Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
Partial differential equations (PDEs): Existence and uniqueness, regularity, boundary conditions, linear and non-linear operators, stability, soliton theory, integrable PDEs, conservation laws, qualitative dynamics.
2
votes
1
answer
217
views
If an estimate is false on $L^{1}$, then it is false for the $\delta$ distribution?
Let $u=\int e^{\dot{\imath}K(x,y)} f(y) dy$. My advisor told me that we can disprove an integrability estimate
$$\|u\|_{L^p}\lesssim \|f\|_{L^{1}}\label{1}\tag{1}$$
by disproving it when $f=\delta$, t …
7
votes
2
answers
398
views
$L^p-L^q$ boundedness of this simple singular oscillatory integral operator
Let $0<\alpha<1$ and define
$$Tf(x):=\int e^{\dot{\imath} x y} \frac{f(y)}{|x-y|^{\alpha}}dy.$$
The Hardy-Littlewood-Sobolev inequality characterizes $L^p-L^q$ boundedness of
$Hf(x):=\int \frac{f(y)} …
5
votes
3
answers
309
views
The integrability of $\widehat{e^{-|x|^a}}$, $a>0$
Let $f_a(x)=e^{-|x|^a}$, $x\in \mathbb{R}^n$. Then $f \in L^p(\mathbb{R}^n)$ for every $1\leq p\leq \infty$. It is also smooth away from the origin and decays faster than any polynomial as $|x|\righta …
1
vote
The integrability of $\widehat{e^{-|x|^a}}$, $a>0$
I would like to thank @Lorenzo Pompili and @Christian Remling for their insightful answers.
I found a straight-forward proof in Lemma 2.1 in
"C. Miao, B. Yuan, B. Zhang, Well-posedness of the Cauchy p …
4
votes
2
answers
303
views
Is this an $L^p-L^{\infty}$ operator?
Let $1\leq p <\infty$ and let $p^{\prime}$ denote its conjugate exponent. Consider the following operator on Schwartz functions:
$$Tf(x)=\int_{0}^{\infty}t^{\frac{n}{2 p^{\prime}}-1}e^{-t}
\int_{|x-y| …
-2
votes
2
answers
316
views
$f\in (W^{1,p}(\Omega)\cap C(\Omega) \cap L^{\infty}(\Omega))\setminus C(\bar{\Omega})$, $f=...
Q1:
Let $p\geq 1$, and let $f\in W^{1,p}(\Omega)\cap C(\Omega)$. Assume also
$f\in L^{\infty}(\Omega)$ and $f=0$ on $\partial \Omega$. Is it true that
$f\in W^{1,p}_{0}(\Omega)$ even if $f\notin C(\ba …
3
votes
1
answer
239
views
Asymptotic behavior of a double oscillatory integral
Let $0<\theta_1,\theta_2<\pi/2$. Suppose $\psi$ is a smooth real-valued function with compact support.
Consider the oscillatory integral
$$I(t):=\int_{0}^{1}\frac{1}{(y-e^{\dot{\imath}\theta_1})
(y-e^ …
0
votes
$f\in (W^{1,p}(\Omega)\cap C(\Omega) \cap L^{\infty}(\Omega))\setminus C(\bar{\Omega})$, $f=...
The answer is No.
There can be $f\in W^{1,p}(\Omega)\cap C(\Omega) \cap L^{\infty}(\Omega)$
such that $f=0$ on $\partial \Omega$ but $f\notin W^{1,p}_{0}(\Omega)$.
If $f\notin C({\bar{\Omega}}) …
5
votes
1
answer
299
views
Maximal operator estimates for the Schrödinger equation
Let $a>0$ and consider the operator
$$Tf(t,x)= \int_{\mathbb{R}^{n}}e^{ i x\cdot \xi} e^{i t \lvert\xi\rvert^{a}} \widehat{f}(\xi) \, d\xi.$$
When $a=2$, the function $Tf$ solves the Cauchy problem as …
0
votes
1
answer
613
views
Does this dyadic sum converge?
Let $a\in (0,1)$ and define
$$J(j):=\int_{0}^{\infty} e^{- 2^{j} s} \frac{s^{a}}{1+s^{2a}} ds,\quad j\in \mathbb{Z}.$$
Note that rescaling $2^{j} s\mapsto s$ shows that
$$J(j)\leq 2^{-j(1+a)}\int_{0} …
0
votes
1
answer
102
views
A simple bilinear estimate
Let $2\leq p,q <\infty$ and fix $0<\alpha<1$ such that
$\frac{1}{p}+\frac{1}{q}\leq 2-\alpha$.
Suppose that $f\in L^{p}([0,1])$ and $g\in L^{q}([0,1])$.
What is the optimal value of $t=t(\alpha,p,q)$ …
5
votes
1
answer
232
views
An asymmetric quadrilinear estimate
Fix $1<p<2$ and let $a_{i}=1-\frac{\theta_{i}}{p^{\prime}}$
where $\theta_{i}\in (0,1/2)$, $i=1,2,3,4$, and $p^{\prime}$ is the conjugate exponent of $p$. Note here that $0<a_{i}=1-\theta_{i}+\frac{\t …
1
vote
1
answer
146
views
Why does failure of boundedness of this operator for $p<q$ implies its failure for $p>q^{\pr...
I am reading the paper "P.Sjolin, Convolution with Oscillating Kernels, Indiana University Mathematics Journal Vol. 30, No. 1 (1981), pp. 47-55" where $L^p-L^p$ boundedness of the operator
$$Tf(x)=\i …
1
vote
1
answer
112
views
A bilinear estimate with a simple one-dimensional oscillatory integral kernel
Let $f\in L^{p}(\mathbb{R})$, $1\leq p\leq 2$.
I am trying to show that
$$\int_{\mathbb{R}}\int_{\mathbb{R}}
\,K(y,z)\,
\frac{f(y)f(z)}{y^{\frac{1}{2\,p^{\prime}}}\,z^{\frac{1}{2\,p^{\prime}}}}\,dy\,d …
3
votes
1
answer
341
views
Is Brascamp-Lieb inequality on the sphere applicable for these functions for some $1\leq p<2$
My question is on Brascamp-Lieb-inequality on the Euclidean sphere (which can be viewed as an analogue of Young's inequality on the sphere) obtained in [1].
(See also this question:
Brascamp-Lieb ine …