User alex a - MathOverflow

Analyticity and the FBI transform

2013-03-25T15:12:20Z

Let $g\in \mathcal{E}'(\mathbb{R} )$ be a distribution on $\mathbb{R}$ having compact support. Define $f\in \mathcal{D}'(\mathbb{R})$ by $$ f = \varphi \ast g , $$ where $\ast $ denotes convolution and $\varphi (t) = \sqrt{t}$ for $t\ge 0$ and 0 otherwise. I want to apply the FBI transform (see here) to see that if $f$ is locally equal to an analytic function, then so is $g$. I can show that $$ |\mathcal{F}_a (f)(t,y)| = |\int \varphi (y-x) \mathcal{F}_a (g) (t,x) e^{ixt}\,dx| $$ but from this it appears difficult to say much about $\mathcal{F}_a (g) (t,x)$.

Any insights are greatly appreciated.

Integral equation

2012-10-03T12:46:54Z

Assume (for definiteness) $g:\mathbb{R} \to \mathbb{R}$ is continuous and that $f$ is defined by $$ f(E) = \int _0^{E-1} \Big ( (E - t)^2 - 1\Big )^{3/2} g(t) \, dt. $$ I'm interested in whether $g$ can be recovered assuming we know $f$.

Does anyone know if this type of integrals have been studied before?

For instance I am familiar to the fact that (Riemann-Liouville) integrals of the form $$ (J^\alpha g)(E) = \frac{1}{\Gamma (\alpha )}\int _0 ^E(E - t)^{\alpha -1}g(t) \, dt $$ can be inverted when $\alpha $ is a half-integer by using identities of the form $J^\alpha \circ J^\beta = J^{\alpha + \beta }$ and then differentiate.

EDIT: I would just like to point out that I'm not necessarily looking for an explicit inversion formula. If the above equation fits into some general theory which concludes that $g$ can be recovered I'm happy.

EDIT II: I have narrowed the problem down into finding $g_0$ (only depending on $t$) with $$ \int _1 ^{E-1} \Big ( (E - t)^2 - 1\Big )^{3/2} g_0(t) \, dt = 1, \qquad E>1. $$ Not sure whether that helps though.

EDIT III: If it helps I actually do know the solution in my particular case is $$ g(t) = \int _{\{h^{-1}(t)\}} \frac{1}{|\nabla h|}\,dS $$ for some $h$ for which the gradient never vanishes on $\{h^{-1}(t)\}$ . Here $dS$ is surface measure. (The reason I still want to solve the equation is that I know $f$ is a certain invariant and I need to show $g$ is also invariant.)

Non-analyticity of convolution

2012-08-31T11:55:33Z

I have posted a similar question in the past but let me make a final try in a simpler framework.

Let $g \in C_0 ^\infty (\mathbb{R})$ be smooth and compactly supported. Define $$ f(x) = \int \big ((x - y)^2 - 1 \big )^{1/2}(x-y) g (y) \,dy $$ where integration is performed over the set where $|y - x|>1$ and $y\in \operatorname {supp}g$.

If $g$ fails to be real analytic at some point $x_0$ can we deduce that also $f$ fails to be real analytic at some point depending on $x_0$, like perhaps $x_0 \pm 1$?

Hopeless Fourier transform?

2012-06-27T13:03:37Z

Let $Y_2(x)$ denote the Bessel function of second kind and second order. I'm interested in the Fourier transform of the reciprocal $1/Y_2(|x|)$ in the distributional sense. I suppose there is not much hope of a closed formula for the transform, but all I need is that the Fourier transform of $1/Y_2(|x|)$ is real analytic away from $\pm 1$ but not real analytic in any neighborhood of $\pm 1$.

What kind of methods/theorems could one use to analyze this question?

Constructing an example of Hamiltonian flow

2012-05-07T09:48:36Z

I have this Hamiltonian flow generated by $$ h(x, \xi ) = V(x) + \sqrt{\xi ^2 + 1}, \quad x, \xi \in \mathbb{R}^3, $$ so the defining equations are \begin{align*}\begin{cases} \frac{dx}{dt} &= \frac{\xi (t)}{ \sqrt{\xi ^2 (t) + 1}} \\ \frac{d \xi }{dt} &= - \nabla _x V(x(t)) \end{cases}, \qquad (x(0), \xi (0)) = (x_0, \xi _0). \end{align*} I want to find an example with $V\in C_0^\infty (\mathbb{R}^3)$ (i.e. compactly supported and smooth) such that the solution $(x(t), \xi (t))$ at some energy $E>1$ (i.e. $h(x_0, \xi _0) = E$ and hence $h(x(t), \xi (t)) = E$) satisfies $\lim _{|t|\to \infty }|x(t)| = \infty $.

I'm also grateful for non compactly supported examples (except for $V = \textrm{const.}$).

From microlocal to local

2012-04-25T11:48:34Z

Assume $u\in L^2(\mathbb{R}^n)$ and let $(x_0, \xi _0) \in T^\ast \mathbb{R}^n = \mathbb{R}^n_x \times \mathbb{R}^n_\xi $. Assume I can find $a\in C^\infty (T^\ast \mathbb{R}^n)$ which is also bounded with all derivatives and $a(x,\xi ) = 1$ in a neighborhood of $(x_0, \xi _0)$. Without loss of generality we may even assume $a\in C _0 ^\infty (T^\ast \mathbb{R}^n)$ and still equal to 1 near $(x_0, \xi _0)$. Assume moreover that we know $$\| \operatorname{Op} [a]u \|_{L^2(\mathbb{R}^n)} \le Ch^s$$ where $$\operatorname{Op} [a]u(x) := \frac{1}{(2\pi h)^{n/2}}\iint \limits_{T^\ast \mathbb{R}^n}e^{\tfrac{i}{h}(x-y)\cdot \xi }a(\tfrac{x+y}{2},\xi )u(y)\, dy\, d\xi .$$ Is this enough to conclude that there is some neighborhood $U$ of $x_0$ such that $$\| u \|_{L^2(U)} \le Ch^s \quad ? $$ If not, what kind of condition would be sufficient?

Invertible matrix perturbation

2012-04-20T12:08:45Z

Let $A$ be an $n\times n$ matrix which depends smoothly on a variable $x\in \mathbb{R}^n$ and such that there are constants $C_\alpha > 0 $ so that $\| \partial ^\alpha A \| \le C_\alpha $ for all multi-indices $\alpha \in \mathbb{N}_0^n$ (i.e. $A$ together with all its derivatives are bounded in matrix norm). Call the set of such matrices $S$. Assume moreover that at some point $x_0 \in \mathbb{R}^n$ we have $A(x_0) = I$ (identity matrix).

I now want to prove that we can find a matrix $B\in S$ that is supported away from $x_0$ and such that $A + B$ is invertible for all $x\in \mathbb{R} ^n$.

My approach was that, since $A$ is smooth, we can find neighborhoods $V_0$ and $V_1$ of $x_0$ with $\overline{V_0}\subset V_1$ such that $$ \|A - I\| \le 1/3 \text{ on } V_0 \quad \text{ and } \quad \|A - I\| \le 2/3 \text{ on } V_1 $$ and then take $B=t\chi I$ for some suitable $t\in \mathbb{R}$ where $\chi $ is smooth and $\chi = 0$ on $V_0$ and $\chi = 1$ outside $V_1$. I tried making $A+B$ within distance 1 to identity but did not succeed.

Any ideas?

Fourier transform

2012-04-03T12:56:02Z

Does anyone know what the Fourier transform (in the sense of distributions) of $$ f(x) = (x^2 - 1)^{1/2}x, \quad |x|\ge 1, $$ and $f(x) = 0$ otherwise, is?

Inverting convolution

2012-03-27T09:10:25Z

I have a relation of the form $$\omega = f\ast \mu \qquad (1)$$ where $\omega $ and $\mu $ are distributions in $\mathcal {D}'(\mathbb{R})$ and $$f(x) = H(x-1)(x^2 - 1)^{1/2}x$$ with $H(x) = 1_{\mathbb{R}_+}$ denoting the Heaviside function.

My goal is to show that $\mathrm{WF}_a (\omega ) = \mathrm{WF}_a (\mu )$ where $\mathrm{WF}_a$ is the analytical wave front set, and for this I would like to invert (1), i.e. solve it for $\mu $. By taking Fourier transforms I get $$ \mu = C \widehat {\Big (\frac{1}{\hat {f}}\Big)} \ast \omega $$ for some constant C. Here I would thus like to conclude that if $\omega $ extends to a holomorphic function near some $x\in \mathbb{R}$ then the same goes for $\mu $.

My problem is I have no clue what the transform of $1/\hat{f}$ is.

EDIT: I have changed from $H(x)$ to $H(x-1)$ in the definition of $f$.

1+A not invertible implies 1+A^n not invertible?

2012-01-03T10:40:18Z

Let $A(z)$ be a compact operator on a Hilbert space, depending on a complex parameter $z$. I want to count the number of points where $1+A(z)$ is not invertible and therefore I want to count zeros of $\det (1+A(z))$. Unfortunately $A(z)$ is not of trace class so the determinant does not make sense. However, for some positive integer power $n$ the operator $A(z)^n$ is of trace class. Can I argue that the points $z$ where $1+A(z)^n$ is not invertible include those where $1+A(z)$ is not invertible?

EDIT: I mean $1+(-1)^{n-1}A(z)^n$ rather than $1+A(z)^n$.

Support of analytic extension

2011-12-16T13:42:05Z

Suppose to start with I have a smooth function $f:\mathbb{R}^n \to \mathbb{R}$, a neighborhood $\Omega $ of $\mathbb{R}^n$ in $\mathbb{C}^n$ and that I assume $f$ to have an analytic extension to $\Omega $. If $f$ has support in $|x|>R$ for some $R>0$, may I, without loss of generality, assume that its extension is supported in $|\operatorname{Re}z|>R$ in $\mathbb{C}^n$? If not, can I assume so for $|\operatorname{Re}z|>R-\varepsilon$ for any $\varepsilon >0 $?

Does such a smooth function exist?

2011-11-24T13:32:04Z

I am looking for a $C^\infty $ function $g:\mathbb{R}^3\to \mathbb{R}^3$ such that $g(x)=0$ for $|x|\le 1$ and $g(x)=x$ for $|x|\ge 2$. Certainly such $g$ can be constructed, but I also want it to satisfy the additional property that for each $j=1,2,3$,

$$\sum _{k=1}^3 \left |\frac{\partial ^2 g_k}{\partial x_j \partial x_k}\right | \le C\sum _{i=1}^3 \left | \frac{\partial g_i}{\partial x_i}\right |$$

for some constant $C>0$.

Can I find $g$?

Elliptic pseudodifferential operator estimate

2011-12-02T10:36:09Z

If $P$ is an elliptic pseudodifferential operator of order 1 in the sense that its principal symbol is invertible, then we have the a priori estimate

$\|u\|_{H^1(U)} \le C (\|Pu\|_{L^2(W)} + \|u\|_{L^2(W)}), \quad (C>0 \text{ a constant})$

for any $\overline{U}\subset W\subset \mathbb{R}^n$ with $W$ bounded. Say I'm only interested in an upper bound of $\|u\|_{L^2(U)}$ . Then the term $\|u\|_{L^2(W)}$ on the right hand side above feels a bit redundant. Can I obtain

$\|u\|_{L^2(U)} \le C\|Pu\|_{L^2(W)} \quad \text{?}$

Differential operator estimate

2011-11-29T10:18:49Z

Suppose I have an a priori estimate of the form

$\|u\|\substack{H^1(\mathbb{R}^n)} \le C \| Pu \|_{L^2(\mathbb{R}^n)},$

where $H^1=W^{1,2}$ is the first order $L^2$ Sobolev space (which I suppose is irrelevant for my question). Moreover $P = \sum a_{(\alpha )}(x) \partial ^{(\alpha )}$ is a partial differential operator (say of order 1). Now, let $\Omega \subset \mathbb{R}^n$. Can I claim that

$\|u \| \substack{H^1(\Omega )}\leq C \| Pu \|_{L^2(\tilde{\Omega })}$

for any $\tilde{\Omega }$ such that $\overline {\Omega } \subset \tilde{\Omega }$ (the bar means closure)?

Answer by Alex A for Does such a smooth function exist?

2011-11-25T12:18:09Z

@GH: Considering one variable, assume $|u'(t)|\le C|u(t)|$ and that $u(t)=0$ for $|t|\le 1$. Now $|u(t)|^2 = u(t)^2$ implies

$$2|u(t)|\frac{d|u(t)|}{dt} = 2u(t)u'(t)$$

whence

$$\frac{d|u(t)|}{dt} \le u'(t)\le C|u(t)|.$$

Now the function $f(t):=|u(t)|$ is non-negative, satisfies $f'(t)\le Cf(t)$. By Grönwall's inequality we obtain

$$f(t)\le f(0)e^{Ct}=0 \quad \text{for }t>0$$

and thus $u(t)=0$ for $t\ge 0$. The case $t<0$ can be treated similarly.

Estimate a surface integral

2011-11-25T10:25:13Z

I hope this is easy. Let $\Sigma \subset \mathbb{R}^n$ be bounded with smooth boundary (it could be the shell region between two balls). Assume moreover that $u$ belongs to the Sobolev space $H^1(\mathbb{R}^n) = W^{1,2}(\mathbb{R}^n)$ and $a:\mathbb{R}^n\to \mathbb{R}_{\ge 0}$ is a smooth bounded and non-negative function.

My question:

Is there any way of estimating the surface integral

$$\int \limits _{\partial \Sigma } a(x)|u(x)|^2\ dS_x$$

by some non-surface integral of these quantities that does not involve putting any derivative on $a$? One derivative on $u$ would be fine.

So for instance, if one could estimate it by $C\|\sqrt{a}u\|_{L^2}^2$ for some constant $C>0$, that would be terrific, but other estimates could also do.

What comes to my mind is some Green formula, or possibly the trace theorem, but I didn't succeed yet.

Non-self-adjoint operator estimate

2011-11-23T18:45:24Z

So, what I have to begin with is a so called complex scaled Dirac operator on $L^2(\mathbb{R}^3)$. It can be written in the form

$$D_\theta = -\frac{1}{a(x)}i\alpha \cdot \nabla + \beta +R (x,\partial _x) + V(x)$$

where $a$ is a complex-valued function such that $\operatorname{Re} a$ is uniformly bounded away from 0 and $\operatorname{Im} a$ is non-negative and supported in $|x|>2$. Next, $\alpha $ is a vector of so called Dirac matrices ($4\times 4$), $\beta $ is the constant matrix $\beta = \mathrm{diag}(I_2, -I_2)$ where $I_2$ is the 2x2 identity matrix.

Moreover $R (x,\partial _x)$ is a first order differential operator with smooth compactly supported coefficients supported in $2<|x|<3$ and $V(x)$ is Hermitian and supported in $|x|<1$.

Now, my wish is to obtain a bound of the form

$$\operatorname{Im} z \|u \|^2 \le C \|(D_\theta - z)u\| \quad \text{for }\operatorname{Im} z >0 \text{ and }\operatorname{Re}z > 1$$

where C is some positive constant. My plan of attack is to consider the quantity $\operatorname{Im}\langle a(x)(D_\theta -z)u,u\rangle $. For instance the claim would readily follow if I can show

$$\operatorname{Im}\langle a(x)(D_\theta -z)u,u\rangle \le -C\operatorname{Im}z\|u\|^2$$

Upon expanding the left hand side above and using the fact that the largest eigenvalue of $\beta $ is 1 I obtain

$$\operatorname{Im}\langle a(x)(D_\theta -z)u,u\rangle \le (1 - \operatorname{Re}z)\langle (\operatorname{Im}a(x))u,u\rangle + \operatorname{Im}\langle a(x)Ru,u\rangle -C\operatorname{Im}z\|u\|^2$$

Clearly the first term on the right is non-positive since $\operatorname{Re}z>1$. My problem is to handle the term $\operatorname{Im}\langle a(x)Ru,u\rangle$. There are essentially two ways to go about -- either one uses the term $(1 - \operatorname{Re}z)\langle (\operatorname{Im}a(x))u,u\rangle $ to absorb it, or one estimate it by $C\|(D_\theta - z)u\|$. The latter approach seems more likely, or at least a mixture of the two. I can get close, for if $\chi $ is a smooth bump function which equals 1 near $2<|x|<|3|$ and 0 for $|x|\le 1$, then I can show

$$\operatorname{Im}\langle a(x)Ru,u\rangle \le C\|(D_\theta - z)\chi u\| \|u\|.$$

But when I in turn want to estimate this from above by $C\|(D_\theta - z)u\|\|u\|$ the error is an annoying commutator term $\|[D_\theta , \chi ]u\|$. Actually I am working in the semi-classical setting and then this term is $\mathcal{O}(h)\|u\|_{\operatorname{supp}(\nabla \chi )}$, but this is not small enough to be absorbed by any other term since $\operatorname{Im}a(x)$ is zero on part of $\operatorname{supp}(\nabla \chi )$ and $\operatorname{Im}z$ is supposed to be even smaller in terms of $h$.

Any new ideas would be greatly appreciated.

Operators and eigenfunctions - is this correct?

2011-11-21T16:42:03Z

Assume we have a linear operator $P_0: D->H$ where $D$ is the domain and $H$ some Hilbert space it is acting on. Assume moreover that the spectrum of $P_0$ is purely essential. If $z$ is a complex number not in the spectrum of $P_0$ then the resolvent $(P_0 - z)^{-1}$ exists as a bounded operator $H->D$, meaning there is a constant (depending on $z$) such that $|u|\leq C|(P_0-z)u|$ where the norm on the left hand side is that of the domain D.

Now perturb $P_0$ by a "potential" $V:R^n->C$ (complex-valued function on Euclidean n-space) and denote $P=P_0+V$ where we assume $V$ has compact support. Assume also that $V$ generates an eigenvalue $z_0$ away from the essential spectrum of $P_0$ (which we may assume is also the essential spectrum of $P$). Thus there is a non-zero $u$ in $D$ with $(P-z_0)u=0$. Let $X$ be the characteristic function of some set (call it $X$ as well) which is disjoint from the support of $V$. Then $|Xu|\leq C|(P_0-z_0)Xu|=C|(P-z_0)Xu|=0$, i.e. the eigenfunction must be 0 wherever the potential is zero.

Am I forgetting something?

Comment by Alex A

2013-03-20T08:26:16Z

Yes, you are right, probably the upper bound is supposed to be of the form $Ce^{C|x|^{n(n-1)}}$, which is fine.

Comment by Alex A

2013-01-21T07:53:50Z

Thank you for your reply Sönke! I know, I have seen counterexamples in other situations. I would really like to know whether anything can be said in this special case.

Comment by Alex A

2012-10-26T14:58:48Z

I must admit I cannot entirely see how to carry over the arguments in your paper to my situation. For instance you are using the fact that the interval of integration is "short", which is not necessarily true in my setting.

Comment by Alex A

2012-10-24T09:28:46Z

Thank you very much, Fedja! I will take a look in that paper of yours immediately. I will also consider changing my user name as I was not aware of this problem. Thanks.

Comment by Alex A

2012-10-09T06:38:20Z

I'm only interested in $E$ belonging to a bounded interval and also in recovering $g$ on this interval. If it makes any difference I know $g\ge 0$ in my particular case.

Comment by Alex A

2012-09-03T07:33:05Z

Thanks a lot for your response. The idea of inversion was my first attempt. I do know the Fourier transform of $(x^2 - 1)^{1/2}x 1_{|x| > 1}$ but I don't get much further after that. Actually I'm looking at this in the framework of distribution theory where $g\in \mathcal{E}'(\mathbb{R})$.

Comment by Alex A

2012-09-02T15:10:11Z

Thank you for your edited comment. I'm familiar with what you are saying. But what I want is more or less the opposite inclusion, i.e. a conclusion of the form; if we start with a point $x_0\in S(g)$, then the point $\varphi (x_0)$ belongs to $S(f)$. I want to know what $\varphi $ is, if there is such a $\varphi $.

Comment by Alex A

2012-09-02T13:18:06Z

Of course such a $g$ will not be everywhere real analytic. But what I'm asking is if those points where it isn't will induce points where $f$ is not real analytic, if we define $f$ by the above convolution.

Comment by Alex A

2012-08-31T14:13:59Z

But if instead we have $f(x) = \int (x-y)^\alpha _+ g(y) dy$ then, since the Fourier transform of $x_+^\alpha $ and $1/(x_+^\alpha \hat{)}$ are known one can deduce that $g$ extends to a holomorphic function near $x_0$ if and only if $f$ extends near $x_0$.

Comment by Alex A

2012-08-08T11:25:30Z

@Johannes: Thanks for those encouraging words! I've been somewhat skeptical because I've seen a more general version of the theorem but that involved a certain growth bound on the Fourier transform of the non compactly supported factor which was not satisfied in my case because of the zeros of the Bessel function. @Alexandre: The analytical singular support of a distribution $f$ is the set of points having no open neighborhood to which the restriction of $f$ is real analytic.

Comment by Alex A

2012-05-09T07:38:59Z

Thanks, maybe it's hard to give general sufficient properties of $V$.

Comment by Alex A

2012-04-30T13:57:39Z

Of course, so obvious once you see it! Thanks!

Comment by Alex A

2012-04-28T20:13:58Z

Is there a similar argument if A is merely assumed to be invertible at the point $x_0$, so there is also the possibility of one eigenvalue being negative at $x_0$?

Comment by Alex A

2012-04-27T08:33:19Z

So are you saying that using a partition of unity I should construct a globally elliptic operator $A$ with $\|Au\| = O(h^s)$ and this way obtain a global estimate of $u$?

Comment by Alex A

2012-04-26T08:17:55Z

Ok, thank you, it's what I was afraid of. Actually I also have that $u$ satisfies $Pu=g$ where $\|g\| \le Ch^s$ and $P$ is elliptic in the complement of the set where $u$ is microlocally $O(h^s)$.