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Let's define the following operator $$\mathcal{J}_\epsilon:=(I-\epsilon\Delta)^{-1},$$ where $\epsilon>0$ and $\Delta$ is the Laplacian. We know that $$\Vert\mathcal{J}_\epsilon f\Vert_{L^p(\Omega)}\leq\Vert f\Vert_{L^p(\Omega)},$$ for every $1<p<\infty$, with $\Omega$ an open subset of $\mathbb{R}^3$. Does the same result hold if $\Omega=\mathbb{R}^3$?

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  • $\begingroup$ Hint: The functions with bounded support in $L^p$ are dense in $L^p$ for $p\lt\infty$. $\endgroup$
    – robjohn
    Commented Sep 24, 2015 at 10:29
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    $\begingroup$ @robjohn This is sufficient to ensure the thesis is true in $\mathbb{R}^3$? $\endgroup$
    – Sal
    Commented Sep 24, 2015 at 12:48
  • $\begingroup$ It is. However, I had thought that $\Omega$ was a bounded open subset of $\mathbb{R}^3$ and it seems that $\Omega$ is just an open subset. Since $\mathbb{R}^3$ is an open subset of $\mathbb{R}^3$, why do we need anything more? $\endgroup$
    – robjohn
    Commented Sep 24, 2015 at 18:08

1 Answer 1

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You may have already had this proven, but I thought I would include it for completeness.


The integral of $e^{-2\pi ix\cdot\xi}$ over a sphere of radius $r$ is $$ \begin{align} 2\pi r\int_{-r}^re^{-2\pi it|\xi|}\,\mathrm{d}t &=2\pi r\int_{-r}^r\cos\left(-2\pi t|\xi|\right)\,\mathrm{d}t\\ &=\frac{2r}{|\xi|}\sin\left(2\pi r|\xi|\right)\tag{1} \end{align} $$ Therefore, the Fourier Transform of $\frac1{1+4\pi^2\epsilon|x|^2}$ is $$ \begin{align} \int_{\mathbb{R}^3}\frac{e^{-2\pi ix\cdot\xi}}{1+4\pi^2\epsilon\,|x|^2}\,\mathrm{d}x &=\int_0^\infty\frac{\frac{2r}{|\xi|}\sin\left(2\pi r|\xi|\right)}{1+4\pi^2\epsilon r^2}\,\mathrm{d}r\tag{2a}\\ &=-\frac1{\pi|\xi|^2}\int_0^\infty\frac{r}{1+4\pi^2\epsilon r^2}\,\mathrm{d}\cos\left(2\pi r|\xi|\right)\tag{2b}\\ &=\frac1{\pi|\xi|^2}\int_0^\infty\cos\left(2\pi r|\xi|\right)\frac{1-4\pi^2\epsilon r^2}{\left(1+4\pi^2\epsilon r^2\right)^2}\,\mathrm{d}r\tag{2c}\\ &=\frac1{2\pi^2|\xi|^2\sqrt\epsilon}\int_0^\infty\cos\left(\frac{|\xi|}{\sqrt\epsilon} r\right)\frac{1-r^2}{\left(1+r^2\right)^2}\,\mathrm{d}r\tag{2d}\\ &=-\frac1{8\pi^2|\xi|^2\sqrt\epsilon}\int_{-\infty}^\infty\exp\left(\frac{i|\xi|}{\sqrt\epsilon} r\right)\left[\frac1{\left(r+i\right)^2}+\frac1{\left(r-i\right)^2}\right]\,\mathrm{d}r\tag{2e}\\ &=-\frac1{8\pi^2|\xi|^2\sqrt\epsilon}2\pi i\frac{i|\xi|}{\sqrt\epsilon}\exp\left(-\frac{|\xi|}{\sqrt\epsilon}\right)\tag{2f}\\ &=\frac1{4\pi|\xi|\,\epsilon}\exp\left(-\frac{|\xi|}{\sqrt\epsilon}\right)\tag{2g}\\[6pt] &=K_\epsilon(x)\tag{2h} \end{align} $$ Explanation:
$\text{(2a)}$: use $(1)$
$\text{(2b)}$: prepare to integrate by parts
$\text{(2c)}$: integrate by parts
$\text{(2d)}$: substitute $r\mapsto\frac{r}{2\pi\sqrt{\epsilon}}$
$\text{(2e)}$: extend the integral to $(-\infty,\infty)$ by symmetry and divide by $2$
$\phantom{\text{(2e):}}$ since the rest of the integrand is even, we can change $\cos(x)$ to $e^{ix}$
$\phantom{\text{(2e):}}$ partial fractions
$\text{(2f)}$: use the contour $[-R,R]\cup Re^{i[0,\pi]}$ as $R\to\infty$ and the singularity at $i$
$\text{(2g)}$: simplify
$\text{(2h)}$: define the kernel $K_\epsilon$

Simple computation shows that $$ \|K_\epsilon\|_{L^1\hspace{-1pt}\left(\mathbb{R}^3\right)}=1\tag{3} $$ for all $\epsilon$. Furthermore, $$ \mathcal{J}_\epsilon f(x)=K_\epsilon\ast f(x)\tag{4} $$ $(3)$, $(4)$, and Young's Inequality guarantee that the $L^p$ norm of $\mathcal{J}_\epsilon$ is at most $1$ for $1\le p\le\infty$.

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