# Heisenberg uncertainty principle and its analogues in higher dimension: via
Wigdersons’ method

Yiyu Tang

(Laboratoire d’analyse et de mathématiques appliquées, Université Gustave
Eiffel)

###### Abstract

The following question was proposed by Avi Wigderson and Yuval Wigderson in
[6]: Is it possible to use the method in [6] to prove Heisenberg uncertainty
principle in higher dimension $\mathbb{R}^{d}$, and get the correct dependence
of the constant on $d$? We answer this question affirmatively, and also prove
some generalizations of Heisenberg uncertainty principle in $\mathbb{R}^{d}$
via Wigdersons’ method.

##  1 Introduction

#####

Recall the definition of Schwartz class:

$\mathcal{S}(\mathbb{R}^{d}):=\\{f\in
C^{\infty}(\mathbb{R}^{d};\mathbb{C}):\rho_{\alpha,\beta}(f)\text{ is finite
for all
}\alpha=(\alpha_{i})_{i=1}^{d},\beta=(\beta_{i})_{i=1}^{d}\in\mathbb{N}^{d}\\},$

where
$\rho_{\alpha,\beta}(f)=\sup_{x\in\mathbb{R}^{d}}|x^{\alpha}\partial^{\beta}f(x)|$.
Here $x^{\alpha}=x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}}\cdots
x_{d}^{\alpha_{d}}$, and
$\partial^{\beta}=\partial_{1}^{\beta_{1}}\partial_{2}^{\beta_{2}}\cdots\partial_{d}^{\beta_{d}}$.

For a Schwartz function $f\in\mathcal{S}(\mathbb{R}^{d})$, its Fourier
transform $\hat{f}$ is defined by
$\hat{f}(\xi):=\int_{\mathbb{R}^{d}}f(x)\mathrm{e}^{-2\pi\mathrm{i}x\cdot\xi}\mathrm{d}x$,
where $x\cdot\xi$ is the inner product in $\mathbb{R}^{d}$. The Heisenberg
uncertainty principle states that for any $f\in\mathcal{S}(\mathbb{R}^{d})$

$\bigg{(}\int_{\mathbb{R}^{d}}|x|^{2}|f(x)|^{2}\mathrm{d}x\bigg{)}\cdot\bigg{(}\int_{\mathbb{R}^{d}}|\xi|^{2}|\hat{f}(\xi)|^{2}\mathrm{d}\xi\bigg{)}\geq\frac{d^{2}}{16\pi^{2}}||f||^{2}_{L^{2}(\mathbb{R}^{d})}\cdot||\hat{f}||^{2}_{L^{2}(\mathbb{R}^{d})},$
(1.1)

and the equality holds if and only if $f$ has the form
$f(x)=C\mathrm{e}^{-\pi\delta|x|^{2}}$, where $\delta>0$.

The classical proof of (1.1) is to write

$d||f||^{2}_{L^{2}}=\int_{\mathbb{R}^{d}}f(x)\overline{f(x)}\sum_{j=1}^{d}\frac{\partial}{\partial
x_{j}}x_{j}\mathrm{d}x,\text{ where }x=(x_{1},x_{2},...,x_{d}).$

For $x\in\mathbb{R}^{d}$, we denote
$|x|:=\sqrt{|x_{1}|^{2}+\cdots+|x_{d}|^{2}}$, and
$\langle\cdot,\cdot\rangle_{\mathbb{R}^{d}}$ is the inner product in
$\mathbb{R}^{d}$. Using integration by parts, Cauchy-Schwarz inequality, and
Plancherel identity:

$\displaystyle d||f||^{2}_{2}$
$\displaystyle=2\cdot\mathrm{Re}\bigg{(}-\int_{\mathbb{R}^{d}}\langle\nabla
f,\overline{xf}\rangle_{\mathbb{R}^{d}}\mathrm{d}x\bigg{)}$ $\displaystyle\leq
2\bigg{|}-\int_{\mathbb{R}^{d}}\langle\nabla
f,\overline{xf}\rangle_{\mathbb{R}^{d}}\mathrm{d}x\bigg{|}$ $\displaystyle\leq
2\int_{\mathbb{R}^{d}}|\nabla f|\cdot|\overline{xf}|\mathrm{d}x$
$\displaystyle\leq 2\bigg{(}\int_{\mathbb{R}^{d}}|\nabla
f|^{2}\mathrm{d}x\bigg{)}^{\frac{1}{2}}\cdot\bigg{(}\int_{\mathbb{R}^{d}}|x|^{2}|f|^{2}\mathrm{d}x\bigg{)}^{\frac{1}{2}}$
$\displaystyle=2\bigg{(}\int_{\mathbb{R}^{d}}\sum_{j=1}^{d}|\widehat{\partial_{j}f}(\xi)|^{2}\mathrm{d}\xi\bigg{)}^{\frac{1}{2}}\cdot\bigg{(}\int_{\mathbb{R}^{d}}|x|^{2}|f|^{2}\mathrm{d}x\bigg{)}^{\frac{1}{2}}$

Finally, just recall a special property of Fourier transform
$\widehat{\partial_{j}f}(\xi)=2\pi\mathrm{i}\cdot\xi_{j}\hat{f}(\xi)$.

###  1.1 An overview of Wigdersons’ method

#####

In [6], A.Wigderson and Y.Wigderson suggested a new method to give a unified
version of various uncertainty principles. This method leads to a very
different proof of (1.1) in dimension $1$. Comparing with the classical proof,
their method did not use any special property of Fourier transform, so it can
be generalized to prove analogues of uncertainty principle for other
operators. In this paper, we only focus on the Fourier transform operator.

Now we give a quick review of Wigdersons’ proof of (1.1) in dimension $1$.
Let’s begin with a basic inequality:
$||\hat{f}||_{L^{\infty}}\leq||f||_{L^{1}}$. Replacing $f$ by $\hat{f}$, we
have $||f||_{L^{\infty}}\leq||\hat{f}||_{L^{1}}$. Multiplying these two
inequalities gives us

$\frac{||f||_{1}\cdot||\hat{f}||_{1}}{||f||_{\infty}\cdot||\hat{f}||_{\infty}}\geq
1,\text{ where }f\in\mathcal{S}(\mathbb{R}^{1})\setminus\\{0\\}.$ (1.2)

We call (1.2) the primary uncertainty principle.

Now for $1<p<\infty$, we have an obvious inequality:
$||f||^{p}_{p}\leq||f||^{p-1}_{\infty}||f||_{1}$, this is just because
$|f(x)|^{p}\leq||f||_{\infty}^{p-1}|f(x)|$. Multiplying both sides by
$||f||_{1}^{p-1}$, we get

$||f||_{1}^{p-1}||f||_{p}^{p}\leq||f||_{\infty}^{p-1}||f||_{1}^{p}.$ (1.3)

The inequality (1.3) implies that

$\frac{||f||_{1}}{||f||_{p}}\geq\bigg{(}\frac{||f||_{1}}{||f||_{\infty}}\bigg{)}^{\frac{p-1}{p}},\text{
and also
}\frac{||\hat{f}||_{1}}{||\hat{f}||_{p}}\geq\bigg{(}\frac{||\hat{f}||_{1}}{||\hat{f}||_{\infty}}\bigg{)}^{\frac{p-1}{p}}\text{
after replacing }f\text{ by }\hat{f}.$ (1.4)

Define $V(f):=\int_{\mathbb{R}}|x|^{2}|f|^{2}$, our goal is to prove that
there exists a constant $C>0$ independent of $f$, so that

$V(f)V(\hat{f})\geq C||f||^{2}_{2}||\hat{f}||^{2}_{2}.$

Now we choose $T:=\frac{1}{8}\big{(}\frac{||f||_{1}}{||f||_{2}}\big{)}^{2}$,
by Cauchy-Schwarz inequality,

$\int_{-T}^{T}|f|=\int_{-T}^{T}1\cdot|f|\leq\sqrt{2T}\cdot\bigg{(}\int_{\mathbb{R}}|f|^{2}\bigg{)}^{\frac{1}{2}}=\frac{1}{2}||f||_{1}.$

The above inequality is equivalent to
$\frac{1}{2}||f||_{1}\leq\int_{|x|>T}|f|$. For the part $|x|>T$, another
application of Cauchy-Schwarz inequality yields that

$\displaystyle\frac{1}{2}||f||_{1}\leq\int_{|x|>T}|f|\leq$
$\displaystyle\bigg{(}\int_{|x|>T}\frac{\mathrm{d}x}{|x|^{2}}\bigg{)}^{\frac{1}{2}}\bigg{(}\int_{|x|>T}|x|^{2}|f|^{2}\mathrm{d}x\bigg{)}^{\frac{1}{2}}$
(1.5) $\displaystyle\leq$ $\displaystyle\sqrt{\frac{2}{T}}\sqrt{V(f)}.$

After a simplification, we get

$\frac{V(f)}{||f||_{2}^{2}}\geq\frac{1}{64}\bigg{(}\frac{||f||_{1}}{||f||_{2}}\bigg{)}^{4}.$
(1.6)

Replacing $f$ by $\hat{f}$ and taking a multiplication, we get, by (1.4) and
(1.2),

$\frac{V(f)}{||f||_{2}^{2}}\cdot\frac{V(\hat{f})}{||\hat{f}||_{2}^{2}}\geq\frac{1}{64^{2}}\bigg{(}\frac{||f||_{1}||\hat{f}||_{1}}{||f||_{2}||\hat{f}||_{2}}\bigg{)}^{4}\geq\frac{1}{64^{2}}\bigg{(}\frac{||f||_{1}||\hat{f}||_{1}}{||f||_{\infty}||\hat{f}||_{\infty}}\bigg{)}^{2}\geq\frac{1}{64^{2}}.$
(1.7)

######  Remark 1.1.

Comparing with the original proof, this new proof seems longer. However, we
should point out that, in the classical proof, we use the identity
$\widehat{\partial_{j}f}(\xi)=2\pi\mathrm{i}\cdot\xi_{j}\hat{f}(\xi)$, which
is a special property of Fourier transform, and not shared by other operators.

In the new proof, essentially, we only use the property that Fourier transform
and its inverse are $L^{1}\rightarrow L^{\infty}$ bounded, and this
boundedness property is less restrictive. So the new proof can be generalized
to many other operators, for example, if a linear operator
$L:\mathcal{S}(\mathbb{R})\rightarrow\mathcal{S}(\mathbb{R})$ satisfies

$||L||_{L^{1}\rightarrow L^{\infty}}\leq 1,\
||L^{\star}Lf||_{\infty}\geq||f||_{\infty},$

then we also have a Heisenberg-type uncertainty principle for $L$:

$V(f)V(Lf)\geq C||f||_{2}^{2}||\hat{f}||_{2}^{2}.$

For more details, see [6], section 4.2 and 4.3.

######  Remark 1.2.

Comparing with the optimal bound $\frac{1}{16\pi^{2}}$, in (1.7), the constant
is $1/64^{2}$, which is much smaller than the optimal bound. It seems that
this method can not be used to prove the sharp bound, since we do not use full
properties of Fourier transform.

###  1.2 Questions proposed by A.Wigderson and Y.Wigderson

#####

In [6], the authors observed that this method can not be used to prove
Heisenberg uncertainty principles in higher dimension $d\geq 2$ directly. This
is because the integral

$\int_{x\in\mathbb{R}^{d}:|x|>T}\frac{\mathrm{d}x}{|x|^{2}}$

is divergent. They asked if this convergence issues represented a real
limitation of the new approach, or if there was a way around them. More
precisely, they asked the following questions:

######  Question ([6], page 255, Open problem).

Is it possible to prove the Heisenberg uncertainty principle in
$\mathbb{R}^{d}$ via the new method? That’s to say, could one use this new
approach to prove the existence of an absolute constant $C=C(d)>0$ independent
of $f$, so that

$\frac{V(f)}{||f||_{2}^{2}}\frac{V(\hat{f})}{||\hat{f}||^{2}_{2}}\geq C,\text{
for all }f\in\mathcal{S}(\mathbb{R}^{d})\setminus\\{0\\}?$

Here $V(f):=\int_{\mathbb{R}^{d}}|x|^{2}|f(x)|^{2}\mathrm{d}x$.

Moreover, we already knew that the best constant $C$ is
$\frac{d^{2}}{16\pi^{2}}$. Could one use this new approach to obtain the
correct dependence on the dimension? That’s to say, is it possible to prove
the existence of an absolute constant $C^{\prime}$, which is independent of
$f$ and $d$, so that

$\frac{V(f)}{||f||_{2}^{2}}\frac{V(\hat{f})}{||\hat{f}||^{2}_{2}}\geq
C^{\prime}d^{2},\text{ for all
}f\in\mathcal{S}(\mathbb{R}^{d})\setminus\\{0\\}?$

###  1.3 Main theorems

#####

The following theorem answers the above two questions affirmatively.

######  Theorem 1.

Via Wigdersons’s method, it’s possible to prove that for every
$f\in\mathcal{S}(\mathbb{R}^{d})\setminus\\{0\\}$, we have

$\frac{V(f)}{||f||_{2}^{2}}\frac{V(\hat{f})}{||\hat{f}||^{2}_{2}}\geq Cd^{2},$

where $C$ is an absolute constant independent of $f$ and $d$, for example, one
may take $C=\frac{1}{10^{10}}$.

Now we consider a generalization of Heisenberg uncertainty principle. For
$p\in(1,\infty)$, we define

$V_{p}(f):=\int_{\mathbb{R}^{d}}|x|^{p}|f(x)|^{p}\mathrm{d}x.$

Then $V_{p}(f)=V(f)$ when $p=2$, and we also have a Heisenberg-type inequality
for $V_{p}$ when $p$ is “small”. This is the following theorem:

######  Theorem 2.

Suppose that $d\geq 1$.

  * •

If $1<p<\frac{2d}{d-1}$, then via Wigdersons’s method, it’s possible to prove
that there exists a constant $C=C(d,p)$, so that

$\frac{V_{p}(f)}{||f||_{p}^{p}}\frac{V_{p}(\hat{f})}{||\hat{f}||^{p}_{p}}\geq
C(d,p),\text{ for every }f\in\mathcal{S}(\mathbb{R}^{d})\setminus\\{0\\}.$

Moreover, if $1<p\leq 2$ is fixed, then as $d$ goes to infinity, the growth of
$C(d,p)$ is $C_{1}(p)\cdot d^{p}$, where $C_{1}(p)$ is a constant independent
of the dimension $d$, and up to a dimension-free constant $C_{1}(p)$, the
growth $d^{p}$ is sharp.

  * •

If $\frac{2d}{d-1}<p<\infty$, then

$\inf_{f\in\mathcal{S}(\mathbb{R}^{d})\setminus\\{0\\}}\frac{V_{p}(f)}{||f||_{p}^{p}}\frac{V_{p}(\hat{f})}{||\hat{f}||^{p}_{p}}=0.$

So there is no uncertainty type theorem when $p>\frac{2d}{d-1}$.

######  Remark 1.3.

It seems that Wigdersons’ method can not be dealt with the critical case
$p=\frac{2d}{d-1}$. Also, this method does not work when $p=1$ or $p=\infty$.
The three cases $p=1$, $p=\frac{2d}{d-1}$ and $p=\infty$ are unknown.

#####

Our final result deals with a theorem of Cowling-Price and its
generalizations. Recall Cowling-Price uncertainty principle states that if
$\theta,\phi\in(0,\infty)$ and $p,q\in[1,\infty]$, then

$\Big{|}\Big{|}|x|^{\theta}f\Big{|}\Big{|}_{p}\cdot\Big{|}\Big{|}|\xi|^{\phi}\hat{f}\Big{|}\Big{|}_{q}\geq
C||f||^{2}_{2},\text{ for all }f\in\mathcal{S}(\mathbb{R})$

if and only if

$\theta>\frac{1}{2}-\frac{1}{p},\
\phi>\frac{1}{2}-\frac{1}{q}\quad\text{and}\quad\theta+\frac{1}{p}=\phi+\frac{1}{q}.$

For more details on this theorem, see [2], Theorem 5.1. Cowling-Price theorem
was also discussed in [1], Lemma 3.3 and Theorem 3.6.

By using Wigdersons’ method, we give another proof of Cowling-Price theorem
(up to the endpoints) and its generalizations in higher dimension.

######  Theorem 3.

Assume that $d\geq 1$. Let $1<p,q<\infty$ and $0<\theta,\phi<\infty$, then
there exists a constant $C$ so that

$\Big{|}\Big{|}|x|^{\theta}f\Big{|}\Big{|}_{p}\cdot\Big{|}\Big{|}|\xi|^{\phi}\hat{f}\Big{|}\Big{|}_{q}\geq
C||f||^{2}_{2}$ (1.8)

for all $f\in\mathcal{S}(\mathbb{R}^{d})$, if and only if

$\frac{\theta}{d}>\frac{1}{2}-\frac{1}{p},\quad\frac{\phi}{d}>\frac{1}{2}-\frac{1}{q}\quad\text{and}\quad\frac{1}{q}+\frac{\phi}{d}=\frac{1}{p}+\frac{\theta}{d}.$
(1.9)

Take $d=1$, we get the Cowling-Price theorem except for two endpoints.

#####

The article is organized as follows:

  * •

In section 2, we prove Theorem 1. Recall that, when $d=1$, we use integral
$\int_{|x|>T}1/|x|^{2}$ in (1.5). In $\mathbb{R}^{d}$, we will use the
integral $\int_{|x|>T}1/|x|^{d+1}$, with a special choice of some parameters,
see (2.1). In the study of the behavior of the constant $C$ as $d$ goes to
infinity, we will use an asymptotic formula of Gamma function frequently:

$\Gamma(x)\sim\sqrt{\frac{2\pi}{x}}\Big{(}\frac{x}{e}\Big{)}^{x},\text{ when
}x\rightarrow\infty.$

  * •

In section 3, we prove Theorem 2. The proof is essentially the same with the
$L^{2}$ case, and we only need to change the choice of some parameters. In
$L^{2}(\mathbb{R}^{d})$ case, we use the property that
$\int_{x\in\mathbb{R}^{d}:|x|\geq 1}1/|x|^{d+1}<\infty$, and in
$L^{p}(\mathbb{R}^{d})$ case we will use the integral
$\int_{x\in\mathbb{R}^{d}:|x|\geq 1}1/|x|^{d+\epsilon}<\infty$, where
$\epsilon>0$ is a parameter, see Proposition 3.1.

  * •

In section 4, we prove Theorem 3. We will see that the real difficulties are
the first two inequalities in (1.9). The conditions (1.9) imply the inequality
(1.8) is parallel to the proof of Theorem 2. For another direction, there are
two different cases, so we should consider them separately, see section 4.2.
In both cases, assuming that (1.9) fails, we will construct a function so that
(1.8) does not hold.

##  2 Proof of Theorem 1

#####

In (1.6), we proved that when $d=1$:

$\frac{V(f)}{||f||_{2}^{2}}\geq\frac{1}{64}\bigg{(}\frac{||f||_{1}}{||f||_{2}}\bigg{)}^{4}.$

With this inequalities in mind, now we try to prove an inequality for general
$d$ of the following form:

$\frac{V(f)}{||f||_{L^{2}(\mathbb{R}^{d})}^{2}}\geq
C(d)\bigg{(}\frac{||f||_{L^{a}(\mathbb{R}^{d})}}{||f||_{L^{2}(\mathbb{R}^{d})}}\bigg{)}^{\kappa},$

where $a$ and $\kappa$ are constants to be determined. Also, we should take
the choice of $T$ into consideration.

###  2.1 Choice of some parameters & Sketch of the proof

#####

We choose $a,r,s\geq 1$ as following:

$a=\frac{2(d+1)}{d+3},\ r=\frac{d+3}{d+1},\ s=\frac{d+3}{2}.$ (2.1)

Notice that $1\leq a<2$ and $1/r+1/s=1$. Also, define

$T:=c_{d}\cdot\bigg{(}\frac{||f||_{a}}{||f||_{2}}\bigg{)}^{\frac{d+1}{d}},$

where $c_{d}$ is a constant to be determined later, and it’s independent of
$f$, see (2.4).

#####

Our goal is to prove, for every
$f\in\mathcal{S}(\mathbb{R}^{d})\setminus\\{0\\}$,

$\displaystyle\frac{V(f)}{||f||_{2}^{2}}\geq\frac{1}{4}\bigg{(}\frac{c_{d}^{2}}{\omega_{d-1}^{2}}\bigg{)}^{\frac{1}{d+1}}\bigg{(}\frac{||f||_{a}}{||f||_{2}}\bigg{)}^{(2+\frac{2}{d})},$
(2.2)

where $\omega_{d-1}$ is the area of $(d-1)$-sphere $\mathbb{S}^{d-1}$. We will
prove this inequality in the next subsection.

#####

Assuming (2.2), then replacing $f$ by $\hat{f}$, we have

$\displaystyle\frac{V(\hat{f})}{||\hat{f}||_{2}^{2}}\geq\frac{1}{4}\bigg{(}\frac{c_{d}^{2}}{\omega_{d-1}^{2}}\bigg{)}^{\frac{1}{d+1}}\bigg{(}\frac{||\hat{f}||_{a}}{||\hat{f}||_{2}}\bigg{)}^{(2+\frac{2}{d})}.$

Taking a multiplication we get

$\displaystyle\frac{V(f)}{||f||_{2}^{2}}\cdot\frac{V(\hat{f})}{||\hat{f}||_{2}^{2}}\geq\frac{1}{16}\bigg{(}\frac{c_{d}^{2}}{\omega_{d-1}^{2}}\bigg{)}^{\frac{2}{d+1}}\bigg{(}\frac{||f||_{a}||\hat{f}||_{a}}{||f||_{2}||\hat{f}||_{2}}\bigg{)}^{(2+\frac{2}{d})}.$

Let $a^{\prime}$ be the conjugate exponent of $a$. Hölder inequality implies
that

$||\hat{f}||_{2}=||f||_{2}\leq||f||^{\frac{1}{2}}_{a}||f||^{\frac{1}{2}}_{a^{\prime}}.$

Also, when $1\leq a<2$, Hausdorff-Young inequality yields that
$||\hat{f}||_{a^{\prime}}\leq||f||_{a}$, replacing $f$ by $\hat{f}$, we have
$||f||_{a^{\prime}}\leq||\hat{f}||_{a}$, and

$\frac{||f||_{a}||\hat{f}||_{a}}{||f||_{2}||\hat{f}||_{2}}\geq\frac{||f||_{a}||\hat{f}||_{a}}{||f||_{a}||f||_{a^{\prime}}}=\frac{||\hat{f}||_{a}}{||f||_{a^{\prime}}}\geq
1.$ (2.3)

Therefore,

$\displaystyle\frac{V(f)}{||f||_{2}^{2}}\cdot\frac{V(\hat{f})}{||\hat{f}||_{2}^{2}}\geq\frac{1}{16}\bigg{(}\frac{c_{d}^{2}}{\omega_{d-1}^{2}}\bigg{)}^{\frac{2}{d+1}}.$

#####

Now only need to study the growth of the quotient
$\big{(}\frac{c_{d}}{\omega_{d-1}}\big{)}^{\frac{2}{d+1}}$, when $d$ goes to
infinity. In subsection 2.3, we will prove that

$\bigg{(}\frac{c_{d}}{\omega_{d-1}}\bigg{)}^{\frac{2}{d+1}}>\frac{1}{C}d,$

where $C$ is a large constant independent of $d$, for example $C=100000$, and
we eventually get

$\displaystyle\frac{V(f)}{||f||_{2}^{2}}\cdot\frac{V(\hat{f})}{||\hat{f}||_{2}^{2}}\geq\frac{1}{C^{\prime}}d^{2}.$

Here one may take $C^{\prime}=10^{10}$.

###  2.2 Proof of (2.2)

To prove (2.2), we use Hölder inequality for the pair $r$ and $s$, notice that
$ar=2$, $as=d+1$ and $1/s=(2-a)/2$ by our choice.

$\begin{aligned} \int_{|x|\leq T}|f|^{a}\leq&\bigg{(}\int_{|x|\leq
T}1^{s}\mathrm{d}x\bigg{)}^{\frac{1}{s}}\cdot\bigg{(}\int_{|x|\leq
T}|f|^{ar}\bigg{)}^{\frac{1}{r}}\\\
=&(T^{d}v_{d})^{\frac{1}{s}}\bigg{(}\int_{|x|\leq
T}|f|^{2}\bigg{)}^{\frac{a}{2}}\\\
\leq&(T^{d}v_{d})^{\frac{2-a}{2}}||f||_{2}^{a}\end{aligned},$

where $v_{d}$ is the volume of the unite ball in $\mathbb{R}^{d}$.

The term $(T^{d}v_{d})^{\frac{2-a}{2}}$ is

$(T^{d}v_{d})^{\frac{2-a}{2}}=(c_{d}^{d}v_{d})^{\frac{2-a}{2}}\bigg{(}\frac{||f||^{d+1}_{a}}{||f||^{d+1}_{2}}\bigg{)}^{\frac{2-a}{2}}$

Notice that $\frac{(d+1)(2-a)}{2}=a$, so

$\bigg{(}\frac{||f||^{d+1}_{a}}{||f||^{d+1}_{2}}\bigg{)}^{\frac{2-a}{2}}=\bigg{(}\frac{||f||_{a}}{||f||_{2}}\bigg{)}^{a}$

Now we choose $c_{d}$ so that

$(c_{d}^{d}v_{d})^{\frac{2-a}{2}}=1/2.$ (2.4)

Particularly, the constant $c_{d}$ depends only on dimension $d$. For this
$c_{d}$, we have

$\displaystyle\int_{|x|\leq T}|f|^{a}\leq\frac{1}{2}||f||_{a}^{a},$

which is equivalent to $\int_{|x|\geq T}|f|^{a}\geq\frac{1}{2}||f||_{a}^{a}$.

#####

Another application of Hölder inequality gives

$\displaystyle\int_{|x|\geq T}|f|^{a}=$ $\displaystyle\int_{|x|\geq
T}\frac{1}{|x|^{a}}|x|^{a}|f(x)|^{a}$ $\displaystyle\leq$
$\displaystyle\bigg{(}\int_{|x|\geq
T}\frac{1}{|x|^{as}}\bigg{)}^{\frac{1}{s}}\cdot\bigg{(}\int_{|x|\geq
T}|x|^{ar}|f(x)|^{ar}\bigg{)}^{\frac{1}{r}}$ $\displaystyle=$
$\displaystyle\bigg{(}\int_{|x|\geq
T}\frac{1}{|x|^{d+1}}\bigg{)}^{\frac{1}{s}}\cdot\bigg{(}\int_{|x|\geq
T}|x|^{2}|f(x)|^{2}\bigg{)}^{\frac{1}{r}}$ $\displaystyle=$
$\displaystyle\bigg{(}\frac{1}{T}\int_{|x|\geq
1}\frac{1}{|x|^{d+1}}\bigg{)}^{\frac{1}{s}}\cdot V(f)^{\frac{1}{r}}$

An easy calculation shows that $\int_{|x|\geq
1}\frac{1}{|x|^{d+1}}=\omega_{d-1}$. Therefore

$\displaystyle\frac{1}{2}||f||_{a}^{a}\leq\int_{|x|\geq T}|f|^{a}\leq$
$\displaystyle\bigg{(}\frac{\omega_{d-1}}{T}\bigg{)}^{\frac{2}{d+3}}\cdot
V(f)^{\frac{d+1}{d+3}}.$

After a simplification, we obtain

$\displaystyle\frac{1}{2^{d+3}}\frac{||f||_{a}^{a(d+3)+\frac{2(d+1)}{d}}}{||f||_{2}^{\frac{2(d+1)}{d}+2(d+1)}}\leq\frac{\omega_{d-1}^{2}}{c_{d}^{2}}\bigg{(}\frac{V(f)}{||f||_{2}^{2}}\bigg{)}^{d+1}$

Notice that $a(d+3)=2(d+1)$. So

$\displaystyle\frac{\omega_{d-1}^{2}}{c_{d}^{2}}\bigg{(}\frac{V(f)}{||f||_{2}^{2}}\bigg{)}^{d+1}\geq\frac{1}{2^{d+3}}\frac{||f||_{a}^{a(d+3)+\frac{2(d+1)}{d}}}{||f||_{2}^{\frac{2(d+1)}{d}+2(d+1)}}=\frac{1}{2^{d+3}}\bigg{(}\frac{||f||_{a}}{||f||_{2}}\bigg{)}^{(d+1)(2+\frac{2}{d})},$

taking a $\frac{1}{d+1}$-power we get

$\displaystyle\frac{V(f)}{||f||_{2}^{2}}\geq\bigg{(}\frac{c_{d}^{2}}{\omega_{d-1}^{2}}\bigg{)}^{\frac{1}{d+1}}\frac{1}{2^{\frac{d+3}{d+1}}}\bigg{(}\frac{||f||_{a}}{||f||_{2}}\bigg{)}^{(2+\frac{2}{d})}\geq\frac{1}{4}\bigg{(}\frac{c_{d}^{2}}{\omega_{d-1}^{2}}\bigg{)}^{\frac{1}{d+1}}\bigg{(}\frac{||f||_{a}}{||f||_{2}}\bigg{)}^{(2+\frac{2}{d})}.$

###  2.3 Asymptotic behavior of the quotient

#####

In this subsection, we study the growth of
$\big{(}\frac{c_{d}^{2}}{\omega_{d-1}^{2}}\big{)}^{\frac{1}{d+1}}$ when $d$
tends to infinity. Recall the choice of $c_{d}$, after a calculation we have

$\displaystyle\bigg{(}\frac{c_{d}}{\omega_{d-1}}\bigg{)}^{\frac{2}{d+1}}=\frac{1}{(v_{d})^{\frac{2}{d(d+1)}}2^{\frac{d+3}{d(d+1)}}\omega_{d-1}^{\frac{2}{d+1}}}$

#####

The first factor

$\displaystyle\frac{1}{(v_{d})^{\frac{2}{d(d+1)}}}=\frac{\Gamma(\frac{d}{2}+1)^{\frac{2}{d(d+1)}}}{\pi^{\frac{1}{d+1}}}\in\bigg{[}\frac{\Gamma(\frac{d}{2}+1)^{\frac{2}{d(d+1)}}}{\sqrt{\pi}},\Gamma(\frac{d}{2}+1)^{\frac{2}{d(d+1)}}\bigg{]}$

Recall the approximation of Gamma function when $x>0$ is large:
$\Gamma(x)\sim\sqrt{\frac{2\pi}{x}}\big{(}\frac{x}{e}\big{)}^{x}$.

$\displaystyle\Gamma(\frac{d}{2}+1)^{\frac{2}{d(d+1)}}\sim$
$\displaystyle\Bigg{(}\sqrt{\frac{4\pi}{d+2}}\bigg{(}\frac{2+d}{2\mathrm{e}}\bigg{)}^{\frac{d}{2}+1}\Bigg{)}^{\frac{2}{d(d+1)}}$
$\displaystyle<$
$\displaystyle\Bigg{(}2\sqrt{\frac{\pi}{d+2}}\bigg{(}\frac{2+d}{\pi}\bigg{)}^{\frac{d}{2}+1}\Bigg{)}^{\frac{2}{d(d+1)}}$
$\displaystyle\leq$ $\displaystyle
2\bigg{(}\frac{2+d}{\pi}\bigg{)}^{\frac{1}{d}}.$

Notice that if $d\geq 1$, then the following two inequalities hold:
$\frac{2+d}{\pi}\leq d$, and $d^{\frac{1}{d}}\leq\mathrm{e}$. So we have

$1\leq\Gamma(\frac{d}{2}+1)^{\frac{2}{d(d+1)}}\leq 2\mathrm{e}.$ (2.5)

The lower bound is trivial because $\Gamma(x)\geq 1$ when $x\geq 1$. Actually,
we cheated a little bit, because the asymptotic formula
$\Gamma(x)\sim\sqrt{\frac{2\pi}{x}}\big{(}\frac{x}{e}\big{)}^{x}$ holds only
when $x$ is large. The inequality (2.5) is true only when $d$ is large.
However, this is really a minor problem, since we’re only interested in the
behavior of the constant when $d$ is large, and we may assume $d$ is already
large at the very beginning.

These bounds implies that the first factor
$\frac{1}{(v_{d})^{\frac{2}{d(d+1)}}}$ lies in the interval
$[\frac{1}{\sqrt{\pi}},2\mathrm{e}]$, which does not give a decay or growth on
dimension $d$. Like the first factor, the 2nd factor also makes no
contribution on the growth because

$\frac{1}{4}\leq\frac{1}{2^{\frac{d+3}{d(d+1)}}}\leq 1.$

#####

Recall the surface area
$\omega_{d-1}=\frac{2\pi^{\frac{d}{2}}}{\Gamma(\frac{d}{2})}$. So for the 3rd
factor, we have

$\displaystyle\frac{1}{\omega_{d-1}^{\frac{2}{d+1}}}=$
$\displaystyle\frac{\Gamma(\frac{d}{2})^{\frac{2}{d+1}}}{(2\pi^{\frac{d}{2}})^{\frac{2}{d+1}}}>\frac{\Gamma(\frac{d}{2})^{\frac{2}{d+1}}}{10}.$

Combining all three factors yields

$\frac{1}{(v_{d})^{\frac{2}{d(d+1)}}2^{\frac{d+3}{d(d+1)}}\omega_{d-1}^{\frac{2}{d+1}}}>\frac{1}{100}\Gamma(\frac{d}{2})^{\frac{2}{d+1}}.$

Finally

$\displaystyle\Gamma(\frac{d}{2})^{\frac{2}{d+1}}\sim\Bigg{(}\sqrt{\frac{4\pi}{d}}\bigg{(}\frac{d}{2e}\bigg{)}^{\frac{d}{2}}\Bigg{)}^{\frac{2}{d+1}}>\bigg{(}\frac{d}{2e}\bigg{)}^{1-\frac{2}{d+1}}=\frac{d}{2e}\bigg{(}\frac{d}{2e}\bigg{)}^{-\frac{2}{d+1}}>\frac{d}{2\mathrm{e}^{3}}.$

So

$\bigg{(}\frac{c_{d}}{\omega_{d-1}}\bigg{)}^{\frac{2}{d+1}}>\frac{1}{10^{5}}d,$

which implies that

$\displaystyle\frac{V(f)}{||f||_{2}^{2}}\cdot\frac{V(\hat{f})}{||\hat{f}||_{2}^{2}}\geq\bigg{(}\frac{c^{2}_{d}}{\omega^{2}_{d-1}}\bigg{)}^{\frac{2}{d+1}}\geq\frac{1}{10^{10}}d^{2}.$

##  3 $L^{p}(\mathbb{R}^{d})$ case

#####

Now we consider the general case $L^{p}(\mathbb{R}^{d})$. Parallel to $L^{2}$,
now we try to prove an inequality of the following form:

$\frac{V_{p}(f)}{||f||_{L^{p}(\mathbb{R}^{d})}^{p}}\geq
C(d)\bigg{(}\frac{||f||_{L^{a}(\mathbb{R}^{d})}}{||f||_{L^{p}(\mathbb{R}^{d})}}\bigg{)}^{\kappa},$
(3.1)

where $a,\kappa$ are constants to be determined. Suppose that (3.1) holds,
replacing $f$ by $\hat{f}$ and take a product,

$\frac{V_{p}(f)}{||f||_{p}^{p}}\cdot\frac{V_{p}(\hat{f})}{||\hat{f}||_{p}^{p}}\geq
C(d)^{2}\Bigg{(}\frac{||f||_{a}||\hat{f}||_{a}}{||f||_{p}||\hat{f}||_{p}}\Bigg{)}^{\kappa}.$

Recall (2.3), we proved
$\frac{||f||_{a}||\hat{f}||_{a}}{||f||_{2}||\hat{f}||_{2}}\geq 1$, when
$a=2(d+1)/(d+3)$. However, here we should prove

$\frac{||f||_{a}||\hat{f}||_{a}}{||f||_{p}||\hat{f}||_{p}}\geq 1,$

where $a$ is a constant to be chosen. This is already known:

######  Theorem 4 ([5], Theorem 1.3.).

If $1<a<p<\infty$ and $\frac{1}{a}+\frac{1}{p}\geq 1$, then

$\frac{||f||_{a}||\hat{f}||_{a}}{||f||_{p}||\hat{f}||_{p}}\geq 1,\text{ for
every }f\in\mathcal{S}(\mathbb{R}^{d})\setminus\\{0\\}.$

This theorem was proved in [5] for $d=1$, but it can be generalized to general
$d$ without any difficulties. We only used Hölder inequality and Hausdorff-
Young inequality in the proof of Theorem 4, and both inequalities are true for
general $d$.

###  3.1 Choice of some parameters

The following proposition and (3.2) are substitutes of (2.1) in $L^{p}$ case.

######  Proposition 3.1.

For every $p\in(1,\frac{2d}{d-1})$, one can always find an $\epsilon>0$ so
that

$\frac{(d+\epsilon)p}{d+\epsilon+p}>1,\text{ and
}p\leq\frac{2(d+\epsilon)}{d-1+\epsilon}.$

###### Proof.

These two inequalities are equivalent to

$\epsilon>\frac{d+p-dp}{p-1},\text{ and }(p-2)\epsilon\leq 2d-p(d-1).$

Notice that $2d-p(d-1)>0$ always holds. There are two cases.

  * •

When $p\in(1,2]$: The second inequality $(p-2)\epsilon\leq 2d-p(d-1)$ is
trivial, because $(p-2)\epsilon$ is negative while $2d-p(d-1)>0$. We only need
to choose an $\epsilon>0$ satisfies the first inequality, for example
$\epsilon=p/(p-1)$.

  * •

When $p\in(2,\frac{2d}{d-1})$: Now we need to prove the existence of
$\delta>0$ so that

$\frac{d+p-dp}{p-1}<\delta\leq\frac{2d-p(d-1)}{p-2}.$

It suffices to prove that the interval
$(\frac{d+p-dp}{p-1},\frac{2d-p(d-1)}{p-2}]$ is non-empty, and the right
endpoint $\frac{2d-p(d-1)}{p-2}>0$. These are just direct calculations.

∎

Now we choose, for $\epsilon$ in Proposition 3.1,

$\displaystyle a$ $\displaystyle:=\frac{p(d+\epsilon)}{d+\epsilon+p},$ (3.2)
$\displaystyle r$
$\displaystyle:=\frac{p}{a}=\frac{d+\epsilon+p}{d+\epsilon},\
s=\frac{d+\epsilon+p}{p},$ $\displaystyle T$
$\displaystyle:=c_{d}\bigg{(}\frac{||f||_{a}}{||f||_{p}}\bigg{)}^{\frac{as}{d}}=c_{d}\bigg{(}\frac{||f||_{a}}{||f||_{p}}\bigg{)}^{\frac{d+\epsilon}{d}},$

where $c_{d}$ is a constant independent of function $f$ to be determined
later, see (3.3). Notice that $1/r+1/s=1$, and $1<r,s<\infty$. By Proposition
3.1, it’s easy to verify

$1<a<p,\text{ and }\frac{1}{a}+\frac{1}{p}\geq 1.$

So the pair $(a,p)$ verifies the hypothesis in Theorem 4. In fact, the
inequality $a<p$ is trivial, and $a>1$ is equivalent to
$(d+\epsilon)p/(d+\epsilon+p)>1$. Also, the second inequality $1/a+1/p\geq 1$
is equivalent to $p\leq 2(d+\epsilon)/(d-1+\epsilon)$.

#####

Similar to the case $L^{2}$, we use Hölder inequality for $p$ and $q$:

$\displaystyle\int_{|x|\leq T}|f|^{a}\leq$ $\displaystyle\bigg{(}\int_{|x|\leq
T}1\mathrm{d}x\bigg{)}^{\frac{1}{s}}\cdot\bigg{(}\int_{|x|\leq
T}|f|^{ar}\bigg{)}^{\frac{1}{r}}$ $\displaystyle\leq$
$\displaystyle(T^{d}v_{d})^{\frac{1}{s}}||f||_{p}^{\frac{p}{r}}$
$\displaystyle\leq$
$\displaystyle(v_{d}c_{d}^{d})^{\frac{1}{s}}\bigg{(}\frac{||f||_{a}}{||f||_{p}}\bigg{)}^{a}||f||_{p}^{\frac{p}{r}}$
$\displaystyle=$ $\displaystyle(v_{d}c_{d}^{d})^{\frac{1}{s}}||f||_{a}^{a}.$

Choose $c_{d}$ so that

$(v_{d}c^{d}_{d})^{\frac{1}{s}}=1/2,$ (3.3)

and we have

$\int_{|x|>T}|f|^{a}\geq\frac{1}{2}\int_{\mathbb{R}^{d}}|f|^{a}.$

#####

For the integral $\int_{|x|>T}|f|^{a}$, again, we use Hölder inequality. In
$L^{2}(\mathbb{R}^{d})$ case, we used $\int_{|x|>T}1/|x|^{d+1}$. Now we
replace it by $\int_{|x|>T}1/|x|^{d+\epsilon}$.

$\displaystyle\int_{|x|\geq T}|f|^{a}=$ $\displaystyle\int_{|x|\geq
T}\frac{1}{|x|^{a}}|x|^{a}|f(x)|^{a}$ $\displaystyle\leq$
$\displaystyle\bigg{(}\int_{|x|\geq
T}\frac{1}{|x|^{as}}\bigg{)}^{\frac{1}{s}}\cdot\bigg{(}\int_{|x|\geq
T}|x|^{ar}|f(x)|^{ar}\bigg{)}^{\frac{1}{r}}$ $\displaystyle=$
$\displaystyle\bigg{(}\int_{|x|\geq
T}\frac{1}{|x|^{d+\epsilon}}\bigg{)}^{\frac{1}{s}}\cdot\bigg{(}\int_{|x|\geq
T}|x|^{p}|f(x)|^{p}\bigg{)}^{\frac{1}{r}}$ $\displaystyle\leq$
$\displaystyle\bigg{(}\frac{\omega_{d-1}}{\epsilon}\bigg{)}^{\frac{1}{s}}\bigg{(}\frac{1}{T}\bigg{)}^{\frac{\epsilon}{s}}V_{p}(f)^{\frac{1}{r}}.$

We have

$\bigg{(}\frac{1}{T}\bigg{)}^{\frac{\epsilon}{s}}=\bigg{(}\frac{1}{c_{d}^{\epsilon}}\bigg{)}^{\frac{1}{s}}\bigg{(}\frac{||f||_{p}}{||f||_{a}}\bigg{)}^{\frac{a\epsilon}{d}}$

So

$\displaystyle\frac{1}{2}||f||_{a}^{a}\leq\int_{|x|\geq T}|f|^{a}\leq$
$\displaystyle\bigg{(}\frac{\omega_{d-1}}{\epsilon
c_{d}^{\epsilon}}\bigg{)}^{\frac{1}{s}}\bigg{(}\frac{||f||_{p}}{||f||_{a}}\bigg{)}^{\frac{a\epsilon}{d}}V_{p}(f)^{\frac{1}{r}}.$

This is equivalent to

$\displaystyle\frac{||f||_{a}^{a+\frac{a\epsilon}{d}}}{||f||_{p}^{\frac{a\epsilon}{d}+\frac{p}{r}}}\leq
2\bigg{(}\frac{\omega_{d-1}}{\epsilon
c_{d}^{\epsilon}}\bigg{)}^{\frac{1}{s}}\bigg{(}\frac{V_{p}(f)}{||f||_{p}^{p}}\bigg{)}^{\frac{1}{r}}.$

By the choice of $p$ and $a$, we have $p/r=a$. Therefore

$\displaystyle\frac{V_{p}(f)}{||f||_{p}^{p}}\geq\frac{1}{2}\bigg{(}\frac{\epsilon
c_{d}^{\epsilon}}{2\omega_{d-1}}\bigg{)}^{r-1}\bigg{(}\frac{||f||_{a}}{||f||_{p}}\bigg{)}^{p(1+\frac{\epsilon}{d})}.$

Replace $f$ by $\hat{f}$,

$\displaystyle\frac{V_{p}(\hat{f})}{||\hat{f}||_{p}^{p}}\geq\frac{1}{2}\bigg{(}\frac{\epsilon
c_{d}^{\epsilon}}{2\omega_{d-1}}\bigg{)}^{r-1}\bigg{(}\frac{||\hat{f}||_{a}}{||\hat{f}||_{p}}\bigg{)}^{p(1+\frac{\epsilon}{d})}.$

Taking a product and use Theorem 4, we get

$\displaystyle\frac{V_{p}(\hat{f})\cdot
V_{p}(f)}{||\hat{f}||_{p}^{p}\cdot||f||_{p}^{p}}\geq\frac{1}{4}\bigg{(}\frac{\epsilon
c_{d}^{\epsilon}}{2\omega_{d-1}}\bigg{)}^{2(r-1)}\bigg{(}\frac{||f||_{a}\cdot||\hat{f}||_{a}}{||f||_{p}\cdot||\hat{f}||_{p}}\bigg{)}^{p(1+\frac{\epsilon}{d})}\geq\frac{1}{4}\bigg{(}\frac{\epsilon
c_{d}^{\epsilon}}{2\omega_{d-1}}\bigg{)}^{2(r-1)}.$

###  3.2 Asymptotic behavior of the quotient

#####

In this subsection, we study the growth of the quotient $\big{(}\frac{\epsilon
c_{d}^{\epsilon}}{2\omega_{d-1}}\big{)}^{r-1}$ when $d$ tends to infinity and
$p$ fixed. Notice that if $p$ is fixed and $1<p<2d/(d-1)$ holds for all $d\geq
1$, then we must have $1<p\leq 2$. The case $p=2$ has been proved. Therefore,
in this subsection, we consider a fixed number $p\in(1,2)$. Moreover, once
$p\in(1,2)$ is fixed, the choice of $\epsilon$ in Proposition 3.1 can be made
independent of $d$, for example, we may take $\epsilon=p/(p-1)>1$. By (3.3),

$\displaystyle\bigg{(}\frac{\epsilon
c_{d}^{\epsilon}}{2\omega_{d-1}}\bigg{)}^{r-1}=\bigg{(}\frac{\epsilon}{2^{\frac{s\epsilon}{d}+1}}\bigg{)}^{r-1}\bigg{(}\frac{1}{\omega_{d-1}v_{d}^{\frac{\epsilon}{d}}}\bigg{)}^{r-1}$

Recall $r=1+\frac{p}{d+\epsilon}$, for the second factor, we have

$\displaystyle\bigg{(}\frac{1}{\omega_{d-1}v_{d}^{\frac{\epsilon}{d}}}\bigg{)}^{r-1}$
$\displaystyle=\frac{1}{\omega_{d-1}^{\frac{p}{d+\epsilon}}v_{d}^{\frac{\epsilon
p}{d(d+\epsilon)}}}$

##### Asymptotic behavior of
$\big{(}\epsilon/2^{\frac{s\epsilon}{d}+1}\big{)}^{r-1}$:

Recall that $\epsilon$ is independent of $d$, we have

$\bigg{(}\frac{\epsilon}{2^{\frac{s\epsilon}{d}+1}}\bigg{)}^{r-1}=\frac{\epsilon^{\frac{p}{d+\epsilon}}}{2^{\frac{\epsilon
r}{d}+(r-1)}}\sim\frac{1}{2^{(\frac{\epsilon}{d}+1)r}}=\frac{1}{2^{\frac{d+\epsilon+p}{d}}}\sim
1.$

##### Asymptotic behavior of $v_{d}^{\frac{\epsilon p}{d(d+\epsilon)}}$:

We have

$\frac{1}{v_{d}^{\frac{\epsilon
p}{d(d+\epsilon)}}}=\frac{\Gamma\Big{(}\frac{d}{2}+1\Big{)}^{\frac{\epsilon
p}{d(d+\epsilon)}}}{\pi^{\frac{p}{2(d+\epsilon)}}}\sim\Gamma\Big{(}\frac{d}{2}+1\Big{)}^{\frac{\epsilon
p}{d(d+\epsilon)}}\sim\bigg{(}\sqrt{\frac{4\pi}{d+2}}\Big{(}\frac{d+2}{2\mathrm{e}}\Big{)}^{\frac{d}{2}+1}\bigg{)}^{\frac{\epsilon
p}{d(d+\epsilon)}}\sim 1.$

##### Asymptotic behavior of $\omega_{d-1}^{\frac{p}{d+\epsilon}}$:

We have

$\displaystyle\frac{1}{\omega_{d-1}^{\frac{p}{d+\epsilon}}}=\frac{\Gamma(d/2)^{\frac{p}{d+\epsilon}}}{(2\pi^{d/2})^{\frac{p}{d+\epsilon}}}\sim\Gamma\Big{(}\frac{d}{2}\Big{)}^{\frac{p}{d+\epsilon}}\sim\bigg{(}\sqrt{\frac{4\pi}{d}}\big{(}\frac{d}{2\mathrm{e}}\big{)}^{\frac{d}{2}}\bigg{)}^{\frac{p}{d+\epsilon}}.$

The following inequalities holds for all $d\geq 1$:

$\sqrt{\frac{2\mathrm{e}}{d}}\big{(}\frac{d}{2\mathrm{e}}\big{)}^{\frac{d}{2}}<\sqrt{\frac{4\pi}{d}}\big{(}\frac{d}{2\mathrm{e}}\big{)}^{\frac{d}{2}}<2\sqrt{\frac{2\mathrm{e}}{d}}\big{(}\frac{d}{2\mathrm{e}}\big{)}^{\frac{d}{2}}.$

Therefore

$\bigg{(}\sqrt{\frac{4\pi}{d}}\big{(}\frac{d}{2\mathrm{e}}\big{)}^{\frac{d}{2}}\bigg{)}^{\frac{p}{d+\epsilon}}\sim\bigg{(}\sqrt{\frac{2\mathrm{e}}{d}}\big{(}\frac{d}{2\mathrm{e}}\big{)}^{\frac{d}{2}}\bigg{)}^{\frac{p}{d+\epsilon}}=\bigg{(}\frac{d}{2\mathrm{e}}\bigg{)}^{\frac{p(d-1)}{2(d+\epsilon)}}\sim
d^{\frac{p}{2}}.$

##### Sharpness of the growth $d^{p}$:

Taking Gaussian $g(x)=e^{-\pi|x|^{2}}$, a direct calculation shows that

$||g||_{p}^{p}=\frac{1}{p^{\frac{d}{2}}},\text{ and
}V_{p}(g)=\frac{\omega_{d-1}}{2(\pi
p)^{\frac{p+d}{2}}}\Gamma\Big{(}\frac{p+d}{2}\Big{)}.$

Therefore,

$\frac{V_{p}(g)}{||g||_{p}^{p}}\frac{V_{p}(\hat{g})}{||\hat{g}||^{p}_{p}}=\frac{1}{(\pi
p)^{p}}\frac{\Gamma\Big{(}\frac{p+d}{2}\Big{)}^{2}}{\Gamma\Big{(}\frac{d}{2}\Big{)}^{2}}\sim
d^{p},$

which shows that up to a dimension-free constant, the growth $d^{p}$ is
optimal.

###  3.3 Sharpness of the range $p$

To prove the sharpness of the range, we need to show that when
$p\in(\frac{2d}{d-1},\infty)$,

$\inf_{f\in\mathcal{S}(\mathbb{R}^{d})\setminus\\{0\\}}\frac{V_{p}(f)}{||f||_{p}^{p}}\frac{V_{p}(\hat{f})}{||\hat{f}||^{p}_{p}}=0.$

We use the functions

$g_{c}=c^{-\frac{d}{2}}\mathrm{e}^{-\pi\frac{|x|^{2}}{c^{2}}}+c^{\frac{d}{2}}\mathrm{e}^{-\pi
c^{2}|x|^{2}},\ c>0.$

Notice that $g_{c}=\widehat{g_{c}}$, and these functions already appeared in
[6] when $d=1$. We estimate the $L^{p}$ norm of $g_{c}$ and $|x|\cdot|g_{c}|$

$\displaystyle\int|g_{c}|^{p}\geq$ $\displaystyle
c^{-\frac{dp}{2}}\int\mathrm{e}^{-\pi\frac{p|x|^{2}}{c^{2}}}\mathrm{d}x+c^{\frac{dp}{2}}\int\mathrm{e}^{-\pi
pc^{2}|x|^{2}}\mathrm{d}x$ $\displaystyle=$ $\displaystyle
c^{-\frac{dp}{2}}\cdot\frac{c^{d}}{p^{\frac{d}{2}}}+c^{\frac{dp}{2}}\cdot\frac{1}{p^{\frac{d}{2}}c^{d}}$
$\displaystyle=$
$\displaystyle\frac{c^{d-\frac{dp}{2}}+c^{\frac{dp}{2}-d}}{p^{\frac{d}{2}}}.$

For $|x|g_{c}$, we have

$\displaystyle\int|x|^{p}|g_{c}|^{p}$ $\displaystyle\leq
2^{p-1}\int|x|^{p}c^{-\frac{dp}{2}}\mathrm{e}^{-\pi\frac{p|x|^{2}}{c^{2}}}+|x|^{p}c^{\frac{dp}{2}}\mathrm{e}^{-\pi
pc^{2}|x|^{2}}$
$\displaystyle\leq\frac{2^{p-1}\omega_{d-1}\Gamma(\frac{d+p}{2})}{2(\pi
p)^{\frac{d+p}{2}}}\bigg{(}c^{d+p-\frac{dp}{2}}+c^{-(d+p)+\frac{dp}{2}}\bigg{)}.$

So

$\frac{\int|x|^{p}|g_{c}|^{p}}{\int|g_{c}|^{p}}\lesssim\frac{c^{d+p-\frac{dp}{2}}+c^{-(d+p)+\frac{dp}{2}}}{c^{d-\frac{dp}{2}}+c^{\frac{dp}{2}-d}}=:h(c),$

where the implicit constant in $\lesssim$ is independent of $c$. Let
$t:=c^{\frac{dp}{2}-d}$, then

$\frac{c^{d+p-\frac{dp}{2}}+c^{-(d+p)+\frac{dp}{2}}}{c^{d-\frac{dp}{2}}+c^{\frac{dp}{2}-d}}=\frac{t^{\frac{2}{(p-2)d}(d+p-\frac{dp}{2})}+t^{-\frac{2}{(p-2)d}(d+p-\frac{dp}{2})}}{t+\frac{1}{t}}$

There are two cases

  * •

The case $d+p-\frac{dp}{2}\geq 0$. Notice that $p>\frac{2d}{d-1}>2$, so we
have $\frac{2}{(p-2)d}(d+p-\frac{dp}{2})\geq 0$. In this case, it’s easy to
check that $p>\frac{2d}{d-1}$ is equivalent to

$\frac{2}{(p-2)d}(d+p-\frac{dp}{2})<1.$

  * •

The case $d+p-\frac{dp}{2}<0$, now we have
$-\frac{2}{(p-2)d}(d+p-\frac{dp}{2})>0$. It is easy to check that $p>1$
implies

$-\frac{2}{(p-2)d}(d+p-\frac{dp}{2})<1.$

So after a change of variable $t=c^{\frac{dp}{2}-d}$, the function $h(c)$ has
the form

$\frac{t^{\alpha}+\frac{1}{t^{\alpha}}}{t+\frac{1}{t}},\ 0\leq\alpha<1.$

Let $t\rightarrow\infty$, we get the infimum which is $0$.

##  4 Cowling-Price theorem in higher dimension

Firstly, we point out that the third condition
$\frac{1}{q}+\frac{\phi}{d}=\frac{1}{p}+\frac{\theta}{d}$ is an easy
consequence of homogeneous consideration. If (1.8) holds for all
$f\in\mathcal{S}(\mathbb{R}^{d})$, then it also holds for $f(cx)$ for all
$c>0$. After a simplification we get

$\frac{c^{\phi+\frac{d}{q}}}{c^{d}c^{\theta+\frac{\theta}{p}}}\Big{|}\Big{|}|x|^{\theta}f\Big{|}\Big{|}_{p}\cdot\Big{|}\Big{|}|\xi|^{\phi}\hat{f}\Big{|}\Big{|}_{q}\geq\frac{1}{c^{d}}||f||^{2}_{2}.$

We see that $d/q+\phi=\theta/p+\theta$ must hold, otherwise, take
$c\rightarrow 0$ or $c\rightarrow\infty$ would give a contradiction. So the
real difficult parts are first two inequalities in (1.9).

###  4.1 Proof of Theorem 3: If part

Similar to the proof of Theorem 2, we should make careful choices of some
parameters. The following proposition is an analogue of Proposition 3.1.

######  Proposition 4.1.

There exists $\delta>0$, so that

$\displaystyle 1+\frac{d}{\phi q}-(1-\frac{1}{p})(1+\frac{d}{\phi
q})\frac{d}{\theta}<\delta<\min\Big{\\{}1+\frac{d}{\phi q},1+\frac{d}{\theta
p},1+\frac{d}{\phi q}-(\frac{1}{2}-\frac{1}{p})(1+\frac{d}{\phi
q})\frac{d}{\theta}\Big{\\}}.$

###### Proof.

We consider two cases: $p\geq 2$ and $1<p<2$.

  * •

When $p\geq 2$, it suffices to prove that there exists a strictly positive
$\delta$ so that

$\displaystyle 1+\frac{d}{\phi q}-(1-\frac{1}{p})(1+\frac{d}{\phi
q})\frac{d}{\theta}<\delta<\min\Big{\\{}1+\frac{d}{\theta p},1+\frac{d}{\phi
q}-(\frac{1}{2}-\frac{1}{p})(1+\frac{d}{\phi q})\frac{d}{\theta}\Big{\\}}.$

For this we only need to show that

$\min\Big{\\{}1+\frac{d}{\theta p},1+\frac{d}{\phi
q}-(\frac{1}{2}-\frac{1}{p})(1+\frac{d}{\phi
q})\frac{d}{\theta}\Big{\\}}>\max\Big{\\{}0,1+\frac{d}{\phi
q}-(1-\frac{1}{p})(1+\frac{d}{\phi q})\frac{d}{\theta}\Big{\\}}.$

We omit details and point out that

$\displaystyle 1+\frac{d}{\theta p}>1+\frac{d}{\phi
q}-(1-\frac{1}{p})(1+\frac{d}{\phi q})\frac{d}{\theta}\text{ holds, if and
only if }\theta+\frac{d}{p}<d+\phi q.$

Notice that $\frac{d}{q}+\phi=\frac{d}{p}+\theta$, so
$\theta+\frac{d}{p}<d+\phi q$ is equivalent to $q>1$, which is one of the
hypothesis. Also,

$\displaystyle 1+\frac{d}{\phi q}-(\frac{1}{2}-\frac{1}{p})(1+\frac{d}{\phi
q})\frac{d}{\theta}>0\text{ holds, if and only if
}\frac{\theta}{d}>\frac{1}{2}-\frac{1}{p},$

which is our hypothesis. Other parts are almost trivial.

  * •

When $1<p<2$, it suffices to prove that there exists a $\delta>0$ so that

$\displaystyle 1+\frac{d}{\phi q}-(1-\frac{1}{p})(1+\frac{d}{\phi
q})\frac{d}{\theta}<\delta<\min\Big{\\{}1+\frac{d}{\theta p},1+\frac{d}{\phi
q}\Big{\\}}.$

We omit the details and point out that we only need to show

$\min\Big{\\{}1+\frac{d}{\theta p},1+\frac{d}{\phi
q}\Big{\\}}>\max\Big{\\{}0,1+\frac{d}{\phi q}-(1-\frac{1}{p})(1+\frac{d}{\phi
q})\frac{d}{\theta}\Big{\\}}.$

The above inequality is equivalent to $p,q>1$.

∎

Once $\delta$ has been chosen, we define $\epsilon,\tilde{\epsilon}$ via

$\frac{\epsilon}{d+\epsilon}=\delta\frac{\phi q}{d+\phi q},\text{ and
}\frac{\tilde{\epsilon}}{d+\tilde{\epsilon}}=\delta\frac{\theta p}{d+\theta
p}.$ (4.1)

Particularly,

$\frac{d}{\epsilon}=\frac{d+\phi q-\delta\phi q}{\delta\phi q}.$

By Proposition 4.1, the numerator $d+\phi q-\delta\phi q>0$, so $\epsilon>0$.
By the same reason, we have $\tilde{\epsilon}>0$.

Now we choose

$\displaystyle a$ $\displaystyle:=\frac{p}{1+\frac{p\theta}{d+\epsilon}},$
$\displaystyle r$ $\displaystyle:=\frac{2}{a}\text{ and }s:=\frac{r}{r-1},$
$\displaystyle T$
$\displaystyle:=c_{d}\bigg{(}\frac{||f||_{a}}{||f||_{2}}\bigg{)}^{\frac{as}{d}},$

where $c_{d}$ is the constant so that $(c_{d}^{d}v_{d})^{1/s}=1/2$.

By Hölder inequality for the pair $(r,s)$ as before, we have

$\frac{1}{2}||f||_{a}^{a}\leq\int_{|x|>T}|f|^{a}.$

Now we choose $b>0$ and $r_{1},s_{1}\in(1,\infty)$ so that

$\displaystyle ar_{1}$ $\displaystyle=p,$ (4.2) $\displaystyle s_{1}$
$\displaystyle=\frac{r_{1}}{r_{1}-1},$ $\displaystyle br_{1}$
$\displaystyle=\theta p.$

Moreover, we have

$bs_{1}=\frac{\theta p}{r_{1}-1}=\frac{\theta p}{\frac{p}{a}-1}=d+\epsilon.$

Trivially, we have $b>0$, and we still need to verify $r_{1}>1$. This
equivalent to $p>a$, which is true by the choice of $a$. Now, writing

$\int_{|x|>T}|f|^{a}=\int_{|x|>T}\frac{1}{|x|^{b}}\cdot|x|^{b}|f|^{a}$

and using Hölder inequality for the pair $(s_{1},r_{1})$, we can prove (recall
$bs_{1}=d+\epsilon$ and $br_{1}=\theta p$)

$\frac{1}{2}||f||_{a}^{a}\leq\bigg{(}\frac{\omega_{d-1}}{\epsilon
T^{\epsilon}}\bigg{)}^{\frac{1}{s_{1}}}\Big{|}\Big{|}|x|^{\theta}f\Big{|}\Big{|}^{a}_{p},$

after a simplification

$\frac{\Big{|}\Big{|}|x|^{\theta}f\Big{|}\Big{|}_{p}}{||f||_{2}}\geq\bigg{(}\frac{\epsilon
c_{d}^{\epsilon}}{2^{s_{1}}\omega_{d-1}}\bigg{)}^{\frac{1}{as_{1}}}\bigg{(}\frac{||f||_{a}}{||f||_{2}}\bigg{)}^{1+\frac{s\epsilon}{ds_{1}}}$

#####

For the Fourier transform side, we choose

$\displaystyle\tilde{a}:=\frac{q}{1+\frac{q\phi}{d+\tilde{\epsilon}}},\quad\tilde{r}:=\frac{2}{\tilde{a}}\text{
and
}\tilde{s}:=\frac{\tilde{r}}{\tilde{r}-1},\quad\tilde{T}:=\widetilde{c_{d}}\bigg{(}\frac{||\hat{f}||_{\tilde{a}}}{||\hat{f}||_{2}}\bigg{)}^{\frac{\tilde{a}\tilde{s}}{d}},$

Now we verify that $a=\tilde{a}$ under our choice of
$\epsilon,\tilde{\epsilon}$. After taking the reciprocal, it suffices to prove

$\frac{1}{q}+\frac{\phi}{d+\tilde{\epsilon}}=\frac{1}{p}+\frac{\theta}{d+\epsilon}.$

Using the hypothesis $1/q+\phi/d=1/p+\theta/d$, we only need to prove
$\frac{\phi}{d+\tilde{\epsilon}}-\frac{\phi}{d}=\frac{\theta}{d+\epsilon}-\frac{\theta}{d}$,
this is equivalent to

$\frac{\phi\tilde{\epsilon}}{d+\tilde{\epsilon}}=\frac{\theta\epsilon}{d+\epsilon}.$

The above equality is easy to prove by combining the hypothesis
$1/q+\phi/d=1/p+\theta/d$ with the choice of $\epsilon$ and
$\tilde{\epsilon}$, see (4.1).

As an immediate corollary of $a=\tilde{a}$, we have

$\tilde{r}=r,\quad\tilde{s}=s.$

The constant $\widetilde{c_{d}}$ satisfies
$(\widetilde{c_{d}}^{d}v_{d})^{1/\tilde{s}}=1/2$, so
$\widetilde{c_{d}}=c_{d}$. Now parallel to previous proof, we obtain

$\frac{1}{2}||\hat{f}||_{\tilde{a}}^{\tilde{a}}=\frac{1}{2}||\hat{f}||_{a}^{a}\leq\int_{|\xi|>\widetilde{T}}|\hat{f}|^{a}.$

Now we choose $\tilde{b}>0$ and
$\widetilde{r_{1}},\widetilde{s_{1}}\in(1,\infty)$ so that

$\displaystyle\tilde{a}\widetilde{r_{1}}$ $\displaystyle=q,$ (4.3)
$\displaystyle\widetilde{s_{1}}$
$\displaystyle=\frac{\widetilde{r_{1}}}{\widetilde{r_{1}}-1},$
$\displaystyle\tilde{b}\widetilde{r_{1}}$ $\displaystyle=\phi q.$

Similar to $bs_{1}=d+\epsilon$, we have
$\tilde{b}\widetilde{s_{1}}=d+\tilde{\epsilon}$. Now we can prove, by the same
method,

$\frac{1}{2}||\hat{f}||_{a}^{a}\leq\bigg{(}\frac{\omega_{d-1}}{\tilde{\epsilon}\big{(}\widetilde{T}\big{)}^{\tilde{\epsilon}}}\bigg{)}^{1/\widetilde{s_{1}}}\Big{|}\Big{|}|\xi|^{\phi}\hat{f}\Big{|}\Big{|}^{a}_{q},$

A simplification shows that

$\frac{\Big{|}\Big{|}|\xi|^{\phi}\hat{f}\Big{|}\Big{|}_{q}}{||\hat{f}||_{2}}\geq\bigg{(}\frac{\tilde{\epsilon}c_{d}^{\tilde{\epsilon}}}{2^{\widetilde{s_{1}}}\omega_{d-1}}\bigg{)}^{1/a\widetilde{s_{1}}}\bigg{(}\frac{||\hat{f}||_{a}}{||\hat{f}||_{2}}\bigg{)}^{s\tilde{\epsilon}/d\widetilde{s_{1}}+1}.$

Take a product we get

$\frac{\Big{|}\Big{|}|x|^{\theta}f\Big{|}\Big{|}_{p}}{||f||_{2}}\frac{\Big{|}\Big{|}|\xi|^{\phi}\hat{f}\Big{|}\Big{|}_{q}}{||\hat{f}||_{2}}\geq\bigg{(}\frac{\epsilon
c_{d}^{\epsilon}}{2^{s_{1}}\omega_{d-1}}\bigg{)}^{\frac{1}{as_{1}}}\bigg{(}\frac{\tilde{\epsilon}c_{d}^{\tilde{\epsilon}}}{2^{\widetilde{s_{1}}}\omega_{d-1}}\bigg{)}^{1/a\widetilde{s_{1}}}\bigg{(}\frac{||f||_{a}}{||f||_{2}}\bigg{)}^{1+\frac{\epsilon
s}{ds_{1}}}\bigg{(}\frac{||\hat{f}||_{a}}{||\hat{f}||_{2}}\bigg{)}^{\tilde{\epsilon}s/d\widetilde{s_{1}}+1}.$

Notice that by our choice of $\epsilon,\tilde{\epsilon}$,

$1+\frac{\epsilon s}{ds_{1}}=1+\frac{\tilde{\epsilon}s}{d\widetilde{s_{1}}}.$
(4.4)

To prove this, it suffices to check that
$\epsilon/s_{1}=\tilde{\epsilon}/\widetilde{s_{1}}$. We recall
$bs_{1}=d+\epsilon$ and $\tilde{b}\widetilde{s_{1}}=d+\tilde{\epsilon}$, so
(4.4) is equivalent to $\epsilon
b/(d+\epsilon)=\tilde{\epsilon}\tilde{b}/(d+\widetilde{\epsilon})$. Also by
(4.2) and (4.3), we have $b=a\theta$ and $\tilde{b}=\tilde{a}\phi$. Finally,
we already proved $a=\tilde{a}$, so (4.4) is equivalent to

$\frac{\epsilon\theta}{d+\epsilon}=\frac{\tilde{\epsilon}\phi}{d+\widetilde{\epsilon}}.$

This equality is easy to verify by the definition of $\epsilon$ and
$\tilde{\epsilon}$, see (4.1).

So, by (4.4), we get

$\displaystyle\frac{\Big{|}\Big{|}|x|^{\theta}f\Big{|}\Big{|}_{p}}{||f||_{2}}\frac{\Big{|}\Big{|}|\xi|^{\phi}\hat{f}\Big{|}\Big{|}_{q}}{||\hat{f}||_{2}}$
$\displaystyle\geq\bigg{(}\frac{\epsilon
c_{d}^{\epsilon}}{2^{s_{1}}\omega_{d-1}}\bigg{)}^{\frac{1}{as_{1}}}\bigg{(}\frac{\tilde{\epsilon}c_{d}^{\tilde{\epsilon}}}{2^{\widetilde{s_{1}}}\omega_{d-1}}\bigg{)}^{1/a\widetilde{s_{1}}}\bigg{(}\frac{||f||_{a}||\hat{f}||_{a}}{||f||_{2}||\hat{f}||_{2}}\bigg{)}^{1+\frac{\epsilon
s}{ds_{1}}}$ $\displaystyle\geq\bigg{(}\frac{\epsilon
c_{d}^{\epsilon}}{2^{s_{1}}\omega_{d-1}}\bigg{)}^{\frac{1}{as_{1}}}\bigg{(}\frac{\tilde{\epsilon}c_{d}^{\tilde{\epsilon}}}{2^{\widetilde{s_{1}}}\omega_{d-1}}\bigg{)}^{1/a\widetilde{s_{1}}},$

where the last inequality we use (2.3).

###  4.2 Proof of Theorem 3: Only if part

#####

Now we consider the converse part. It suffices to prove that the theorem is
not true if

$\frac{\theta}{d}\leq\frac{1}{2}-\frac{1}{p}\quad\text{or}\quad\frac{\phi}{d}\leq\frac{1}{2}-\frac{1}{q}.$

There are two cases. The first case is
$\frac{\theta}{d}=\frac{1}{2}-\frac{1}{p}$, and
$\frac{\phi}{d}=\frac{1}{2}-\frac{1}{q}.$ The second case is
$\frac{\theta}{d}<\frac{1}{2}-\frac{1}{p}$, and
$\frac{\phi}{d}<\frac{1}{2}-\frac{1}{q}.$ Notice that due to the homogeneous
condition,

$\frac{1}{q}+\frac{\phi}{d}=\frac{1}{p}+\frac{\theta}{d},$

other cases is impossible, for example, we do not need to consider
$\frac{\theta}{d}=\frac{1}{2}-\frac{1}{p}$, and
$\frac{\phi}{d}<\frac{1}{2}-\frac{1}{q}$.

####  4.2.1 The case $\frac{\theta}{d}=\frac{1}{2}-\frac{1}{p}$, and
$\frac{\phi}{d}=\frac{1}{2}-\frac{1}{q}.$

#####

Our goal is to construct an $f$ so that

$|x|^{\theta}f\in L^{p}(\mathbb{R}^{d})\text{ and }|\xi|^{\phi}\hat{f}\in
L^{q}(\mathbb{R}^{d}),\text{ but }f\notin L^{2}(\mathbb{R}^{d}).$

For this, we need a theorem of Wainger. The following version is a special
case of Wainger’s theorem, where we choose $l=0$, and $b(t)$ in slowly varying
class([4], Definition 2, page 7) is a logarithmic power.

######  Theorem 5 ([4], Theorem 1, page 8).

Let $b(t)=\log^{-\frac{1}{2}}(t+2)$. Let $\psi(t)$ be a smooth function with
the following properties

  * •

$\psi(t)=0$ when $t\leq\frac{1}{2}$.

  * •

$\psi(t)=1$ when $t\geq 1$.

  * •

$0\leq\psi(t)\leq 1$ for all $t\in\mathbb{R}$.

Assume that $-\infty<\gamma<d$, and $\gamma\neq 0,-2,-4,...$ Define

$F_{\epsilon}(x):=\mathcal{F}^{-1}\Big{(}\frac{\psi(|\xi|)}{|\xi|^{\gamma}}b(|\xi|)\mathrm{e}^{-\epsilon|\xi|}\Big{)},$

where $\mathcal{F}^{-1}$ is the Fourier inversion. Then

  * (1)

$F(x):=\lim_{\epsilon\rightarrow 0^{+}}F_{\epsilon}(x)$ exists for all $x\neq
0$, and $F(x)$ is $C^{\infty}$-differentiable when $x\neq 0$.

  * (2)

For any integer $r$, we have $|F_{\epsilon}(x)|=\mathcal{O}(|x|^{-r})$
uniformly in $\epsilon\geq 0$, as $|x|\rightarrow\infty$.

When $x$ is near $0$, we have

  * (3)

$F(x)=C(\gamma,d)\cdot b\big{(}\frac{1}{|x|}\big{)}|x|^{\gamma-d}+E(x)$, where
$C(\gamma,d)$ is a non-zero constant and $E(x)$ is the error term.

  * (4)

The error term satisfies
$|E(x)|=\mathcal{O}(1)+o\Big{(}b\big{(}1/|x|\big{)}|x|^{\gamma-d}\Big{)}$ as
$x\rightarrow 0$.

Also, if $F(x)\in L^{1}(\mathbb{R}^{d})$, then $F_{\epsilon}(x)$ is dominated
by an integrable function, and

$\widehat{F}(\xi)=\frac{\psi(|\xi|)}{|\xi|^{\gamma}}b(|\xi|).$

To apply this theorem, we define our $f$ via

$f(x):=F(x)=\lim_{\epsilon\rightarrow 0^{+}}F_{\epsilon}(x),\text{ with
}\gamma=\frac{d}{2}.$

However, the function $f$ may not in Schwartz class. This is a minor problem
since we can approximate $f$ by elements in $\mathcal{S}(\mathbb{R}^{d})$ via
a density argument.

##### The function $f$ is not $L^{2}$ integrable:

When $x$ is near $0$, function $f$ behaves like

$\frac{1}{|x|^{\frac{d}{2}}\log^{\frac{1}{2}}(\frac{1}{|x|}+2)}\sim\frac{1}{|x|^{\frac{d}{2}}\log^{\frac{1}{2}}(\frac{1}{|x|})},\text{
as }x\rightarrow 0.$

An easy calculation yields that the function

$x\mapsto\frac{1}{|x|^{\frac{d}{2}}\log^{\frac{1}{2}}(|x|^{-1})}$

is not $L^{2}$ integrable near $0$. So $f\notin L^{2}(\mathbb{R}^{d})$.

##### The function $|x|^{\theta}|f|$ is $L^{p}$ integrable:

We notice that singularity appears only when $x\rightarrow 0$ and
$|x|\rightarrow\infty$, so we only need to prove the $L^{p}$-integrability of
$|x|^{\theta}f$ when $x$ is near $0$, and when $|x|$ is large. When $x$ is
near $0$, similar to the proof of $f\notin L^{2}$, the function
$|x|^{\theta}|f|$ behaves like

$\frac{1}{|x|^{\frac{d}{2}-\theta}\log^{\frac{1}{2}}(\frac{1}{|x|})},\text{ as
}x\rightarrow 0.$

Taking $L^{p}$-norm, and using the hypothesis
$\frac{\theta}{d}=\frac{1}{2}-\frac{1}{p}$, we obtain an integral

$\int_{x\in\mathbb{R}^{d}:|x|\ll
1}\frac{\mathrm{d}x}{|x|^{d}\log^{\frac{p}{2}}(\frac{1}{|x|})}$ (4.5)

Here $|x|\ll 1$ means that $|x|$ is much less than $1$. Notice that
$\frac{\theta}{d}=\frac{1}{2}-\frac{1}{p}$ and $\theta>0$, we must have $p>2$.
Therefore, (4.5) is finite, and $|x|^{\theta}f$ is $L^{p}$ integrable near
$x=0$.

The $L^{p}$ integrability of $|x|^{\theta}f$ when $|x|\rightarrow\infty$ is
easier. By Wainger’s theorem, for every integer $r$, as
$|x|\rightarrow\infty$,

$|F_{\epsilon}(x)|=\mathcal{O}(|x|^{-r}),\text{ uniformly in }\epsilon\geq 0.$

So $f(x)=\mathcal{O}(|x|^{-r})$ also holds. Just take $r$ large, we see that
$|x|^{\theta}|f|$ is $L^{p}$ integrable when $|x|\gg 1$. Here $|x|\gg 1$ means
that $|x|$ is much larger than $1$.

#####

Finally, we prove $|\xi|^{\phi}\hat{f}$ is $L^{q}$ integrable. If we can prove
$f\in L^{1}$, then by Wainger’s theorem

$\hat{f}(\xi)=\frac{\psi(|\xi|)}{|\xi|^{\frac{d}{2}}\log^{\frac{1}{2}}(|\xi|+2)}.$

The function $|\xi|^{\phi}\hat{f}\in L^{q}$ is similar to the proof of
$|x|^{\theta}f\in L^{p}$. Notice that $|\xi|^{\phi}\hat{f}$ has no singularity
at $x=0$, since $\psi(t)=0$ when $t\leq 1/2$. So it suffices to prove the
$L^{1}$-integrability of $f$. When $x$ is near $0$, function $f$ behaves like

$\frac{1}{|x|^{\frac{d}{2}}\log^{\frac{1}{2}}\Big{(}\frac{1}{|x|}\Big{)}},$

and it’s easy to show that

$\int_{x\in\mathbb{R}^{d}:|x|\ll
1}\frac{\mathrm{d}x}{|x|^{\frac{d}{2}}\log^{\frac{1}{2}}\Big{(}\frac{1}{|x|}\Big{)}}<\infty.$

When $|x|\gg 1$, the $L^{1}$-integrability of $f$ is easier, just notice that
$f(x)=\mathcal{O}(|x|^{-r})$ for every $r\in\mathbb{N}$.

####  4.2.2 The case $\frac{\theta}{d}<\frac{1}{2}-\frac{1}{p}$, and
$\frac{\phi}{d}<\frac{1}{2}-\frac{1}{q}.$

We’re going to use the Rudin-Shapiro construction, which is similar to the
1-dimensional case in [2] (see [2], counterexample II). To avoid complicated
notations, we only focus on the case $d=2$. After proving the two-dimensional
case, we will give a brief explanation on general $d$.

Choose a non-negative smooth function $f$ which is supported in
$[1/10,9/10]^{2}$ and $||f||_{\infty}=1$, we define
$\\{f_{k}\\}^{\infty}_{k=0},\\{g_{k}\\}^{\infty}_{k=0}$,
$\\{p_{k}\\}^{\infty}_{k=0},\\{q_{k}\\}^{\infty}_{k=0}$ inductively:

  * •

When $k=0$, define $f_{0}=g_{0}=p_{0}=q_{0}:=f$.

  * •

Assume that the four sequences have been defined for $0,1,...,k$, now we
define

$\displaystyle f_{k+1}$
$\displaystyle:=f_{k}+\Big{[}g_{k}(x-(2^{k},0))+p_{k}(x-(0,2^{k}))+q_{k}(x-(2^{k},2^{k}))\Big{]}$
$\displaystyle g_{k+1}$
$\displaystyle:=f_{k}+\Big{[}g_{k}(x-(2^{k},0))-p_{k}(x-(0,2^{k}))-q_{k}(x-(2^{k},2^{k}))\Big{]}$
$\displaystyle p_{k+1}$
$\displaystyle:=f_{k}-\Big{[}g_{k}(x-(2^{k},0))-p_{k}(x-(0,2^{k}))+q_{k}(x-(2^{k},2^{k}))\Big{]}$
$\displaystyle q_{k+1}$
$\displaystyle:=f_{k}-\Big{[}g_{k}(x-(2^{k},0))+p_{k}(x-(0,2^{k}))-q_{k}(x-(2^{k},2^{k}))\Big{]}$

Notice that

$\displaystyle\widehat{f_{k+1}}$
$\displaystyle=\widehat{f_{k}}+\mathrm{e}^{-2\pi\mathrm{i}2^{k}\xi_{1}}\widehat{g_{k}}+\mathrm{e}^{-2\pi\mathrm{i}2^{k}\xi_{2}}\widehat{p_{k}}+\mathrm{e}^{-2\pi\mathrm{i}2^{k}(\xi_{1}+\xi_{2})}\widehat{q_{k}}$
$\displaystyle\widehat{g_{k+1}}$
$\displaystyle=\widehat{f_{k}}+\mathrm{e}^{-2\pi\mathrm{i}2^{k}\xi_{1}}\widehat{g_{k}}-\mathrm{e}^{-2\pi\mathrm{i}2^{k}\xi_{2}}\widehat{p_{k}}-\mathrm{e}^{-2\pi\mathrm{i}2^{k}(\xi_{1}+\xi_{2})}\widehat{q_{k}}$
$\displaystyle\widehat{p_{k+1}}$
$\displaystyle=\widehat{f_{k}}-\mathrm{e}^{-2\pi\mathrm{i}2^{k}\xi_{1}}\widehat{g_{k}}+\mathrm{e}^{-2\pi\mathrm{i}2^{k}\xi_{2}}\widehat{p_{k}}-\mathrm{e}^{-2\pi\mathrm{i}2^{k}(\xi_{1}+\xi_{2})}\widehat{q_{k}}$
$\displaystyle\widehat{q_{k+1}}$
$\displaystyle=\widehat{f_{k}}-\mathrm{e}^{-2\pi\mathrm{i}2^{k}\xi_{1}}\widehat{g_{k}}-\mathrm{e}^{-2\pi\mathrm{i}2^{k}\xi_{2}}\widehat{p_{k}}+\mathrm{e}^{-2\pi\mathrm{i}2^{k}(\xi_{1}+\xi_{2})}\widehat{q_{k}}$

By parallelogram law, for $a,b,c,d\in\mathbb{C}$

$|a+b+c+d|^{2}+|a+b-c-d|^{2}+|a-b+c-d|^{2}+|a-b-c+d|^{2}=4(|a|^{2}+|b|^{2}+|c|^{2}+|d|^{2}).$
(4.6)

Let $(a,b,c,d)=\Big{(}\widehat{f_{k}},\
\mathrm{e}^{-2\pi\mathrm{i}2^{k}\xi_{1}}\widehat{g_{k}},\
\mathrm{e}^{-2\pi\mathrm{i}2^{k}\xi_{2}}\widehat{p_{k}},\
\mathrm{e}^{-2\pi\mathrm{i}2^{k}(\xi_{1}+\xi_{2})}\widehat{q_{k}}\Big{)}$, we
get

$|\widehat{f_{k+1}}|^{2}+|\widehat{g_{k+1}}|^{2}+|\widehat{p_{k+1}}|^{2}+|\widehat{q_{k+1}}|^{2}=4\Big{(}|\widehat{f_{k}}|^{2}+|\widehat{g_{k}}|^{2}+|\widehat{p_{k}}|^{2}+|\widehat{q_{k}}|^{2}\Big{)}.$
(4.7)

Furthermore, the function $f_{k}$ is supported in $[0,2^{k}]^{2}$. When $0\leq
m,n\leq 2^{k}-1$,

$f_{k}\Big{|}_{[m,m+1]\times[n,n+1]}=\pm f(x-(m,n)),$

that is to say, in every cubes
$Q_{m,n}:=[m,m+1]\times[n,n+1]\subset[0,2^{k}]^{2}$, function $f_{k}$ is
obtained, up to a sign, via a translation of $f$ along the vector $(m,n)$.
Therefore,

$||f_{k}||_{2}^{2}=2^{2k}||f||_{2}^{2},\text{ and
}||f_{k}||_{\infty}=||f||_{\infty}=1.$ (4.8)

#####

Now we calculate the $L^{p}$ norm of $|x|^{\theta}f_{k}$:

$\displaystyle\bigg{(}\int_{\mathbb{R}^{d}}|x|^{\theta
p}|f_{k}|^{p}\mathrm{d}x\bigg{)}^{\frac{1}{p}}=$
$\displaystyle\bigg{(}\int_{[0,2^{k}]^{2}}|x|^{\theta
p}|f_{k}|^{p}\mathrm{d}x\bigg{)}^{\frac{1}{p}}$ $\displaystyle\leq$
$\displaystyle||f_{k}||_{\infty}\bigg{(}\int_{[0,2^{k}]^{2}}|x|^{\theta
p}\mathrm{d}x\bigg{)}^{\frac{1}{p}}$ $\displaystyle\leq$
$\displaystyle\bigg{(}\int_{|x|\leq\sqrt{d}2^{k}}|x|^{\theta
p}\mathrm{d}x\bigg{)}^{\frac{1}{p}}$ $\displaystyle\leq$ $\displaystyle
C(d,\theta,p)\cdot 2^{k(\frac{d}{p}+\theta)}.$

Here $C(d,\theta,p)$ is a constant independent of $k$, here $d=2$.

#####

By (4.7), and recall $f_{0}=g_{0}=p_{0}=q_{0}=f$, we have

$|\widehat{f_{k}}|^{2}\leq|\widehat{f_{k}}|^{2}+|\widehat{g_{k}}|^{2}+|\widehat{p_{k}}|^{2}+|\widehat{q_{k}}|^{2}=4^{k+1}|\hat{f}|^{2},$
(4.9)

which implies that $|\widehat{f_{k}}|\leq 2^{k+1}|\hat{f}|$. The $L^{q}$ norm
of $|\xi|^{\phi}\hat{f}$ can be controlled:

$\bigg{(}\int_{\mathbb{R}^{d}}|\xi|^{\phi
q}|\widehat{f_{k}}|^{q}\mathrm{d}\xi\bigg{)}^{\frac{1}{q}}\leq
2^{k+1}\bigg{(}\int_{\mathbb{R}^{d}}|\xi|^{\phi
q}|\widehat{f}|^{q}\mathrm{d}\xi\bigg{)}^{\frac{1}{q}}=2^{k+1}\Big{|}\Big{|}|\xi|^{\phi}\hat{f}\Big{|}\Big{|}_{q}.$

#####

Now we test (1.8) by $f_{k}$, recall that
$||f_{k}||_{2}^{2}=2^{2k}||f||^{2}_{2}$.

$\displaystyle C\cdot
2^{2k}||f||^{2}_{2}\leq\Big{|}\Big{|}|x|^{\theta}f_{k}\Big{|}\Big{|}_{p}\cdot\Big{|}\Big{|}|\xi|^{\phi}\widehat{f_{k}}\Big{|}\Big{|}_{q}\leq
2^{k+1}\Big{|}\Big{|}|\xi|^{\phi}\hat{f}\Big{|}\Big{|}_{q}\cdot
C(d,\theta,p)\cdot 2^{k(\frac{d}{p}+\theta)}.$ (4.10)

After a simplification, inequality (4.10) is reduced to

$2^{k}\leq C_{d,\theta,p,\phi,q,f}\cdot 2^{k(\frac{d}{p}+\theta)}.$ (4.11)

The constant $C_{d,\theta,p,\phi,q,f}$ is independent of $k$, and the above
inequality should hold for all $k$. This is impossible, because when $d=2$,

$\text{the
hypothesis}\quad\frac{\theta}{d}<\frac{1}{2}-\frac{1}{p}\quad\text{implies}\quad\frac{d}{p}+\theta<1.$

Let $k$ goes to infinity will give a contradiction. Therefore, the inequality
(1.8) can not hold for $f_{k}$ when $k$ is large.

#####

For general $d\geq 1$, we need to construct $2^{d}$ families of functions

$\\{f_{1,k}\\}_{k=0}^{\infty},\ \\{f_{2,k}\\}_{k=0}^{\infty},...,\
\\{f_{2^{d},k}\\}_{k=0}^{\infty}.$

One may think, when $d=2$,

$\displaystyle\\{f_{1,k}\\}_{k=0}^{\infty}$
$\displaystyle=\\{f_{k}\\}_{k=0}^{\infty},\
\\{f_{2,k}\\}_{k=0}^{\infty}=\\{g_{k}\\}_{k=0}^{\infty},$
$\displaystyle\\{f_{3,k}\\}_{k=0}^{\infty}$
$\displaystyle=\\{p_{k}\\}_{k=0}^{\infty},\
\\{f_{4,k}\\}_{k=0}^{\infty}=\\{q_{k}\\}_{k=0}^{\infty}.\ $

The reason for using $2^{d}$ families is because a cube in $\mathbb{R}^{d}$,
for example, $[0,2)^{d}$ has $2^{d}$ dyadic children. Each of its children has
the form

$I_{1}\times I_{2}\times\cdots\times I_{d},\text{ where }I_{i}=[0,1)\text{ or
}[1,2).$

Another ingredient is a parallelogram law for $2^{d}$ numbers.

######  Proposition 4.2.

For every $d\geq 1$, there exists a choice of the signs
$\\{\epsilon_{i,j}\\}_{1\leq i\leq 2^{d},\\\ 2\leq j\leq 2^{d}}$(notice that
$i$ starts from $1$, while $j$ starts from $2$), with $\epsilon_{i,j}=\pm 1$
for all $i,j$, so that for every $a_{1},a_{2},...,a_{2^{d}}\in\mathbb{C}$, we
have

$\Big{|}a_{1}+\epsilon_{1,2}a_{2}+\cdots+\epsilon_{1,2^{d}}a_{2^{d}}\Big{|}^{2}+\cdots+\Big{|}a_{1}+\epsilon_{2^{d},2}a_{2}+\cdots+\epsilon_{2^{d},2^{d}}a_{2^{d}}\Big{|}^{2}=2^{d}(|a_{1}|^{2}+|a_{2}|^{2}+\cdots|a_{2^{d}}|^{2}).$
(4.12)

###### Proof.

We prove it by induction, when $d=1$, this is just the ordinary parallelogram
law

$|a_{1}+a_{2}|^{2}+|a_{1}-a_{2}|^{2}=2(|a_{1}|^{2}+|a_{2}|^{2}).$

Also, we already saw the case $d=2$ in (4.6). Assume that the proposition is
true for $1,2,...,d$, now we prove the case $d+1$. The left hand side of
(4.12) has exactly $2^{d}$ terms, and we may write it as

$\sum_{i=1}^{2^{d}}\Big{|}a_{1}+\epsilon_{i,2}a_{2}+\cdots+\epsilon_{i,2^{d}}a_{2^{d}}\Big{|}^{2}=\sum_{i=1}^{2^{d}}\Big{|}a_{1}+\sum_{j=2}^{2^{d}}\epsilon_{i,j}a_{j}\Big{|}^{2}.$
(4.13)

For each term
$|a_{1}+\epsilon_{i,2}a_{2}+\cdots+\epsilon_{i,2^{d}}a_{2^{d}}|^{2}$ in
(4.13), we define two new terms

$\displaystyle\Big{|}(a_{1}+\epsilon_{i,2}a_{2}+\cdots+\epsilon_{i,2^{d}}a_{2^{d}})+(a_{2^{d}+1}+\epsilon_{i,2}a_{2^{d}+2}\cdots+\epsilon_{i,2^{d}}a_{2^{d+1}})\Big{|}^{2}+$
(4.14)
$\displaystyle\Big{|}(a_{1}+\epsilon_{i,2}a_{2}+\cdots+\epsilon_{i,2^{d}}a_{2^{d}})-(a_{2^{d}+1}+\epsilon_{i,2}a_{2^{d}+2}\cdots+\epsilon_{i,2^{d}}a_{2^{d+1}})\Big{|}^{2}.$

This is exactly our choice of the signs in $d+1$ case and finishes the proof.
In fact, by parallelogram law,

$(\ref{parallelogram law for 2^{d+1} numbers, a form of general
term})=2\Big{(}|a_{1}+\epsilon_{i,2}a_{2}+\cdots+\epsilon_{i,2^{d}}a_{2^{d}}|^{2}+|a_{2^{d}+1}+\epsilon_{i,2}a_{2^{d}+2}\cdots+\epsilon_{i,2^{d}}a_{2^{d+1}}|^{2}\Big{)}.$

So after taking a sum, we get

$\displaystyle\sum^{2^{d}}_{i=1}(\ref{parallelogram law for 2^{d+1} numbers, a
form of general term})=$ $\displaystyle
2\times\sum^{2^{d}}_{i=1}|a_{1}+\epsilon_{i,2}a_{2}+\cdots+\epsilon_{i,2^{d}}a_{2^{d}}|^{2}+2\times\sum^{2^{d}}_{i=1}|a_{2^{d}+1}+\epsilon_{i,2}a_{2^{d}+2}\cdots+\epsilon_{i,2^{d}}a_{2^{d+1}}|^{2}$
$\displaystyle=$ $\displaystyle 2\cdot
2^{d}(|a_{1}|^{2}+\cdots+|a_{2^{d}}|^{2})+2\cdot
2^{d}(|a_{2^{d}+1}|^{2}+\cdots+|a_{2^{d+1}}|^{2})$ $\displaystyle=$
$\displaystyle 2^{d+1}(|a_{1}|^{2}+\cdots+|a_{2^{d+1}}|^{2}).$

Here, in the second-to-last equality, we used the induction hypothesis for
$d$. ∎

Choose a smooth function $f$ supported in $[1/10,9/10]^{d}$ and
$||f||_{\infty}=1$. We define, inductively,

  * •

When $k=0$, the functions $f_{1,0}=f_{2,0}=...=f_{2^{d},0}:=f$,

  * •

Assume that $2^{d}$ sequences have been defined for $0,1,...,k$, now we define

$\displaystyle f_{i,k+1}(x)$
$\displaystyle:=f_{1,k}(x)+\sum_{j=2}^{2^{d}}\epsilon_{i,j}f_{j,k}(x-\mathbf{x_{j,k}}),\text{
where }1\leq i\leq 2^{d}.$

As $j$ runs over $2$ to $2^{d}$, the point $\mathbf{x_{j,k}}\in\mathbb{R}^{d}$
runs over all $2^{d}-1$ integer points of the form

$\mathbf{x_{j,k}}=(x_{1},x_{2},...,x_{d}),\text{ each }x_{i}=0\text{ or
}2^{k},\text{ and at least one }x_{i}\neq 0.$

It’s easy to see when $d=2$, our $\mathbf{x_{j,k}}$ are exactly the three
points $(2^{k},2^{k})$, $(0,2^{k})$, and $(2^{k},0)$. The subsequent
calculation is the same as for $d=2$, so we omit it here. Just replace (4.8)
and (4.9) by

$||f_{1,k}||^{2}_{2}=2^{dk}||f||_{2}^{2},\text{ and
}|\widehat{f_{1,k}}|^{2}\leq 2^{d(k+1)}|\hat{f}|^{2}$

respectively.

## References

  * [1] Folland G.B., Sitaram A. The uncertainty principle: A mathematical survey. The Journal of Fourier Analysis and Applications 3, 207–238 (1997).
  * [2] Michael G. Cowling, John F. Price. Bandwidth Versus Time Concentration: The Heisenberg–Pauli–Weyl Inequality. SIAM Journal on Mathematical Analysis, Vol. 15, Iss. 1 (1984).
  * [3] Linzhe Huang, Zhengwei Liu and Jinsong Wu. Quantum smooth uncertainty principles for von Neumann bi-algebras. https://arxiv.org/abs/2107.09057, Submitted on 19th July 2021.
  * [4] Stephen Wainger. Special trigonometric series in $k$-dimensions. Memoirs of the American Mathematical Society, 59(1965), 102 pp.
  * [5] Tang, Y. A Simple Observation on Heisenberg-Like Uncertainty Principles. Journal of Mathematical Sciences, 268, 810–815 (2022).
  * [6] Y. Wigderson and A. Wigderson, The uncertainty principle: variations on a theme. Bulletin of the American Mathematical Society, Vol. 58, No. 2, April 2021, 225-261.


# Multi-scale simulation of thrombus formation at LVAD inlet cannula
connection: Importance of Virchow’s triad

Rodrigo Méndez Rojano
Meinig School of Biomedical Engineering
Cornell University
Ithaca, NY 14850
<EMAIL_ADDRESS>
Mansur Zhussupbekov
Meinig School of Biomedical Engineering
Cornell University
Ithaca, NY 14850
<EMAIL_ADDRESS>
James F. Antaki
Meinig School of Biomedical Engineering
Cornell University
Ithaca, NY 14850
<EMAIL_ADDRESS>

###### Abstract

As pump thrombosis is reduced in current-generation ventricular assist devices
(VAD), adverse events such as bleeding or stroke remain at unacceptable rates.
Thrombosis around the VAD inlet cannula (IC) has been highlighted as a
possible source of stroke events. Recent computational fluid dynamics (CFD)
studies have attempted to characterize the thrombosis risk of different IC-
ventricle configurations. However, purely CFD simulations relate thrombosis
risk to ad-hoc criteria based on flow characteristics, with little
consideration of biochemical factors. This study investigates the genesis of
IC thrombosis including two elements of the Virchow’s triad: Endothelial
injury and Hypercoagulability. To this end a multi-scale thrombosis simulation
that includes platelet activity and coagulation reactions was performed. Our
results show significant thrombin formation in stagnation regions
($|\textbf{u}|<0.002$ m s-1) close to the IC wall. In addition, high shear-
mediated platelet activation was observed over the leading-edge tip of the
cannula which mirrors the thrombus deposition pattern observed clinically. The
current study reveals the importance of biochemical factors to the genesis of
thrombosis at the ventricular-cannula junction which can inform clinical
decisions in terms of anticoagulation/antiplatelet therapy and guide engineers
to develop more robust designs.

_Keywords_ Multi-Scale Thrombosis Modeling, Ventricular Assist Devices, Inlet
Cannula, tissue factor, coagulation

##  1 Introduction

Current-generation ventricular assist devices (VADS) have virtually eliminated
the incidence of pump thrombosis [1]. However, adverse events related to
coagulation such as bleeding or stroke remain at unacceptable rates [2, 3, 4].
Thrombosis around the VAD inlet cannula (IC) has been recently highlighted as
a possible source of ischemic stroke events in patients receiving a Left
Ventricular Assist Device (LVAD) [5, 6]. In fact, the potential role of the
ventricular cannula in thrombosis related complications has motivated
numerical, in-vitro, and clinical studies for over two decades [7, 8, 9, 10,
11, 12, 13, 14].

The majority of studies have concentrated in the intraventricular flow
dynamics, specifically in the region of the ventricular apex. Early flow
visualization studies in-vitro [8, 9] and in isolated hearts revealed the
existence of stagnation and recirculation regions related to the shape and
positioning of the cannula tip [13]. More recently, particle image velocimetry
studies were conducted on the EVAHEART (EVAHEART, Inc., Houston, TX) cannula
by May-Newman et al. [15, 16] and postulated that systematic washout may
reduce the risk of thrombosis.

Chivukula et al. [17, 18, 19] studied the thrombosis risk in IC-LV
configurations using a computational fluid dynamics (CFD) approach. Using a
Lagrangian approach, Chivukula and coworkers assessed the influence of the
cannula angle, the cannula protrusion height, and the ventricle size on the
thrombosis risk. In a similar vein, Liao et al. [20] studied the influence of
cannula geometry on blood washout and hemolysis index. Recently, Ghodrati et
al. [7, 6] performed transient simulations in patient-specific, cannulated
left ventricles to investigate the influence of the insertion site position
and orientation, with the goal of relating stagnation regions and non-
physiological wall shear stress to thrombosis risk. Most recently, Rossini et
al., employed 2D color doppler velocimetry in seven LVAD patients with dilated
cardiomyopathy to investigate anomalies in intraventricular flow that may
increase the risk of pump-related thrombosis [14].

Although the aforementioned numerical and experimental studies have provided
great insight into the IC thrombogenesis risk, they all have relied upon ad-
hoc, qualitative criteria to assess thrombosis risk. None, to the best
knowledge of the authors, have considered the biochemical elements that
contribute to the formation and stabilization of the thrombus.

This study aimed to investigate the genesis of IC thrombosis considering
Virchow’s triad, employing a multi-scale thrombosis simulation that includes
cellular and biochemical transport, as well as coagulation reactions. Our
approach considered biochemical transport, platelet deposition, cleaning, and
activation as well as coagulation reactions triggered by tissue factor. The
aim of the current work is to provide a better insight into the mechanisms
that drive thrombosis at the IC and identify the biochemical mechanisms that
promote thrombus growth.

##  2 Methods

A set of chemo-fluidic multi-scale CFD simulations using the model of Wu et
al. [21] were performed considering a VAD IC within an anatomically correct
dilated left ventricle (LV). Details of the ventricular geometry have been
described previously in [22].

###  2.1 Thrombosis Model

The thrombosis model of Wu et al. [21] was used for the computations. In this
framework, the Navier-Stokes (NS) equations for blood and scalar equations for
its biochemical constituents are solved in a coupled manner. Mechanical and
chemical activation of platelets can be induced by shear stress or the
combined effect of adenosine diphosphate (ADP), thrombin (TB), and thromboxane
A2 (TxA2). Platelets can deposit onto artificial surfaces or on previously
deposited platelets forming a thrombus. Deposited platelets and thrombus
periphery are susceptible to cleaning by local wall shear stress. The growing
thrombus can alter the flow field by means of a hindrance term in the momentum
equation in regions where the platelet volume fraction is high. (See Fig. 1)
For a more detailed description of the present thrombosis model and thrombosis
modeling in general, the reader is referred to [21, 23].

Figure 1: Thrombosis model diagram describing the platelet-based model of Wu
et al. [21]. In addition, TF-driven thrombin (TB) production is depicted.

###  2.2 Baseline Geometry and boundary conditions

Figure 2A shows the computational domain and mesh used in the simulations. A
generic VAD cannula tip was placed in-silico at the apex, along the long axis
of an anatomically correct dilated human LV. The outer and inner diameter of
the IC are 20 mm and 14 mm, respectively. The insertion depth of the IC in the
LV was about 20 mm. At the atrium, a conduit of 30 mm diameter and 30 mm
height was added to promote a developed velocity profile and represent the
out-flow jet typical to this configuration [6]. A uniform velocity boundary
condition was prescribed at the inlet of the entrance region (corresponding to
constant flow rate of 5 L min-1). Non-slip boundary condition was set at the
walls of the LV and the inflow cannula. For the biological scalar species,
physiological concentration of platelets was prescribed $3\times 105$
platelets $\mu$L-1 at the inlet with 1% of background platelet activation. A
reactive boundary condition was used at the inflow cannula surface allowing
both platelet deposition and cleaning on the walls.

Figure 2B shows the corresponding unstructured mesh with 840k hexahedral and
tetrahedral elements that was used in the simulations. The time step used in
the simulations was 0.0001 s which ensured that CFL $<$ 1\. A mesh convergence
study was performed using progressively finer meshes of 500k, 840k, and 2.5
million elements, which indicated that the computed IC wall shear stress was
sufficiently accurate with 840k elements. The fluid was assumed Newtonian with
a kinematic viscosity of $3.333\times 10^{-6}$ m2 s-1 and a density of 1050 kg
m-3. The solution of the flow and scalar equations was performed using the
open-source CFD software OpenFOAM www.openfoam.org.

Figure 2: A) Computational domain showing the dilated LV and the LVAD inflow
cannula inserted at the apical position. B) Middle cut visualization of the
unstructured mesh visualization used in the thrombosis simulations.

The species considered in the thrombosis model are the following: activated
platelets (AP), resting platelets (RP), prothrombin (PT), thrombin (TB),
antithrombin (AT), thromboxane A2 (TxA2), and adenosine diphosphate (ADP). A
specific reaction term is solved for each of the species according to their
biochemical interactions. For an extensive description of the reaction terms
and diffusion coefficients of the biochemical species, see [21]. The initial
conditions and specific model parameters used in the present study are listed
in Table 1.

Parameter | Value
---|---
Inlet AP concentration [Plt m-3] | $3.0\times 10^{12}$
Inlet RP concentration [Plt m-3] | $3.0\times 10^{14}$
PT concentration [M] | $1.1\times 10^{6}$
ATIII concentration [M] | $2.844\times 10^{6}$
RP-boundary(wall) deposition rate, karpd [m s-1] | $4.0\times 10^{-5}$
AP-boundary(wall) deposition rate, kaapd [m s-1] | $4.0\times 10^{-4}$
Table 1: Initial conditions and thrombosis model parameters. See Wu et al.
[21] for definitions of these terms and their use in the species convection-
diffusion-reaction equations.

###  2.3 Coagulation reactions triggered by Tissue Factor

During VAD implantation, a coring procedure is performed at the apex of the LV
to insert the IC. This procedure can potentially result in exposed myocardium
within the ventricle at the interface with the cannula tip. Consequently,
tissue factor (TF) is exposed, which may initiate a cascade of enzymatic
reactions that leads to thrombin production which participate in various
feedback pathways that contribute to the formation and stabilization of
thrombus. Accordingly, an “injury” region adjacent to the ventricle-cannula
interface was introduced to the domain (See Fig. 3). Over this region, a
Dirichlet boundary condition with a constant concentration value of TF = 0.8
nmol m-3 was prescribed.

Figure 3: TF exposure boundary condition region (red) at the junction of the
cannula and the myocardial surface of dilated LV.

To model the dynamics of thrombin formation, a simplified set of five
additional reactions were incorporated into the existing thrombosis model (See
Fig. 4). This reduced-order scheme economizes computational time by using a
minimal number of biochemical species [24].

Figure 4: Reduced biochemical scheme for thrombin production triggered by
Tissue Factor (TF) based on [24]. Factor VIIa activates prothrombin (II), then
thrombin (IIa) activates factor V in a feedback loop and is consumed by
antithrombin (ATIII).

This approach has been used in literature to study the biochemistry of
coagulation along with complex fluid dynamics. For the current application,
the hallmark model of Hockin et al. [25] was used as a reference to calibrate
the reduced model. Least square optimization was performed with the Python
module S-timator (https://webpages.ciencias.ulisboa.pt/~aeferreira/stimator/)
to identify parameters that best fit the numerical data of Hockin et al. (See
Table 2). Figure 5 shows the thrombin formation curve for both the reduced
kinetic model and the data from [25].

Parameter | value
---|---
k1 |  $2.28378\times 10^{-5}$ s-1
km2 |  $30.1908\times 10^{3}$ nmol m-3
kcat2 | 1.09473 s-1
km3 |  $7.90731\times 10^{3}$ nmol m-3
kcat3 |  $0.00178566$ s-1
km4 |  $202.974\times 10^{3}$ nmol m-3
kcat4 |  $1.17018$ s-1
kcat5 |  $7.24351\times 10^{-9}$ nmol m-3 s-1
II0 |  $1610.98\times 10^{3}$ nmol m-3
AT0 |  $3576.03\times 10^{3}$ nmol m-3
V0 |  $20\times 10^{3}$ nmol m-3
Table 2: Kinetic parameters and coagulation factors and initial conditions.
Figure 5: Thrombin production in batch reactor simulation. Current model
results are compared with the results published by Hockin et al.[25].

Excess of calcium and phospholipids are assumed for both the batch reaction
and free flow cases. The thrombin produced by the coagulation reactions was
coupled with the Wu platelet-based model by means of thrombin-mediated
platelet activation. Thus, this new source of thrombin contributes to platelet
activation in regions where the thrombin activation threshold is reached.

##  3 Results

###  3.1 Intraventricular Flow Field

Figure 6 A shows the streamlines inside the LV. Relatively low flow velocity
regions can be observed around the IC. Streamlines near the cannula leading
edge show important velocity gradients which results in elevated mechanical
platelet activation. Figure 6 B shows the velocity field magnitude in the
coronal plane, highlighting the presence of the mitral jet. Due to the ideal
position of the cannula (parallel to the septal wall), the mitral jet plays an
important role in washout of platelet agonists near the cannula. In general
terms, the velocity field compares well qualitatively with previous numerical
and experimental studies [9, 15, 6]. Two regions of low shear stress (< 0.005
Pa) appeared at the root of the IC (see Fig. 7 A). Biochemical species
advected or diffused inside these regions (stasis regions) may promote
platelet activation, further platelet deposition and hence thrombus formation.
In regions where sufficient blood wash out is achieved the concentration of
agonist is lower limiting further thrombus growth.

Figure 6: A) Path lines inside the dilated LV B) Middle cut velocity magnitude
field.

###  3.2 Platelet activity

The shear stress magnitude within most of the ventricle was less than 10 Pa
which did not provoke platelet activation. However, at the inner leading edge
of the cannula tip, shear rate of up to 7000 s-1 was observed, resulting in
mechanical shear activation (See in Fig. 7 A). For the first 150 seconds,
platelets accumulated along the radial rim of the IC as shown in Fig. 7 B.
However, following 150 seconds, platelet volume fraction reached a plateau,
due the equilibrium between deposition and shear cleaning (washing). These
simulated results resembled the thrombus deposition pattern observed
clinically from a cannula explanted from a Heatmate-3 patient (See Fig. 7 C).

Figure 7: LV IC A) shear rate field (s-1), B) Volume fraction of deposited
platelets (at 150s), HM3 clinical explant (courtesy of Dr. B. P. Griffith).

Most activated platelets and the agonists released in the process (ADP and
TxA2) are aspirated inside the VAD (see Fig. 8 A – 8C). Few activated
platelets enter a stasis region at the base of the cannula near the junction
with the endocardium. The middle coronal plane visualization reveals increased
mechanical platelet activation at the cannula leading edge and the inner
cannula surface.

Figure 8: Coronal plane visualization of key scalar fields in simulation at
150s, (a) activated platelets, (b) ADP, (c) Thromboxane, and (d) mechanical
platelet activation rate. Activated platelets at cannula-ventricular junction
(red circles) lead to aggregation. Mechanical activation initiated at leading
edge of cannula tip (black circles).

###  3.3 Thrombin production

Simulated thrombin synthesis due to coagulation reactions and activated
platelets resulted in super-critical thrombin concentration in the low-
velocity regions ($|\textbf{u}|<0.002$ m s-1) as shown in Fig. 9 A and B.
Figure 9 B shows the regions where thrombin concentration exceeds 1 nmol m-3.
This means that the characteristic flow residence time was large enough for
coagulation reactions to take place and promote prothrombin transformation. As
expected, the regions in which significant concentration of thrombin occurs
coincides with low wall shear stress regions that have poor flow wash out. The
origin thrombin produced in these regions was coagulation reactions rather
than upstream platelet activation by mechanical shear. The simulated thrombin
concentration pattern was found to resemble neo intimal tissue from explanted
cannula of the EVAHEART Fig. 9 C and thrombus deposited on the HVAD cannula
Fig. 9 D. This could be explained by the critical role of thrombin in the
cleavage of fibrin.

Figure 9: A) Velocity magnitude threshold $|\textbf{u}|<0.002$ m s-1. B)
Thrombin production due to tissue factor exposure; concentration threshold
thrombin > 1 nmol m-3 at 150 s. C) Neo intima tissue growth around EVAHART
cannula [26]. D) Thrombus growth in HVAD IC [5].

Figure 10 A illustrates the region adjacent to the cannula tip corresponding
to thrombin concentration greater than 1 nmol m-3. To investigate if the
production of thrombin would continue following a brief exposure of TF, an
additional simulation was performed in which the TF boundary condition was set
to zero after 50 s. In this case, thrombin production was substantially
reduced and the accumulated thrombin and enzymes near the cannula tip region
were rapidly cleared (See Fig. 10 B).

Figure 10: A) Thrombin threshold region (1 nmol m${-3}$) at 150 seconds using
a constant TF Dirichlet boundary condition. B) Thrombin threshold region (1
nmol m-3) at 150 seconds Dirichlet TF boundary condition turned off after 50
seconds.

##  4 Discussion

Numerical simulation of thrombosis has become a viable tool to predict
thrombus formation in blood wetted medical devices. Advances in VAD technology
have reduced the incidence of pump thrombosis; nonetheless, ischemic stroke
remains a significant adverse event in VAD patients. Recent studies have
pointed to the IC as a possible source of emboli and subsequent stroke [6, 5].
CFD studies of left ventricles and IC have identified stagnation regions and
flow patterns which are related to an elevated thrombosis risk. However, most
CFD analyses rely on qualitative assessment of the velocity and shear fields,
which are somewhat arbitrary or ad-hoc. To the best knowledge of the authors,
no prior study has explicitly simulated the genesis of thrombus, which is a
complex process involving biochemical reactions and transport of platelets and
agonists, particularly thrombin. In this study, the three components of the
Virchow’s triad were considered: flow, endothelial injury, and coagulation
dynamics. In terms of platelet activation, our results suggest that non-
physiological shear stress rates present at the inner radius of the cannula
activates platelets that deposit downstream on the interior surface,
resembling thrombosis patterns found on cannulas of explanted VADs.

At the junction between the LV apex and the IC, TF-driven thrombin formation
was sustained until the end of the simulation (150 s). Thrombin accumulated
mainly in low-velocity regions ($|\textbf{u}|<0.002$ m s-1), which is
concordant with criteria used in previous CFD studies. For example, Ghodrati
et al. (2020) specified a velocity magnitude of $0.005$ m s-1 as a theshold
for thrombosis risk [6]. Thrombin formed in these regions can lead to fibrin
polymerization and activation of platelets, which further amplifies thrombin
generation. Agonists accumulated via this positive feedback loop may be
transported up and over the cannula wall, whereupon pre-sensitized platelets
(due to high shear) may become activated and deposit downstream. As more
attention is given to prevention of blood stasis regions during LVAD support
[13], it is imperative to build stasis criteria upon biochemical mechanisms
that are less arbitrary than a prescribed cut-off value. A brief exposure of
TF was explored in the simulations showing that thrombin formation can be
limited by shutting down the influx of TF. This phenomenon suggest that the
thrombin formation process can be altered by suppressing the initial step of
the coagulation cascade, which might be an interesting anticoagulation target.
Platelet deposition in low wall shear stress zones near the base of the
cannula continued to grow slowly over the course of the simulation. However,
the rate of growth was limited by the reduced platelet transport towards this
region. In the long run, the accumulation of pro-coagulant species such as
activated platelets or thrombin towards stagnation regions could be a
potential source of thrombus at the base of the cannula. However, the
computational cost of this simulation prevented us from reaching that
timepoint.

Despite the insightful results of these simulations, we acknowledge several
limitations of the current study. First, the surface properties (material
composition, chemistry, texture) of the cannula itself was not explicitly
prescribed. Instead, this study focused on the effect of exposed tissue factor
at the IC insertion site due to surgical wound. Another limitation was the
absence of pulsatility and LV wall motion, both of which may play a role in
the dynamics of species transport and thrombus washout. However, the combined
computational cost of fluid-structure interaction, pulsatility (unsteady
flow), and the multi-constituent thrombosis model would be prohibitive.
Nevertheless, simulations on a static ventricle reflect a worst-case scenario,
corresponding to a severe dilated cardiomyopathy [17, 18, 19, 20]. These
simulations also did not account for fibrin polymerization which might be
important in thrombus stabilization/embolization mechanisms. This is a topic
of an ongoing investigation. Lastly, the current study considers the IC
positioned along the long axis of the ventricle; however, cannula misalignment
is common in LVAD patients, which could exacerbate the thrombogenic processes
described here. Extension of these studies to account for sub-optimal cannula
positioning and various anticoagulation strategies is a focus of ongoing work.

In conclusion, the genesis of thrombus deposition at the ventricular-cannula
junction relies on a complex interaction of hemodynamic, hematological, and
biochemical factors – elements of Virchow’s triad. Accordingly, this study
aimed to elucidate the importance of introducing the synthesis and transport
of biochemical species into a simulation in combination with intraventricular
fluid dynamics. We hope that continued investigation along these lines will
contribute to the utility of in-silico studies in guiding the design of
ventricular apical cannula and postoperative anticoagulation management, which
can help reduce the incidence of embolization and attendant adverse events in
VAD patients.

##  5 Acknowledgments

This research was supported by NIH R01 HL089456 and NIH R01 HL086918.

##  6 Disclosure

The authors declare no conflict of interest.

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