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Let $0<b<\infty$ and $1\le p<\infty$. Show that

$$\bigg(\int_{0}^{\infty}\bigg(\frac{1}{x}\int_{0}^{x}|f(t)|dt\bigg)^{p}x^{p-b-1}dx\bigg)^{\frac{1}{p}}\le\frac{p}{b}\bigg(\int_{0}^{\infty}|f(t)|^{p}t^{p-b-1}dt\bigg)^{\frac{1}{p}}$$

I know how to show it when the term $x^{p-b-1}$ is $1$ (so, $b=p-1$).

But how to proceed when the inequality has the weight $x^{p-b-1}$?

user1992
  • 1,536

3 Answers3

2

On the multiplicative group $\left((0,\infty),\dfrac{dt}{t}\right)$, use the convolution inequality $\|F\ast G\|_{L^{p}((0,\infty),dt/t)}\leq\|F\|_{L^{p}((0,\infty),dt/t)}\|G\|_{L^{p}((0,\infty),dt/t)}$, where $F(x)=|f(x)|x^{1-b/p}$ and $G(x)=x^{-b/p}\chi_{[1,\infty)}$, one has \begin{align*} \|G\|_{L^{1}((0,\infty),dt/t)}&=\int_{1}^{\infty}\dfrac{1}{t^{b/p+1}}dt=\dfrac{p}{b},\\ \|F\|_{L^{p}((0,\infty),dt/t)}&=\left(\int_{0}^{\infty}|f(t)|^{p}t^{(1-b/p)p}\dfrac{dt}{t}\right)^{1/p}\\ &=\left(\int_{0}^{\infty}|f(t)|^{p}t^{p-b-1}\right)^{1/p}. \end{align*} On the other hand, \begin{align*} F\ast G(x)&=\int_{0}^{\infty}|f(x/t)|(x/t)^{1-b/p}t^{-b/p}\chi_{[1,\infty)}(t)\dfrac{dt}{t}\\ &=\int_{1}^{\infty}|f(x/t)|(x/t)x^{-b/p}\dfrac{dt}{t}\\ &=x^{-b/p}\int_{0}^{x}|f(u)|u\cdot\dfrac{du}{u}\\ &=x^{-b/p}\int_{0}^{x}|f(u)|du, \end{align*} where we have used the transformation $u=x/t$, then $du/u=-dt/t$.

Further, \begin{align*} \|F\ast G\|_{L^{p}((0,\infty),dx/x)}&=\left(\int_{0}^{\infty}\left(\int_{0}^{x}|f(t)|dt\right)^{p}x^{-b}\dfrac{dx}{x}\right)^{1/p}\\ &=\left(\int_{0}^{\infty}\left(\int_{0}^{x}|f(t)|dt\right)^{p}x^{-b-1}dx\right)^{1/p}. \end{align*}

user284331
  • 56,315
2

I did as a following way :

\begin{align} \bigg(\int_{0}^{\infty}\bigg(\frac{1}{x}\int_{0}^{x} |f(t)|dt\bigg)^{p}x^{p-b-1}dx\bigg)^{\frac{1}{p}}&=\bigg(\int_{0}^{\infty}\bigg(\int_{0}^{x} |f(t)|dt\bigg)^{p}x^{-b-1}dx\bigg)^{\frac{1}{p}}\\ &=\bigg(\int_{0}^{\infty}\bigg(\int_{0}^{x} x^{\frac{-b-1}{p}}|f(t)|dt\bigg)^{p}dx\bigg)^{\frac{1}{p}}\\ &=\bigg(\int_{0}^{\infty}\bigg(\int_{0}^{1} x^{\frac{-b-1}{p}+1}~|f(xt)|~dt\bigg)^{p}dx\bigg)^{\frac{1}{p}}\\ &\le \int_{0}^{1}\bigg(\int_{0}^{\infty} x^{p-b-1}|f(xt)|^{p}dx\bigg)^{\frac{1}{p}}dt\\ &=\int_{0}^{1}\bigg(\int_{0}^{\infty} (xt)^{p-b-1}\frac{1}{t^{p-b-1}}|f(xt)|^{p}dx\bigg)^{\frac{1}{p}}dt\\ &=\int_{0}^{1}\bigg(\int_{0}^{\infty} \omega^{p-b-1}\frac{1}{t^{p-b-1}}|f(\omega)|^{p}~\frac{d\omega}{t}\bigg)^{\frac{1}{p}}dt\\ &=\int_{0}^{1}\bigg(\int_{0}^{\infty} \omega^{p-b-1}\frac{1}{t^{p-b}}|f(\omega)|^{p}~d\omega\bigg)^{\frac{1}{p}}dt\\ &=\int_{0}^{1}\frac{1}{t^{1-\frac{b}{p}}}~dt\bigg(\int_{0}^{\infty} \omega^{p-b-1}|f(\omega)|^{p}~d\omega\bigg)^{\frac{1}{p}}\\ &=\frac{p}{b}\bigg(\int_{0}^{\infty} \omega^{p-b-1}|f(\omega)|^{p}~d\omega\bigg)^{\frac{1}{p}}\\ \end{align}, where $\omega=xt$ and hence $dx=\frac{d\omega}{t}$ and the first inequality holds by the integral form of Minkowski's inequality.

user1992
  • 1,536
1

Using alternate measure spaces simplifies the answer a bit: $$ \begin{align} \left(\int_0^\infty\left(\frac1x\int_0^x|f(t)|\,\mathrm{d}t\right)^px^{p-b-1}\mathrm{d}x\right)^{\frac1p} &=\left(\int_0^\infty\left(\int_0^1|f(xt)|\,\mathrm{d}t\right)^px^{p-b-1}\mathrm{d}x\right)^{\frac1p}\tag1\\ &\le\int_0^1\left(\int_0^\infty|f(xt)|^px^{p-b-1}\mathrm{d}x\right)^{\frac1p}\,\mathrm{d}t\tag2\\ &=\int_0^1\left(\int_0^\infty|f(x)|^px^{p-b-1}\mathrm{d}x\right)^{\frac1p}t^{\frac bp-1}\,\mathrm{d}t\tag3\\ &=\frac pb\left(\int_0^\infty|f(x)|^px^{p-b-1}\mathrm{d}x\right)^{\frac1p}\tag4 \end{align} $$ Explanation:
$(1)$: substitute $t\mapsto xt$
$(2)$: apply $(7)$ with $\mathrm{d}\mu(x)=x^{p-b-1}\mathrm{d}x$ and $\mathrm{d}\lambda(t)=\mathrm{d}t$
$(3)$: substitute $x\mapsto x/t$
$(4)$: evaluate the integral in $t$

Proof of Minkowski $$ \begin{align} \left(\int_X\left(\int_Tf(x,t)\,\mathrm{d}\lambda(t)\right)^p\,\mathrm{d}\mu(x)\right)^{1/p} &=\sup_{\|g\|_q=1}\int_X\int_Tf(x,t)\,\mathrm{d}\lambda(t)\,g(x)\,\mathrm{d}\mu(x)\tag5\\ &\le\int_T\sup_{\|g\|_q=1}\int_Xf(x,t)\,g(x)\,\mathrm{d}\mu(x)\,\mathrm{d}\lambda(t)\tag6\\ &=\int_T\left(\int_Xf(x,t)^p\,\mathrm{d}\mu(x)\right)^{1/p}\,\mathrm{d}\lambda(t)\tag7\\ \end{align} $$ Explanation:
$(5)$: use the converse of Hölder's Inequality
$(6)$: using a different $g$ for each $t$ gives a greater supremum
$(7)$: use the converse of Hölder's Inequality

robjohn
  • 353,833