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This question is on a point in D.R. Adams paper "A Sharp Inequality of J. Moser for Higher Order Derivatives". Precisely the lemma says:
Given $a(s,t)$ be a non negative measureable function on $(-\infty,\infty)\times [0,\infty)$ such that $$ a(s,t)\leq 1\;\text{ when }\;0<s<t\label{1}\tag{1} $$ and $$ \sup_{t>0}\left(\int_{-\infty}^0+\int_t^{\infty} a(s,t)^q ds\right)^{\frac{1}{q}}=b<\infty\label{2}\tag{2} $$ Then there is a constant $c_0=c_0(p,b)$ such that if $\phi\geq 0$ and $$ \int_{\mathbb{R}}\phi(s)^p ds\leq 1\label{3}\tag{3} $$ then $$ \int_{0}^{\infty} e^{-F(t)} dt\leq c_0 $$ where $F(t)=t-\left(\int_{\mathbb{R}}a(s,t)\phi(s) ds\right)^q$ and $q=\frac{p}{p-1}$.

  1. He claims that $$ \int_{0}^{\infty}e^{-F(t)}dt = \int_{\infty}^{\infty}|E_{\lambda}|e^{-\lambda}d\lambda $$ where $E_{\lambda}:=\{t\geq 0:F(t)\leq\lambda\}$. I don't see how to prove it. I was thinking of the Layer cake representation but I couldn't prove it as the integral in R.H.S is over $\mathbb{R}$ not in $(0,\infty)$.

  2. Furthermore he claims that if $t\in E_{\lambda}$ then $$ t-\lambda\leq((b^q+t)^{\frac{1}{q}}(1-L(t)^p)^{\frac{1}{p}}+bL(t))^q $$ where $L(t)=\left(\int_s^{\infty}\phi(s)^p ds\right)^{\frac{1}{p}}$ (here the 1st term of RHS comes out from using Holder inequality and equations \eqref{1}, \eqref{2}, but the last portion i.e $+bL(t)$ is not coming rather some weaker estimate is coming.

Any help would be very much helpful to understand the thing going on.

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2 Answers 2

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For the second, rewrite $F(t) \leq \lambda$ as $$t - \lambda \leq \left(\int_{\mathbb{R}} a(s,\,t)\phi(s)\,ds\right)^q,$$ which reduces the problem to showing that $$\int_{\mathbb{R}} a(s,\,t)\phi(s)\,ds \leq (b^q + t)^{1/q}(1-L^p(t))^{1/p} + bL(t).$$ To verify this inequality write the left side as $$\int_{-\infty}^t a(s,\,t)\phi(s)\,ds + \int_t^{\infty}a(s,\,t)\phi(s)\,ds,$$ and apply Holder's inequality to each term. The second term is clearly bounded by $bL(t)$. For the first, use that $\int_{-\infty}^t \phi^p \leq 1-L^p(t)$ (this uses $\int_{\mathbb{R}}\phi^p \leq 1$) and that $\int_{-\infty}^t a^q(s,\,t) = \int_{-\infty}^0 a^q + \int_0^t a^q \leq b^q + t$ (using the definition of $b$ for the first, and the bound of $1$ on $a$ in the relevant interval for the second).

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  • $\begingroup$ Thank you Professor Mooney for all your efforts. $\endgroup$
    – User1723
    Commented Oct 17, 2022 at 4:29
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$\newcommand\la\lambda$To answer your first question, write $$\int_{-\infty}^\infty|E_\la|e^{-\la}\,d\la =\int_{-\infty}^\infty e^{-\la}\,d\la\int_0^\infty dt\,1(F(t)\le\la) =\int_0^\infty dt\,\int_{-\infty}^\infty e^{-\la}\,d\la\,1(F(t)\le\la) =\int_0^\infty dt\,\int_{F(t)}^\infty e^{-\la}\,d\la =\int_0^\infty dt\,e^{-F(t)},$$ as desired.

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    $\begingroup$ Damm just Fubini thinking about so many things $\endgroup$
    – User1723
    Commented Oct 16, 2022 at 18:32

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