Hardy inequality

Question

I'm looking for a reference or proof about the classical Hardy's inequality:

$$\int_0^{+\infty} \left(\int_0^xf(t) \,dt\right)^p x^\alpha \, dx \leq C_1 \int_0^{+\infty}f(x)^px^{\alpha+p} \, dx \text{ for } \alpha<-1 \tag{$*$} $$

and

$$\int_0^{+\infty}\left(\int_x^{+\infty}f(t)\,dt\right)^p x^\alpha \, dx \leq C_2 \int_0^{+\infty} f(x)^px^{\alpha+p}\,dx \text{ for } \alpha>-1 \tag{$**$} $$

for some constant $C_1,C_2>0$ independent of $f.$

I looked up wikipedia where there is a similar inequality but I'm not satisfied with the proof given. I also looked up the encyclopedia of mathematics but there a similar inequality is given without proof.

I looked up the classic reference [G.H. Hardy, J.E. Littlewood, G. Pólya, "Inequalities" , Cambridge Univ. Press (1934)], where I found in the chapter named "Hilbert's inequality [...]" at page 245 the identity (330), which is the same as the one I'm interested in except for the fact that $\alpha=-r$ and so the hypothesis becomes $r<1,r>1$.

The discussion preceding the identity, which supposedly subsumes its proof, seems quite lengthy and involved, and I feel there should be a more direct and easy proof.

I would like a precise reference or even some proof written down here.

Answer 1

The convolution inequalities can be rephrased as \begin{align*} \left(\int_{0}^{\infty}\left(\int_{0}^{x}|f(t)|dt\right)^{p}x^{-b-1}dx\right)^{1/p}\leq\dfrac{p}{b}\left(\int_{0}^{\infty}|f(t)|^{p}t^{p-b-1}dt\right)^{1/p}, \end{align*} and \begin{align*} \left(\int_{0}^{\infty}\left(\int_{x}^{\infty}|f(t)|dt\right)^{p}x^{b-1}dx\right)^{1/p}\leq\dfrac{p}{b}\left(\int_{0}^{\infty}|f(t)|^{p}t^{p+b-1}dt\right)^{1/p} \end{align*} for $0<b<\infty$ and $1\leq p<\infty$.

For the first one, consider $h(x)=|f(x)|x^{1-b/p}$ and $k(x)=x^{-b/p}\chi_{[1,\infty)}(x)$ and the convolution under the multiplicative group $(0,\infty)$ with respect to the measure $d(\cdot)/(\cdot)$, we have \begin{align*} h\ast k(x)&=\int_{0}^{\infty}h\left(\dfrac{x}{t}\right)k(t)\dfrac{dt}{t}\\ &=\int_{0}^{\infty}\left|f\left(\dfrac{x}{t}\right)\right|\left(\dfrac{x}{t}\right)^{1-b/p}t^{-b/p}\chi_{[1,\infty)}(t)\dfrac{dt}{t}\\ &=\int_{1}^{\infty}\left|f\left(\dfrac{x}{t}\right)\right|\left(\dfrac{x}{t}\right)^{1-b/p}t^{-b/p}\dfrac{dt}{t}\\ &=\int_{0}^{x}|f(u)|u^{1-b/p}\left(\dfrac{x}{u}\right)^{-b/p}\dfrac{du}{u}\\ &=\int_{0}^{x}|f(u)|x^{-b/p}du. \end{align*} We also have \begin{align*} \|h\ast k\|_{L^{p}}&=\left(\int_{0}^{\infty}(h\ast k)(x)^{p}\dfrac{dx}{x}\right)^{1/p}\\ &=\left(\int_{0}^{\infty}\left(\int_{0}^{x}|f(u)|\dfrac{du}{u}\right)^{p}x^{-b}\dfrac{dx}{x}\right)^{1/p}, \end{align*} which is exactly the left-sided of the first inequality.

Now we exploit to the convolution inequality $\|h\ast k\|_{L^{p}}\leq\|h\|_{L^{p}}\|k\|_{L^{1}}$, so that \begin{align*} \|k\|_{L^{1}}&=\int_{0}^{\infty}k(t)\dfrac{dt}{t}\\ &=\int_{0}^{\infty}t^{-b/p}\chi_{[1,\infty)}(t)\dfrac{dt}{t}\\ &=\int_{1}^{\infty}t^{-b/p-1}dt\\ &=\dfrac{p}{b}, \end{align*} plugging this to the convolution inequality, the first inequality follows.

For the second inequality, perform the similar trick to the functions $h(x)=|f(x)|x^{1+b/p}$ and $k(x)=x^{b/p}\chi_{(0,1]}$.

These are included in Loukas Grafakos book Classical Fourier Analysis as exercises with tips, how does he manage to see such the tricks work is somewhat magical, I have no answer.

Edit:

For the second inequality, we have \begin{align*} \|k\|_{L^{1}}=\int_{0}^{\infty}t^{b/p}\chi_{(0,1]}(t)\dfrac{dt}{t}=\int_{0}^{1}t^{b/p-1}dt=\dfrac{p}{b}, \end{align*} so the constant term in the right-sided inequality is obtained.

We also have \begin{align*} h\ast k(x)&=\int_{0}^{\infty}\left|f\left(\dfrac{x}{t}\right)\right|\left(\dfrac{x}{t}\right)^{1+b/p}t^{b/p}\chi_{(0,1]}(t)\dfrac{dt}{t}\\ &=\int_{0}^{1}\left|f\left(\dfrac{x}{t}\right)\right|\left(\dfrac{x}{t}\right)^{1+b/p}t^{b/p}\dfrac{dt}{t}\\ &=\int_{x}^{\infty}\left|f\left(u\right)\right|u^{1+b/p}\left(\dfrac{x}{u}\right)^{b/p}\dfrac{du}{u}\\ &=\int_{x}^{\infty}\left|f\left(u\right)\right|x^{b/p}\dfrac{du}{u}, \end{align*} so \begin{align*} \|h\ast k\|_{L^{p}}&=\left(\int_{0}^{\infty}\left(\int_{x}^{\infty}|f(u)|\dfrac{du}{u}\right)^{p}x^{b}\dfrac{dx}{x}\right)^{1/p}\\ &=\left(\int_{0}^{\infty}\left(\int_{x}^{\infty}|f(u)|\dfrac{du}{u}\right)^{p}x^{b-1}dx\right)^{1/p}. \end{align*}