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Let $K(x):=|x|^{-\alpha}$ be a function on $\mathbb{R}^{n}\setminus\{0\}$ with $0<\alpha<n$. Suppose $\phi$ is a $C^{\infty}_{c}(\mathbb{R}^n)$ function such that $$\text{(i)}\quad \text{supp}\phi \s …

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 …

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| …

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 …

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}}) …

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 …

Fix $\theta\in (-\pi/2,\pi/2)$ and let $a>0$. Suppose that $f:\mathbb{C}\rightarrow \mathbb{C}$ is analytic in $S:=\{z\in \mathbb{C}: |\arg{z}|<\pi/2\}$ and $$|f(z)|\sim |z|^{-a},\qquad |z|\to \infty, …

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 $t=t(\alpha,p,q)$ …

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 …

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 …

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 …