Hardy-Littlewood strong type estimate, final equation of proof

Question

I am reading https://en.wikipedia.org/wiki/Hardy–Littlewood_maximal_function proof of strong type estimate and I understand everything but this last equation:

$$2Cp \int_0^\infty \int_{\lvert f \rvert >\frac{1}{2}}t^{p-2} \lvert f \rvert dxdt=C_p \lVert f \rVert_p^p$$ where $C_p$ depends only on $p$ and $d$ where $f$ domain is $R^d$

How can I see it? I think it has to have something in common with the Cavalieri principle : $$\int_{R^d}\lvert f \rvert ^p dx=p\int_0^\infty t^{p-1}m(f>t)dt$$

where $m$ is $d$ dimensional Lebesgue measure, but I can't derive this particular equation.

Answer 1

I will try to extend my comment to an answer. I will call this integral as $I$. Use Fubini's or Tonelli's Theorem to interchange the integrals, we get

$$I=2C_p\int_{\mathbb{R}^n}|f|\int_0^{\infty}t^{p-2}\chi_{ \{|f|>\frac{t}{2}\} }dtdx.$$

Now put $t^{p-1}=2^{p-1}u$

$=>(p-1)t^{p-2}dt=2^{p-1}du$. We get

$$I=\frac{2^pC_p}{p-1}\int_{\mathbb{R}^n}|f|dx\int_0^{\infty}\chi_{ \{|f|^{p-1}>u \} }du.$$

Use Layer cake representation of a function i.e.,

$$|f(x)|=\int_0^{\infty}\chi_{ \{|f|>t \} }dt.$$

We have,

$$I=\frac{2^pC_p}{p-1}\int_{\mathbb{R}^n}|f|^pdx.$$

Answer 2

Substituting this into our integral gives:

$$ \|Mf\|_p^p \leq p \int_0^\infty t^{p-1} \left( \frac{2(5^d)}{t} \int_{|f| > \frac{t}{2}} |f|\, dx \right) dt. $$

This simplifies to:

$$ = 2(5^d)p \int_0^\infty \int_{|f| > \frac{t}{2}} t^{p-2} |f|\, dx\, dt. $$

$$ = 2(5^d)p \int_0^\infty \int_{\mathbb{R}^d} \chi_{\{x:|f(x)| > \frac{t}{2}\}}(x) t^{p-2} |f|\, dx\, dt. $$

Now, we change the order of integration again (applying Tonelli's theorem):

$$= 2(5^d)p\int_{\mathbb{R}^d} \int_0^\infty \chi_{\{x:|f(x)| > \frac{t}{2}\}}(x) t^{p-2} |f| \, dt dx. $$

Observe that $$\chi_{\{x:|f(x)| > \frac{t}{2}\}}(x) = \chi_{\{t:|f(x)| > \frac{t}{2}\}}(t)$$

then,

$$= 2(5^d)p\int_{\mathbb{R}^d} \int_0^\infty \chi_{\{t:|f(x)| > \frac{t}{2}\}}(t) t^{p-2} |f| \, dt \, dx. $$

$$ = 2(5^d)p \int_{\mathbb{R}^d} \left( \int_0^{2|f(x)|} t^{p-2}\, dt \right) |f(x)|\, dx. $$

The inner integral evaluates to:

$$ \int_0^{2|f(x)|} t^{p-2}\, dt = \frac{1}{p-1} (2|f(x)|)^{p-1}. $$

Thus, our expression becomes:

$$ = \frac{2 (5^d)p}{p-1} \int_{\mathbb{R}^d}|f(x)| (2|f(x)|)^{p-1}\, dx. $$

Simplifying:

$$ = \frac{2^{p}(5^d)p}{p-1} \int_{\mathbb{R}^d}|f(x)|^p\, dx. $$

Recognizing the right-hand side as $ \|f\|_p^p $, we get:

$$ \|Mf\|_p^p \le \frac{2^{p} (5^d) p}{p-1} \|f\|_p^p. $$

Finally, defining $ C_p = \frac{2^{p}(5^d)p}{p-1} $, we obtain the desired inequality:

$$ \|Mf\|_p \le (C_p)^{1/p} \|f\|_p. $$

Answer 3

The maximal inequality $$ \lambda(\{M\,f>t\})\leq\frac{3^d}{t}\|f\|_1 $$ for all $f\in L_1$ plays a role here, as well as the fact that $\|Mf\|_\infty\leq\|f\|_\infty$ for all $f\in L_\infty$. This last inequality is easily established by noticing that $$\frac{1}{|B(x;r)|}\int_{B(x;r)}|f(x)|\,dx\leq\|f\|_\infty$$ for all $r>0$.

For $f\in L_p$, $1<p<\infty$, one can use the trick of splitting $f=g+h$ where $g\in L_1$ and $h\in L_\infty$ and use the $L_1$ and $L_\infty$ results along with applications of Fubini's theorem (Cavalieri's principle if you will). This technique is common in getting string bounds in maximal inequalities. One may introduce some additional parameters to find optimal bounds. Here is the general idea at work (from Rudin, W., Real and Complex Analysis, 3rd edition, McGrawHill, 1986, pp. 173-174):

If $1<p<\infty$, then for $0<c<1$ and $t>0$, let \begin{align} g_t= f\mathbb{1}_{\{f>ct\}},\qquad h_t=f\mathbb{1}_{\{f\leq ct\}}. \end{align} Clearly $h_t\in L_\infty$ and, from Chebyshev's and Hölder's inequalities, we also have that $g_t\in L_1$. Hence, $M\,f\leq M\, g_t + M\, h_t\leq M\, g_t + ct$, and so, $\{M\,f>t\}\subset\{M\,g_t>(1-c)t\}$. By Hardy--Littlewood's inequality \begin{align} \lambda(\{M\,f>t\})\leq \lambda(\{M\,g_t>(1-c)t\})\leq \frac{3^d}{t(1-c)}\|g_t\|_1= \frac{3^d}{t(1-c)}\int_{\{f>ct\}} f(x)\,dx \end{align} An application of Fubini's theorem shows \begin{align*} \|M\,f\|^p_p &= p\int^\infty_0t^{p-1}\lambda(Mf>t)\,dt\\ &\leq \tfrac{3^dp}{1-c} \int^\infty_0 t^{p-2}\int_{\{f> ct\}}f(x)\,dx\,dt\\ &=\tfrac{3^dp}{1-c}\int_{\mathbb{R}^d}f(x)\int^{f/c}_0t^{p-2}\,dt\,dx = \tfrac{3^dp c^{1-p}}{(p-1)(1-c)}\int_{\mathbb{R}^d} f^p(x)\,dx \end{align*}

One can minimize the bound $\tfrac{3^dp c^{1-p}}{(p-1)(1-c)}$, as a function of $c$ to get an optimal bound $C_p=(3^dpp'e)^{1/p}$, where $\frac1p+\frac{1}{p'}=1$.