User shanlin huang - MathOverflow

Answer by Shanlin Huang for Meromorphic Functions as Distributions

2013-04-20T17:22:09Z

It's interesting to think distributiions as the boundary of analytic functions, which I believe is originated from Saito. You may find the following result helpful

Let I be an open interval on $\mathbb{R}$, and $$ Z={z\in \mathbb{C};\Re z\in I,0<\Im z <\gamma} $$ is a one sided complex neighborhood. If $f$ is analytic such that there exsists a non-negtive integer $N$ $$ |f(z)|\leq C|\Im z|^{-N},\quad z\in Z $$ Then $f(x+i0)$ exsists as a distribution and is of order $N+1$. And the condition can not be relax very much, in fact, if we assume that $f(x+i0)\in \mathcal{D'}^n$(distribution with order $n$),then we have $$ |f(z)|\leq C|\Im z|^{-n-1} $$

Answer by Shanlin Huang for Nonintegrable inverse powers as distributions

2013-04-15T04:45:42Z

The related topic here is the homogeneous distribution on $\mathbb{R}^n\0$ and its extension to $\mathbb{R}^n$. In your case $T_{f}$ is a homogeneous distribution on $\mathbb{R}^n\0$ of degreee $-n$. And it's always possible to extend it to a distribution on $\mathbb{R}^n$,which may not be homogeneous any more. In fact,you can define the extension $\dot{T_{f}}$ as follows $$ \dot{T_{f}}(\phi)=T_{f}(\psi R_{-n}\phi), \quad \phi\in C_{0}^{\infty}({\mathbb{R}^n})\\ R_{-n}\phi=<t_{+}^{-1},\phi(tx)>,\quad x\neq 0 $$ where $\psi$ is a fixed function in $C_{0}^{\infty}({\mathbb{R}^n}\0)$ satisfies that $\int_{0}^{+\infty}\frac{\psi(tx)}{t}dt=1$.

Now one can see that when restricting test functions in $C_{0}^{\infty}({\mathbb{R}^n}\0)$, then $\dot{T_{f}}=T_{f}$, and the extension depends on the choice of $\psi$. However, all such extension has the above form. And in one dimensin, it's particularly clear. You can find them in chaper $3$ of Hormander's book "The Analysis of Linear Parital Differential Operators"

Is there a good way to estimate the Fourier transform of $\frac{1}{\lambda-iP(\xi)}$

2013-01-30T02:51:54Z

Assume that P is a real valued strong elliptic polynomial, then what do we know about the following $$ K(\lambda,x)=\int{\frac{e^{ix\xi}}{\lambda-iP(\xi)}}d\xi,\quad \lambda\in \mathbb{R}\0 $$ The reason I want to know about it is that I need some properties of the resolvent $(\lambda-iP(D))^{-1}$. It is not hard to show that the resolvent is bounded on $L^p$, a more detailed analysis can obtain $L^p-L^q$ estimates for some {p,q}. However, I want to know if there is a pointwise estimate of its kernel, which will allow me to do more. Any reference is appreciated.

what's the motivation of Weyl calculus ?

2012-05-01T08:06:55Z

In the pseudo-differential operator theory, we can define a pseudo-differential operator by $$a(x,D)u=(2\pi)^{-n}\int{a(x,\xi)e^{i\langle x-y,\xi \rangle}u(y)dyd\xi}$$ with $a(x,\xi)$ belong to some particular function space (denoted by $S^m$).In the Weyl calculus one adopts the symmetric compromise $$a^{w}(x,D)u=(2\pi)^{-n}\int{a((\frac{x+y}{2}),\xi)e^{i\langle x-y,\xi \rangle}u(y)dyd\xi}$$ again defined in the weak sense. From this one can see that the adjoint of $a^w$ is equal to $\bar a^{w}$. In particular, $a^w$ is its own adjoint when a is real valued. Is this convenience making Weyl calculus more applicable for physics? In mathematics, are there other reasons to the motivation of Weyl calculus? Furthermore, Can anyone show some problems which are solved by using this tool?

Why is symplectic geometry so important in modern PDE ?

2012-05-27T15:59:21Z

First, we recall that symplectic manifold is a smooth manifold, $M$, equipped with a closed nondegenerate differential 2-form, $\omega$, called the symplectic form. The study of symplectic manifolds is called symplectic geometry. In Hörmander's classic book ALPDO (The analysis of partial differential operators Ⅰ-Ⅳ) he wrote: symplectic geometry pervades a large part of the modern theory of linear partial differential operators with variable coefficients. And he had devoted the entire chapter ⅩⅩⅠ to discuss it.

Now, with some basic background (its origins in the Hamiltonian formulation of classical mechanics), I want to know further that why it would make such a important role in modern analysis (such as fourier integral operators)?

Thanks in advance.

Fourier transform of a particular function

2012-11-13T09:27:40Z

In order to estimate the fundamental solution of some particular types of differential operators,I need estimates on some kind of oscillatory integrals.For simplicity, consider the Fourier transform of the following function on $\mathbb{R}^4$ $$ f(x)=(1+x_{1}^2+x_{2}^2+x_{3}^2+x_{4}^2+x_{1}^2x_{2}^2+x_{1}^2x_{3}^2+x_{1}^2x_{4}^2+x_{2}^2x_{3}^2+x_{2}^2x_{4}^2+x_{3}^2x_{4}^2)^{\alpha} $$ where $-1<\Re \alpha<0$.

Is there any way to estimate the decay of $\hat{f}(\xi)$ for large $\xi$ ?

I have tried to use a dyadic decomposition (write $\mathbb{R}^4$ as the union of disjoint rectangles)to treat the singularities,and then use integrating by parts.But it seems a little messy.I don't know if there were some papers already dealing with such kind of integrals, so I'm very apprieciated that if someone can show me.

Resonance of Schrödinger operator

2012-10-23T09:15:23Z

Consider the dispersive estimates for the Schrödinger flow $$ e^{itH}P_{c},\quad H=-\Delta+V \quad \text{on}\quad \mathbb{R}^n,n\ge 1 $$ where $P_{c}$ is the projection onto the continuous spectrum of $H$, and we will be most concerned with whether it has the form $$ \|e^{itH}P_{c}\|_{L^1\to L^{\infty}}\leq C |t|^{-\frac{n}{2}} $$ In order to get this estimate, some decay and regularity condition must be put on the potential $V$, an important assumption is that zero is neither an eigenvalue nor a resonance.

If $0$ is a eigenvalue, then it's easy to see that the above estimates may fail. My question is then if zero is a resonance but not an eigenvalue, why will the estimates above go wrong?

Zero is said to be a resonance in the sense that if the operator $(I-V\Delta^{-1})^{-1}$ is bounded on $L^1$ (why not on $L^2$ ?),see the paper of Vodev,I found this is less illuminating for me, so I want to know if there are some better understanding of this definition to make it more intuitive.

Edit As Terry and Delio have commented,the key point is the asymptotic expansions of the resolvents around the zeero energy.for odd dimension,with $\Im z>0$ ,one can write $$ (-\Delta+V-z)^{-1}=\frac{A_{-1}}{z}+\frac{A_{-\frac12}}{z^{\frac12}}+A_{0}+O(z) $$ (for even dimension,the $log z$ terms are included)where $A_{-1}$ is the projection onto the eigenspace of $H$,and $A_{-\frac12}$ is related to both eigenspace and resonance functions. So in order to get the optimal decay ($t^{-\frac{n}{2}}$)for large t,one need $A_{-1}=A_{-\frac12}=0$ ,that is zero is neither an eigenvalue nor a resonance .

Fourier transform and spectrum of PDOs in $L^p$

2012-09-27T11:41:42Z

Let $K$ be a compact subset in $\mathbb{R}^n$ with $m(K)=0$, Suppose $supp\hat{u}\subset K$ for some $u\in L^p$,where $2\leq p\leq \frac{2n}{n-1}$,can we get $u\equiv 0$ ?

Motivation: If $K$ is a compact non-degenerate hypersurface,then it's well known that $u(x)\leq C|x|^{-\frac{n-1}{2}}$,hence $u\in L^p$ for any $p>\frac{2n}{n-1}$,but if we restrict $p$ in $[2,\frac{2n}{n-1}]$,then is it possible that $u\equiv 0$ ?

Another related question: Consider a linear partial differential operator $P(D)$ with constant coefficients in $L^p$,it has been proven that $P(D)$ has no eigenvalue when $1\leq p<\frac{2n}{n-1}$,the bound for $p$ is best possible,in fact consider $-\triangle$ act on $L^p$, $p>\frac{2n}{n-1}$,it's known that $\sigma(-\triangle)=[0,\infty)$ for all $p$,we now care about its point spectrum(eigenvalue) and it's easy to check that $u=\hat{d\mu}$(where $d\mu$ is the surface measure on {$|\xi|=s^2$})is the eigenfunction of $-\triangle$ with eigenvalue $s^2$,hence,in this case,$\sigma_{p}(-\triangle)=(0,\infty)$.But for $p<\frac{2n}{n-1}$,$\sigma_{p}(-\triangle)=\emptyset$,then what I am particularly interested in is the case when $p=\frac{2n}{n-1}$.Is its point spectrum also empty ?

Edit For the first question,it suffice to show that $\mathcal{F(S_{K})}$ is dense in such $L^p$,where $S_{K}$={$\varphi\in C_{0}^{\infty}(\mathbb{R}^n)$,$\phi(x)=0$,when $x\in K$ },this is obvious when K consists of only discrete points,but for more complicated K(such as a low-dimension surface),I don't know how to deal with it right now.

Answer by Shanlin Huang for Sobolev-type inequality.

2012-10-20T02:36:11Z

This is the standard Hardy-Littlewood-Sobolev inequality(or the theorem of fractional integration).A more direct approach is write $$ \int{f(x-y)|y|^{\alpha-n}dy}=\int_{|y|<R}+\int_{|y|\ge R} $$ For the second term on the RHS,using Holder inequality,and easy to see that it's dominated by $\|f\|_{L^p}R^{-\frac{q}{n}}$ . For the first term,one can use the majorizationgiven by the maximal function M,and to see that $$ |f\ast |y|^{\alpha-n}|(x)\leq C(M(f)(x)\cdot R^{\alpha}+\|f\|_{L^p}\cdot R^{-\frac{q}{n}}) $$ Choosing a proper constant R to make the two terms above be equal,and then the desired inequality hold by intergration(note that the maximal operator is bounded on $L^p$ for $1<p<\infty$ ).

what's the idea behind Carleman estimate

2012-10-16T13:56:55Z

A standard Carleman-type estimate is of the form $$ \sum_{|\alpha|<m}{\tau^{2(m-|\alpha|-1)}\int{|D^{\alpha}u|^{2}e^{2\tau\phi}}dx}\leq K\int{|Pu|^{2}e^{2\tau\phi}dx},\quad u\in C_{0}^{\infty} $$ where $\phi$ is some weight function.This formula turn to be very useful in the study of uniqueness of Cauchy problem,and many mathematicians have considered this(such as Calderon,Hormander,Kenig,Sogge,and Tataru...)

For a first look at this inequality,I'm wondering whether the weight fuction makes a essential role,and besides, what's the original idea of it?Are there some very simple but illuminated examples to show the the reasonableness of the Carleman estimates ?

Well,one example in my mind is the first order operator $P=D+ix$,then it's easy to see that $P^*=D-ix$ ,and $$ P^*P-I=PP^*+I=-\frac{d^2}{dx^2}+x^2 $$ which is the so-called harmonic oscillator,then we have $$ 2\|u\|_{L^2}\leq \|Pu\|_{L^2},\quad u\in C_{0}^{\infty} $$ But in this simple example,there is no need to put a weight function,anyhow, from the proof,I guess the decomposition $P=\frac{P+P^*}{2}+\frac{P-P^*}{2}$ may be one of the general idea.

Answer by Shanlin Huang for Fourier transform of $e^{it|\xi|^{\alpha}}$

2012-10-13T08:40:57Z

We consider $t=1$ for simplicity and wright $K_{\alpha}=\mathcal{F}(\eta(|\xi|) e^{i|\xi|^\alpha})+\mathcal{F}((1-\eta )e^{i|\xi|^\alpha})$,where $\eta\in C^{\infty}(\mathbb{R})$,and $\eta=0$,near 0,$\eta(t)=1$,when $t\ge 1$,the second term in the RHS is smooth and has good behaviour at $\infty$,so we look at the first term,in A.Miyachi's paper "On some singular fourier multipliers"see http://repository.dl.itc.u-tokyo.ac.jp/dspace/bitstream/2261/6297/1/jfs280206.pdf it has a thoroughly analysis on it.When $0<\alpha<\frac{1}{2}$,we have $K\in C^{\infty}(\mathbb{R}^{n}\backslash{0})$ and $$ K_{\alpha}=C|x|^{\frac{n(\alpha-1)}{1-2\alpha}}e^{iB|x|^{-\frac{2\alpha}{1-2\alpha}}}+o(|x|^{\frac{n(\alpha-1)}{1-2\alpha}})\quad \text{as}|x|\to 0 $$

When $\alpha>\frac{1}{2}$,$K$ is smooth throughout $\mathbb{R}^{n}$,and $$ K_{\alpha}=C|x|^{\frac{n(\alpha-1)}{1-2\alpha}}e^{iB|x|^{-\frac{2\alpha}{1-2\alpha}}}+o(|x|^{\frac{n(\alpha-1)}{1-2\alpha}})\quad \text{as}|x|\to\infty $$ In this case we can see that unlike $\alpha=1$, for $\alpha>1$,$K_{\alpha}$ has decay of $|x|^{-\frac{n(\alpha-1)}{2\alpha-1}}$ at $\infty$.

Fourier transform of $e^{it|\xi|^{\alpha}}$

2012-10-12T08:48:23Z

Consider the fourier transform of $e^{it|\xi|^{2\alpha}}$ ($\alpha>0$)in $\mathbb{R}^n$,let $K_{\alpha}=\mathcal{F}(e^{it|\xi|^{\alpha}})$,so $K$ is a tempered distribution.Now I want to know if there is a explicit expression of $K$,for the simpliest case,namely $\alpha=1$,it's well known that $$ K_1=(4\pi it)^{-\frac{n}{2}}e^{-\frac{|x|^2}{4it}} $$ Another special case is $\alpha=\frac{1}{2}$,since we know that $\mathcal{F}e^{-t|\xi|}=C_{n}\frac{t}{(t^{2}+|\xi|^2)^{\frac{n+1}{2}}}$,where $t>0$,let $t=-it$,so at least formally, $$ K_{\frac{1}{2}}=C_{n}\frac{-it}{(|\xi|^2-t^{2})^{\frac{n+1}{2}} }$$

My question is how about general $\alpha$ ?,so far I have known that when $0<\alpha<\frac{1}{2}$,$\alpha=\frac{1}{2}$,$\alpha>\frac{1}{2}$,the singularity of $K$ lies at $0$,$t=|x|$,$\infty$ respectively.

What's the use of Malgrange preparation theorem?

2012-10-07T14:03:25Z

The Malgrange preparation theorem,which is the $C^{\infty}$ version of the classical Weierstrass preparation theorem,says that if $f(t,x)$ is a $C^{\infty}$ function of $(t,x)\in\mathbb{R}^{n+1}$ near $(0,0)$ which satisfies $$ f=\frac{\partial{f}}{\partial{t}}=\cdots =\frac{\partial^{k-1}{f}}{\partial{t^{k-1}}}=0\quad \frac{\partial^{k}{f}}{\partial{t^{k}}}\ne 0\quad\text{at}(0,0) $$ Then there exists a factorization $$ f(t,x)=c(t,x)(t^{k}+a_{k-1}(x)t^{k-1}+\cdots +a_{0}(x)) $$ where $a_j$ and $c$ are $C^{\infty}$ functions near $0$ and $(0,0)$ respectively,$c(0,0)\ne 0$ and $a_{j}(0)=0$.As a corollary, there is a division thereom just like the Weierstrass formula.However, unlike the analytic case,this factorization is not unique.The result is said to be highly non-trival even when $k=1$,the difficulty is then the zeros may be lost,For example,$t^{2}+x$ has two real zeros when $x<0$ but none when $x>0$.The proof can be seen in Theorem 7.5.6 in Hormander's The Analysis of linear partial differential operators.

My question is What's the use of Malgrange preparation theorem in mathematics?Is this a verey useful formula in analysis ? Can anyone take some examples to apply this theorem?(In hormander's book,this is used in the method of Stationary Phase).

A quick google search shows that there is also a algebraic version which can be restated as a theorem about modules over rings of smooth, real-valued germs.

Answer by Shanlin Huang for fractional Leibniz formula

2012-09-16T09:20:45Z

Denote by $D^{\alpha}=(-\triangle)^{\frac{\alpha}{2}}$,then we have $$\|D^{\alpha}(f\cdot g)\| \leq C(\|D^{\alpha+s}(f)\|_{p_1}\|D^{-s}(g)\|_{q_1}+\|D^{\alpha+t}(f)\|_{p_2}\|D^{-t}(f)\|_{q_2})$$ where $\alpha$,s,t are positive real numbers,and $\frac{1}{p}=\frac{1}{p_{i}}+\frac{1}{q_{i}}$ with $i=1,2$. The proof can be seen in [Exact smoothing properties of Schrödinger semigroups]（ http://www.jstor.org/stable/10.2307/25098514.）

fractional Leibniz formula

2012-04-24T14:18:19Z

Let $T=(-\triangle)^{\frac{1}{2}}$,Can we have similar estimates below hold in $L^p$ ? $\| T^{\alpha}(fg)-(T^{\alpha}f)g-f(T^{\alpha}g) \|_p \leq \|T^{\alpha-1}f\|_p \|T^{\alpha-1}g\|_p$, where $\alpha>0$,p>1. If we really have such fractional Leibniz formula holds,we can then estimate the fractional integration by parts also.

Almost analytic continuation

2012-09-07T15:55:20Z

Let $f\in S^{\alpha}$ for some $\alpha \in \mathbb{R}$(which means that f is smooth and satisfies $|D^{\beta}f|\leq C(1+|x|)^{\alpha-\beta}$),a function $\tilde{f}$ on $\mathbb{C}$ is called an almost analytic continuation of f if it satisfies

(1) $\tilde{f}$ is a smooth function on $\mathbb{C}$ and $\tilde{f}=f$ for $x\in \mathbb{R}$;

(2) For any $N\geq 0$ $$|\partial_{\bar{z}}\tilde{f}(z)|\leq C_{N}\langle z \rangle^{\alpha-1-N}|\Im z|^{N},\qquad z\in \mathbb{C}$$ where $\partial_{\bar{z}}\tilde{f}(x+iy)=(\partial_{x}+i\partial_{y})\tilde{f}(x+iy)$

Obviously,this is a weaker assumption than the analytic continuation of a function(a somehow similar spirit of this is the definition of almost orthogonal operators).One of its application is in functional calculus,if A is a selfadjoint operator in a hilbert space,and $f\in S^{-\epsilon}$,$\epsilon>0$,then we have $$f(A)=\frac{1}{2\pi i}\int_{C}(\partial_{\bar{z}}\tilde{f}(z))(A-z)^{-1}dzd\bar{z}$$ This formular is sometimes useful when various types of $L^{p}-L^{q}$ estimates are concerned in wave or schrodinger equation.

My first question is that is there an explicit way to construct almost analytic continuation for functions in $ S^{-\epsilon}$,$\epsilon>0$, it seems non-trivial even when $f\in C_{0}^{\infty}(\mathbb{R})$.

Another question I'm interested is about application.In some cases,the analyticity may be too strong to obtain so that we could only hope for the weak analyticity or almost analyticity instead.so are there exactly any other use of this in mathematics(such as complex analysis or function theory)?

Is this kernel space of finite dimension ?

2012-08-25T15:52:53Z

Assume that $P \in \Psi^{m}(X)$ (X is a $C^{\infty}$ manifold)is properly supported and has a real principal part p which is homogeneous of degree m.I'm interested in the existence theorem(at least locally) for the equation $Pu=f$,according to the abstract functional analysis,we should first learn the kernel of the adjoint $P^{\ast}$,that is $$N(K)=\lbrace v \in \epsilon'(K),P^{\ast}v=0 \rbrace$$

Here K is a compact subset of X(since only locally existence is concerned)such that no complete bicharacteristic curve is contained in K.Under this condition,is N(K) a finite dimensional subspace of $C_{0}^{\infty}(K)$ ? If it is, then how to prove this ? Or are there other conditions (about the domain or the operator) to make sure the finiteness of the kernel ?

*EDIT*According to the condition on the domain,it can be shown that actually $N(K)\subset C^{\infty}(K)$,but then I don't know how to prove that the dimension is finite(i.e. the unit ball is precompact).

Convolution operators defined by compactly supported distribtion

2012-08-29T03:25:04Z

Let $\mu_{n}$ be the unit measure over $S^{n-1}$,and consider the convolution operator$$Tf=\mu_{n}\ast f,\quad f\in \mathcal{S}$$ then,it's well-known that T can be extend to a bounded operator on $L^{1}$.My question is whether it's bounded from $H^{1}$ to $H^{1}$ ?

The question is equivallent to say that whether $m(\xi)=\hat{\mu_{n}}$ is a $H^1$ multiplier(I also care about $\frac{d}{dr}m$,where $r=|\xi|$)or not.The theorem related to this is that(the most convenient one I know so far) if $\mathcal{F}^{-1}m$ has compact support,and $$|m(\xi)|\leq (1+|\xi|)^{-b},\quad b>0$$ then m is a $H^p$ multiplier,where $\frac{1}{p}-\frac{1}{2}=\frac{b}{n}$.In our case $p=1$,$b=\frac{n}{2}$.On the other hand,we also know that $\hat{\mu_{n}}=J_{\frac{n}{2}-1}(|\xi|)|\xi|^{-\frac{n}{2}+1}$,so $\hat{\mu_{n}}\sim |\xi|^{-\frac{n-1}{2}}$ when $|\xi|$ large.the problem is that this theorem cann't be applied since we need the decay of $|\xi|^{-\frac{n}{2}} $,and I don't know how to do with it.So are there other mathods to prove and disprove it?

what's the role of $H^{p}(\mathbb{R}^{n})$ in modern (harmonic) analysis?

2012-08-28T14:08:37Z

The classical theory of $H^p$,due to it's heavy dependence on the complex function theory(such as Blaschke products), seemed to have an insurmountable obstacle barrying its extension to several variables.However, C.Fefferman and E.Stein's remarkable paper'$H^{p}$ spaces of several variables' showed that $H^{p}$ classes can be characterized without any recourse to analytic functions, conjugacy of harmonic functions, etc. Thus $H^{p}$ classes have an intrinsic real variable meaning of their own. Another surprising result they got was that the predual of BMO(functions of bounded mean oscillation) was exactly $H^1$.

Well,what I'm particularly interested is its applications in mordern analysis.For instance,From Ferfferman's work,I know that $H^1$ is sometimes a proper subsitute for $L^1$,this can be seen from the CZOs(Calderon-Zygmund operators),which are bounded from $H^1$ to itself,but not on $L^{1}$. This is useful when evaluating some singular integral operators through complex interpolation.

Sometimes it's also very convenient to prove a bounded function to be $L^p$ multipliers through $H^1$,for example $m(\xi)=\psi(\xi)e^{i|\xi|^{a}}|\xi|^{-b}$($b>0$,$a>0$,$a\neq 1$),where $\psi \in C^{\infty}$ is 0 nere 0,and 1 when $|\xi|$ large.Then m is a $L^{p}$ multiplier iff $n|\frac{1}{2}-\frac{1}{p}|\leq \frac{b}{a}$

My question is what's the role of $H^{p}$ in modern (harmonic) analysis,and how people get useful results by choosing $H^p$.
I would appreciate any good examples, as well as some historical outlines on the topics development

What do we know about the semigroup $e^{it\sqrt{-\Delta}}$

2012-08-12T13:40:44Z

I'm very interested in the properties of the semigroup $e^{it\sqrt{-\Delta}}$, it may has some fundamental differences(such as the kernel) with the well-known schrodinger semigroup $e^{it\Delta}$.

Any properties (or references or books) that related this semigroup are appreciated.

Thanks!

Wave equation v.s.Schrödinger equation

2012-08-03T03:35:39Z

The motivation of comparison of this two kind of operators is that,$$\partial_{tt}-\Delta=(\partial_{t}-i\sqrt{-\Delta})(\partial_{t}+i\sqrt{-\Delta})$$ From the above that a wave operator can be seen as the product of two (general type) schr$\ddot{o}$dinger operators.Indeed,compared with other kinds of differential operators(elliptic,parabolic),these two operators seem to share more in common.Such as they don't have the usual global regularity property,and the use of harmonic analysis has made a great success in these two areas.

So I'm very curious to know the intersting links or fundemental differences between this two operator in order to understand them better.

*Edit:*In the above,I'm refering to a general dispersive equation:$u_{t}-i\phi({D})u=0$.When $\phi=-|\xi|^{2}$,it's the free schr$\ddot{o}$dinger equation.

Answer by Shanlin Huang for Fundamental Solutions with compact support (distributions)

2012-07-29T00:20:42Z

For a constant coefficient partial differential operator P(D), the fundamental solution of P can never belong to $\epsilon'(\mathbb{R}^{n})$,i.e.have compact support.

In fact,assume we have $P(D)u=f$,where u is a distribution,then u have compact support $\Leftrightarrow$ $\frac{f}{P(\xi)}$ is analytic(The result can be found in Hormander's ALPDO, volume 1,ch7.)

Now,if we have $$P(D)u=\delta$$,obviously $\frac{1}{P(\xi)}$ is never an analytic function for a polynomia P.So the fundamental solution of P can not be compact supported.

how to prove the range of a closed linear operator is closed ?

2012-05-22T06:05:41Z

The closed range theorem tells us that given two banach spaces X,Y,and a closed densely defined linear operator T：$X \to Y$. We have the following equivalence $R(T)$ is closed in $Y \iff R(T^{*})$ is closed in $X^{*} \iff R(T)=N(T^{*})^{\perp}$ . This gives a complete characterisation, but I don't know whether it's convenient or not for application.

Question 1: Can anyone pose some examples to use this theorem ?

When we consider the case that $T$ is a bounded operator, it's true that if the range $TX$ has finite codimension in $Y$, then $TX$ is closed. So it seems that there are lots of bounded operators whose range is not closed. For instance, when $X=L^{p},1 \leq p<2$, $Y=L^{p'}$, where p' is the dual index. $T$ represents the fourier transform $\mathcal{F}$. Then

Question 2: Is $\mathcal{F}L^{p}$ closed in $L^{p'}$? I believe the range is not closed. BTW, I only know that the map is not surjective, if it is not closed, then we can see that actually the range is much smaller than $L^{p'}$

when a pseudo-differential operators to be compact?

2012-05-14T16:48:45Z

In the theory of Pseudo-differential operators,when a symbol $a(x,\xi)\in S^{0}$,then the operator $a(x,D)$ defined by$$a(x,D)u=\int{e^{ix\xi}a(x,\xi)\widehat{u}}d \xi$$ is $L^2$ bounded.$ $ My question is when add what condition (as least as possible) to the symbol $a(x,\xi)$ can assure the operator $a(x,D)$ to be compact on $L^2$ ?Or is there a equivalent condition of this ?I guess some decayed assumption on $a(x,\xi)$ (about $\xi$) is necessary.but I'm not sure.some references about this are also appreciated

Added:There is a equivalent condition of a pseudo-differential operators to be compact.Assume that $g \leq g^{\sigma}$,that g is $\sigma$ temperate.and that m is $\sigma$,g temperate.then the operators $a^{w}(x.D)$with $a\in S(m,g)$are compact (bounded) in $L^{2}$ if and only if $m \to 0 $ at$\infty$ (m is bound).When we let g to be the metric $|dx|^{2}+|d \xi|^{2}/(1+|\xi|^2)$,and $m=(1+|\xi|^{2})^{\frac{\mu}{2}}.$then the class $S(m,g)$ become the usual$S^{\mu}$,Is it implying that when $\mu<0$,then $a^{w}(x.D)$ is compact in $L^{2}$ without further assuming the kernel to be compact supported ?

More precisely, considering the symbol $a(x,\xi)=V(x)(1+|\xi|^2)^{-1}$(it appears in the scattering problem of the schr\ddot{o}dinger operators),now we have the differential condition (rather than the integral condition)$\partial^{\alpha}{V} \leq (1+|x|)^{-\beta-|\alpha|}$,with $|\beta|>\frac{1}{2}$. Then does the operator $a(x,D)$ compact in $L^{2}$ ?

I try to prove it by decomposing both the $\xi-space$ and $x-space$ with a partition of unity as the almost orthogonality method. But it seems without the use of the decay of $x-space$,the compactness wouldn't be obtained(just think about the case $(1-\triangle)^{-1}$.

Answer by Shanlin Huang for Square roots of the Laplace operator

2012-05-08T16:05:13Z

Denote integral operator as follows:$$Lu=P.V.\int_{\mathbb{R}^n}\frac{u(x) - u(y)}{\|x - y\|^{n + 1}}\ dy$$.when $u\in \varphi$.It can also been writen as(after changing variable y to -y and then taking average)a more symmetric form $$Lu=\int_{\mathbb{R}^n}\frac{u(x+y) +u(x-y)- 2u(x)}{\|y\|^{n + 1}}\ dy$$.Thus the sigularity can be cancelled. We are looking for its symbol (or multipler),that is $$\mathcal {F}(Lu)(\xi)=m(\xi)\mathcal {F}u$$and we want to prove that m(\xi) is exactly $c_{n}|\xi|$,where $c_{n}$ is a constant determinated later which is the answer of the second question of Tom Leinster. First for $u\in \varphi$,we have $\frac{u(x+y) +u(x-y)- 2u(x)}{\|x - y\|^{n + 1}}\in L^{1}{\mathbb{R}^n}$.By fubini theorem(we exchange the integral in y with fourier transform in x),we obtain $$\mathcal {F}(Lu)=\int_{\mathbb{R}^n}\frac{\mathcal {F}(u(x+y) +u(x-y)- 2u(x))}{\| y\|^{n + 1}}\ dy.= \int_{\mathbb{R}^n}\frac{e^{i\xi\cdot y}+e^{-i\xi \cdot y}-2}{\|y\|^{n + 1}}\ dy \mathcal {F}(u)$$ hence,in order to get the desired result,it suffices to show that $\int_{\mathbb{R}^n}\frac{e^{i\xi y}+e^{-i\xi y}-2}{\|y\|^{n + 1}}\ dy=c_{n}|\xi|$. Define $$I(\xi)=\int_{\mathbb{R}^n}\frac{1-cos(y \xi)}{\|y\|^{n + 1}}\ dy$$.Since $I(\xi)$ is rotatonally invariant.so $I(\xi)=I(|\xi|e_{1})$ where $e_{1}$ denotes the first direction vector.So $$I(\xi)=I(|\xi|e_{1})=|\xi|\int\frac{1-cos(z_{1})}{\|z\|^{n + 1}}dz$$.so we take $c_{n}=\int\frac{1-cos(z_{1})}{\|z\|^{n + 1}}\ dz$.And we have $(-\triangle)^{1/2}=\frac{1}{c_{n}}L$ as desired.

(sharp)Garding's inequality and inequality with lower bounds

2012-05-07T01:41:52Z

The origin of Garding's inequality was an effort to solve Dirichlet's problem for linear elliptic operators of high even order.Let $$P(x,D)= \sum a_{\alpha}(x)D^{\alpha}$$ with principal part $$P_{2m}(D)= \sum_{\alpha=2m} a_{\alpha}(x)D^{\alpha}$$ have the property that $$ReP_{2m}(x,\xi)\geq c|\xi|^{2m}$$ for some $c>0$.It is further assumed that the coefficients $a_{\alpha}(x)$ have continious and bounded derivatives.Under this circumstances,one have for all $\epsilon>0$,$u\in C_{0}^{m}$,$$Re(Pu,u)\geq (c-\epsilon)\|u\|^{m,2}-b_{\epsilon}\|u\|^{2}$$. The original Garding's inequality seems to have been used for the first time by Leray in his 1954 lectures for an existence proof for Cauchy's problem for strongly hyperbolic systems.It was used again by Garding(1956) for a proof of the same result using only functional analysis. There is a sharp form of this inequality due to Hormander which says that if the symbol $P(x,\xi)\in S^{2m}$,$ReP(x,\xi) \geq 0$. Then $$Re(P(x,D)u,u)\geq-b\|u\|_{m-\frac{1}{2}}^{2}$$. A much more precise statement in the scalar case is known as the Fefferman-Phong inequality. Here my first question is what's the application of the sharp Garding's inequality? I once saw that Fefferman had wrote a paper named "sharp Garding's inequality and the uncertainty principle". I'm really surprised to see that this two inequalities can have connections.so i become more interested in inequalities with lower bounds.Can anyone list some inequalities appeared in PDE or other banches of math with a lower bound?Maybe some of them can have interesting links.

The fourier transform of homogeneous distribution and related topics

2012-04-25T13:26:09Z

When we have a distribuion $u\in \mathcal{D}'(R^n)$,and the restriction to $R^{n}\backslash{0}$ is homogeneous of degree a,we have $u \in \varphi'$ and $\widehat u$ is of degree(-n-a) in $R^{n}\backslash{0}$ .So it's easy to get when $-n< a <0$, $\widehat |x|^a=c_{a,n}|x|^{-n-a}$. my question is what's the result when $a>0$ ?Indeed, it's homogeneous of degree -n-a in$R^n$ and when a=n=1,we have $\widehat|x|=C p.v.|x|^{-2}$,where C is some constant. Futhermore,when we consider the wavefront set,it has the following:if u is a homogeneous distribuion in $R^{n}\backslash{0}$ ,then $(x,\xi) \in WF(u)\Leftrightarrow (\xi,-x) \in WF(\widehat u)$,where $x \neq 0$ , $\xi \neq 0$. Are there other interesting properties related to homogeneous distribuion ?

Does these commutator estimates bound in $L^{2}$

2012-04-23T14:08:03Z

According to the basic rules of symbolic caculus,$[a(x,D),x_{j}]=-ia^{j}[x,D]$.So we have $[(1-\triangle)^{\frac{1}{2}},x_i]=\partial_i(1-\triangle)^{-\frac{1}{2}}$ which is $L^2$ bounded. It's also true that the $[(1-\triangle)^{\frac{1}{2}},\langle x \rangle]$ is $L^2$ bounded.Can we write the explicit expression of this commutator? More generally,how to show that $[(1-\triangle)^{\frac{\alpha}{2}},\langle x \rangle^{\alpha}]$ ($\alpha \leq1$)is $L^2$ bounded? it is obviously true when $\alpha<0$ ?

what if we use $(-\triangle)^{\alpha}$ instead which the symbol of it is not smooth at 0?

Higher order fractional laplacian

2012-04-20T12:39:35Z

when consider the fractional laplacian $(-\triangle)^\alpha$,is there an essential difference between $0<\alpha<1$ and $\alpha>1$ ? As far as I'm concern,the higer order laplacian ($\alpha>1$ ) ,unlike the lower case, has little connection with stochastic process.(lack of positivy.) Since most paper i have met is the case of $0<\alpha<1$.And i wonder how things change when considering the higer order laplacian?

Answer by Shanlin Huang for Do these kernel functions satisfy the semi-group property?

2012-04-20T16:22:27Z

in fact,when $0<\alpha<1$,it has been shown that the estimate $G_{\alpha}(t,x)\leq \frac{t}{{t^{1/\alpha}+x^2}^{n/2+\alpha}}$,where $x \in R^n$.and when $\alpha>1$,similar result holds by similar proof.see miyajima: "gaussian estimates of order $\alpha$ and $L^p$ spectral independence of generators of $C_0 Semigroups$" Positivity 11 (2007),15-39

Comment by Shanlin Huang

2013-05-06T15:49:17Z

$(-\Delta)^{s}$ is positive if and only if $0<s\les 1$, and they generates positive heat semigroup $e^{-t(-\Delta)^{s}}$.

Comment by Shanlin Huang

2013-02-02T11:04:37Z

since the spectrum of A will not be discrete, it can not be a compact operator due to Riesz-Shauder theory.

Comment by Shanlin Huang

2013-01-31T02:29:48Z

Dear Anatoly Kochubei, I couldn't get it through the internet, neither in my school library. Would you show me what the detailed reslult is? Thanks very much.

Comment by Shanlin Huang

2013-01-24T12:42:14Z

@David: Thanks very much

Comment by Shanlin Huang

2012-11-23T16:09:29Z

notice that $\hat{H}(x)=p.v.(\frac{1}{x})$,and $\hat{xH}=D\hat{H}$,then you will find the answer easily

Comment by Shanlin Huang

2012-11-23T09:41:44Z

I found that C.Sogge's [Lectures on eigenfunctions][1] is also a wonderful reference [1]: mathnt.mat.jhu.edu/sogge/zju/…

Comment by Shanlin Huang

2012-11-23T03:40:53Z

If n equal to 1 or 3,then the inverse fourier transform of $(\xi^{2}-z)^{-1}$ is $c_{n}\frac{e^{-\sqrt{z}|x|}}{|x|}$,for other values of dimension,it can be an expression in terms of bessel functions. For instance,it can be found in stein's book "Singular Integrals and Differentiability Properties of Functions"

Comment by Shanlin Huang

2012-11-14T11:17:08Z

Thanks very much for this useful result.I think I should ask how does $\hat{f}$ behave near zero rather than at $\infty$.

Comment by Shanlin Huang

2012-11-13T15:07:33Z

@Alexandre Eremenko,I want to know how $\hat{f}(\xi)$ decay for large $\xi$,just as Renardy said,the problem is the slow decay of the functon $f$ at $\infty$

Comment by Shanlin Huang

2012-11-08T11:54:48Z

Dear fedja,thank you very much for pointing out that.Indeed,it's true only when we had assumed that the commutator $[V,(\mu-H_{0})^{-1}]$is bounded.But I wonder if there are other aproaches to deal with this.

Comment by Shanlin Huang

2012-10-23T10:12:22Z

@ Delio Mugnolo it's the projection onto the continuous spectrum of $H$

Comment by Shanlin Huang

2012-10-21T12:23:00Z

Thanks,one can get the positive answer when adding the condition $m(K_{\delta})<C\delta$,where $K_{\delta}$ is the set {$x\in \mathbb{R}^n$,dist(x,K)$\leq \delta$},for the case $2<p<\frac{2n}{n-1}$,you can see [this paper][1] [1]: projecteuclid.org/…

Comment by Shanlin Huang

2012-10-16T14:31:51Z

Thanks,it has been fixed

Comment by Shanlin Huang

2012-10-16T14:08:50Z

What's wrong with my latex ?it refused to work

Comment by Shanlin Huang

2012-10-12T14:25:38Z

@Abdelmalek Abdesselam ：sorry for that,see springerlink.com/content/h4g567q026617364 Proposition 2.1 for the transform of $e^{-|\xi|^\alpha}$