2
$\begingroup$

Let $(X,\mu)$ be a measure space and let $s>0$

Let $f$ be a measurable function on $X$ .Show that $0<p<q<\infty $ we have $$\int_{|f|\leq s}|f\,|^{q}\,d\mu\leq\frac{q}{q-p}s^{q-p}\lVert f\,\rVert_{L^{p,\infty}}^p$$

Here's my done :

If $\lVert f\,\rVert_{L^{p,\infty\,}}=\infty$ then there is nothing to prove.So assume $\lVert f\,\rVert_{L^{p,\infty\,}}<\infty$ Then we have

\begin{align} \int_{|f|\leq s}|f\,|^{q}\,d\mu&=q\int_{0}^{s} \alpha^{q-1}\omega(\alpha)\,d\alpha\\ &\leq q\int_{0}^{s} \alpha^{q-1-p}\lVert f\,\rVert_{L^{p,\infty}}^p\,\,\,d\alpha\\ &=\frac{q}{q-p}s|_{0}^{q-p}\lVert f\,\rVert_{L^{p,\infty}}^p\\ &=\frac{q}{q-p}s^{q-p}\lVert f\,\rVert_{L^{p,\infty}}^p \end{align}

where $\omega(\alpha)$ is the distribution function of $f$.

If you have the time,please check my proof for validity with me. Any advice would appreciate.Thanks for considering my request.

$\rule{18cm}{2pt}$

I take note below this line that anyone can ignore it or give your valuable suggestion.

\begin{align} \int_{|f|\leq s}|f(x)|^{q}\,d\mu(x)&= q\int_{0}^{\infty}\alpha^{q-1}\chi_{\{\alpha\,:\, 0<\alpha\leq s\}}\omega(\alpha)\,d\alpha\\ &=q\int_{0}^{s}\alpha^{q-1}\chi_{\{\alpha\,:\, 0<\alpha\leq s\}}\omega(\alpha)\,d\alpha \,+ q\int_{s}^{\infty}\alpha^{q-1}\chi_{\{\alpha\,:\, 0<\alpha\leq s\}}\omega(\alpha)\,d\alpha\\ &=q\int_{0}^{s}\alpha^{q-1}\chi_{\{\alpha\,:\, 0<\alpha\leq s\}}\omega(\alpha)\,d\alpha \,+ 0\\ &=q\int_{0}^{s}\alpha^{q-1}\omega(\alpha)\,d\alpha \end{align}

$\endgroup$
1
$\begingroup$

Okay, or if we let $E=\{x\in X: |f(x)|\leq s\}$, then \begin{align*} \int_{|f|\leq s}|f|^{q}d\mu&=\int_{E}|f|^{q}d\mu\\ &=q\int_{0}^{\infty}\alpha^{q-1}\omega_{E}(\alpha)d\alpha\\ &=q\int_{0}^{\infty}\alpha^{q-1}\chi_{0\leq\alpha<s}\omega_{E}(\alpha)d\alpha\\ &=q\int_{0}^{s}\alpha^{q-1}\omega(\alpha)d\alpha, \end{align*} where $\omega_{E}(\alpha)=\{x\in E:|f(x)|>\alpha\}$, if $\alpha\geq s$, then $\omega_{E}(\alpha)=0$, if $\alpha<s$, then $\omega_{E}(\alpha)=\omega(\alpha)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.