier Analysis, it reads:

Given a function g on $\mathbb{R}^+ = (0, \infty)$, the Hardy operator acting on $g$ is defined by $$ Tg(t) = \frac{1}{t}\int_0^t g(s)\,ds,\quad t\in\mathbb{R}^+. $$ If $g\in L^1(\mathbb{R}^+)$ is non-negative, then, since $Tg$ is continuous, one can show that $$|E(\lambda)| = \frac{1}{\lambda}\int_{E(\lambda)}g(t) dt, $$ where $E(\lambda) = \{t\in\mathbb{R}^+: Tg(t) \ge \lambda\}$.

How to proof the second equation above?

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2 Answers 2

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I don't have any clever way, but I do it in brute force. My way is to assume first that $g=\chi_{(a,b)}$, and then I guess that one can generalize to any step functions, and since step functions are dense in $L^{1}$, then the assertion should go through.

Now assume that $1-\dfrac{a}{b}\leq\lambda<1$, then $|E(\lambda)|=b-\dfrac{a}{1-\lambda}+\dfrac{b-a}{\lambda}-b=\dfrac{b-a-b\lambda}{\lambda(1-\lambda)}$ and one check that \begin{align*} \dfrac{1}{\lambda}\int_{E(\lambda)}\chi_{(a,b)}(t)dt&=\dfrac{1}{\lambda}\int_{a/(1-\lambda)}^{b}\chi_{(a,b)}(t)dt+\dfrac{1}{\lambda}\int_{b}^{(b-a)/\lambda}\chi_{(a,b)}(t)dt\\ &=\dfrac{1}{\lambda}\int_{a/(1-\lambda)}^{b}\chi_{(a,b)}(t)dt\\ &=\dfrac{b-a-b\lambda}{\lambda(1-\lambda)}. \end{align*} For $\lambda\geq 1$ or $\lambda<1-a/b$ is treated similarly.

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Since $g(t) \in L^1(\mathbb{R}^+)$, then $G(t) = \int_0^t g(s) \,\mathrm{d}s$ is monotonically increasing and bounded. Obviously, $G(t)$ is continous. Denote $y(t) = \lambda t$. Then we have $$E(\lambda) = \{t\in\mathbb{R}^+: Tg(t) \ge \lambda\} = \{t \in \mathbb{R}^+: G(t) \ge y(t) \}.$$ and $$ \lim_{t\to \infty} y(t) - G(t) = +\infty.$$ It is easy to see $t \in E(\lambda)$ has an upper bound. Consider that $G(t)$ and $y(t)$ is contious funtions, $E(\lambda)$ can be expressed as the union of finite closet intervals, i.e., $$E(\lambda) = \cup_{i=1}^N[a_i, b_i].$$ A plot will make it more clear, see below.

On interval $[a_i, b_i]$, we have $$G(b_i) - G(a_i) = \lambda (b_i - a_i),$$ i.e., $$ \frac{1}{\lambda} \int_{a_i}^{b_i} g(s) \,\mathrm{d}s = b_i - a_i.$$ Sum the above equality from $i = 1$ to $i = N$, we can obtain the result.

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