User alex a - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T14:58:27Z http://mathoverflow.net/feeds/user/19433 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/125538/analyticity-and-the-fbi-transform Analyticity and the FBI transform Alex A 2013-03-25T15:12:20Z 2013-03-25T15:12:20Z <p>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 <a href="http://en.wikipedia.org/wiki/FBI_transform" rel="nofollow">here</a>) to see that if $f$ is locally equal to an analytic function, then so is $g$. I can show that <code>$$|\mathcal{F}_a (f)(t,y)| = |\int \varphi (y-x) \mathcal{F}_a (g) (t,x) e^{ixt}\,dx|$$</code> but from this it appears difficult to say much about $\mathcal{F}_a (g) (t,x)$. </p> <p>Any insights are greatly appreciated. </p> http://mathoverflow.net/questions/108704/integral-equation Integral equation Alex A 2012-10-03T12:46:54Z 2012-10-19T13:06:33Z <p>Assume (for definiteness) $g:\mathbb{R} \to \mathbb{R}$ is continuous and that $f$ is defined by <code>$$f(E) = \int _0^{E-1} \Big ( (E - t)^2 - 1\Big )^{3/2} g(t) \, dt.$$</code> I'm interested in whether $g$ can be recovered assuming we know $f$. </p> <p>Does anyone know if this type of integrals have been studied before? </p> <p>For instance I am familiar to the fact that (Riemann-Liouville) integrals of the form <code>$$(J^\alpha g)(E) = \frac{1}{\Gamma (\alpha )}\int _0 ^E(E - t)^{\alpha -1}g(t) \, dt$$</code> 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. </p> <p>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. </p> <p>EDIT II: I have narrowed the problem down into finding $g_0$ (only depending on $t$) with <code>$$\int _1 ^{E-1} \Big ( (E - t)^2 - 1\Big )^{3/2} g_0(t) \, dt = 1, \qquad E&gt;1.$$</code> Not sure whether that helps though. </p> <p>EDIT III: If it helps I actually do know the solution in my particular case is <code>$$g(t) = \int _{\{h^{-1}(t)\}} \frac{1}{|\nabla h|}\,dS$$</code> for some $h$ for which the gradient never vanishes on <code>$\{h^{-1}(t)\}$</code>. 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.) </p> http://mathoverflow.net/questions/106026/non-analyticity-of-convolution Non-analyticity of convolution Alex A 2012-08-31T11:55:33Z 2012-09-02T14:53:10Z <p>I have posted a similar question in the past but let me make a final try in a simpler framework. </p> <p>Let $g \in C_0 ^\infty (\mathbb{R})$ be smooth and compactly supported. Define <code>$$f(x) = \int \big ((x - y)^2 - 1 \big )^{1/2}(x-y) g (y) \,dy$$</code> where integration is performed over the set where $|y - x|>1$ and $y\in \operatorname {supp}g$. </p> <p>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 $x_0 \pm 1$? </p> http://mathoverflow.net/questions/100774/hopeless-fourier-transform Hopeless Fourier transform? Alex A 2012-06-27T13:03:37Z 2012-06-27T13:03:37Z <p>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 $\pm 1$. </p> <p>What kind of methods/theorems could one use to analyze this question? </p> http://mathoverflow.net/questions/96200/constructing-an-example-of-hamiltonian-flow Constructing an example of Hamiltonian flow Alex A 2012-05-07T09:48:36Z 2012-05-08T12:00:10Z <p>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 <code>\begin{align*}\begin{cases} \frac{dx}{dt} &amp;= \frac{\xi (t)}{ \sqrt{\xi ^2 (t) + 1}} \\<br> \frac{d \xi }{dt} &amp;= - \nabla _x V(x(t)) \end{cases}, \qquad (x(0), \xi (0)) = (x_0, \xi _0). \end{align*}</code> 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 $\lim _{|t|\to \infty }|x(t)| = \infty$. </p> <p>I'm also grateful for non compactly supported examples (except for $V = \textrm{const.}$).</p> http://mathoverflow.net/questions/95151/from-microlocal-to-local From microlocal to local Alex A 2012-04-25T11:48:34Z 2012-04-26T08:06:58Z <p>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 <code>$$\| \operatorname{Op} [a]u \|_{L^2(\mathbb{R}^n)} \le Ch^s$$</code> where <code>$$\operatorname{Op} [a]u(x) := \frac{1}{(2\pi h)^{n/2}}\iint \limits_{T^\ast \mathbb{R}^n}e^{\tfrac{i}{h}(x-y)\cdot \xi }a(\tfrac{x+y}{2},\xi )u(y)\, dy\, d\xi .$$</code> Is this enough to conclude that there is some neighborhood $U$ of $x_0$ such that <code>$$\| u \|_{L^2(U)} \le Ch^s \quad ?$$</code> If not, what kind of condition would be sufficient?</p> http://mathoverflow.net/questions/94628/invertible-matrix-perturbation Invertible matrix perturbation Alex A 2012-04-20T12:08:45Z 2012-04-20T16:23:56Z <p>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). </p> <p>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 $x\in \mathbb{R} ^n$. </p> <p>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. </p> <p>Any ideas?</p> http://mathoverflow.net/questions/92998/fourier-transform Fourier transform Alex A 2012-04-03T12:56:02Z 2012-04-03T13:46:03Z <p>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?</p> http://mathoverflow.net/questions/92357/inverting-convolution Inverting convolution Alex A 2012-03-27T09:10:25Z 2012-03-28T09:24:46Z <p>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. </p> <p>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 <code>$$\mu = C \widehat {\Big (\frac{1}{\hat {f}}\Big)} \ast \omega$$</code> 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 $\mu$. </p> <p>My problem is I have no clue what the transform of $1/\hat{f}$ is. </p> <p>EDIT: I have changed from $H(x)$ to $H(x-1)$ in the definition of $f$. </p> http://mathoverflow.net/questions/84796/1a-not-invertible-implies-1an-not-invertible 1+A not invertible implies 1+A^n not invertible? Alex A 2012-01-03T10:40:18Z 2012-01-03T12:27:50Z <p>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? </p> <p>EDIT: I mean $1+(-1)^{n-1}A(z)^n$ rather than $1+A(z)^n$. </p> http://mathoverflow.net/questions/83619/support-of-analytic-extension Support of analytic extension Alex A 2011-12-16T13:42:05Z 2011-12-16T13:42:05Z <p>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 $\varepsilon >0$? </p> http://mathoverflow.net/questions/81802/does-such-a-smooth-function-exist Does such a smooth function exist? Alex A 2011-11-24T13:32:04Z 2011-12-09T12:22:12Z <p>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 $j=1,2,3$, </p> <p>$$\sum _{k=1}^3 \left |\frac{\partial ^2 g_k}{\partial x_j \partial x_k}\right | \le C\sum _{i=1}^3 \left | \frac{\partial g_i}{\partial x_i}\right |$$</p> <p>for some constant $C>0$. </p> <p>Can I find $g$?</p> http://mathoverflow.net/questions/82444/elliptic-pseudodifferential-operator-estimate Elliptic pseudodifferential operator estimate Alex A 2011-12-02T10:36:09Z 2011-12-02T11:39:52Z <p>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 </p> <p><code>$\|u\|_{H^1(U)} \le C (\|Pu\|_{L^2(W)} + \|u\|_{L^2(W)}), \quad (C&gt;0 \text{ a constant})$</code></p> <p>for any $\overline{U}\subset W\subset \mathbb{R}^n$ with $W$ bounded. Say I'm only interested in an upper bound of <code>$\|u\|_{L^2(U)}$</code>. Then the term $\|u\|_{L^2(W)}$ on the right hand side above feels a bit redundant. Can I obtain </p> <p><code>$\|u\|_{L^2(U)} \le C\|Pu\|_{L^2(W)} \quad \text{?}$</code> </p> http://mathoverflow.net/questions/82155/differential-operator-estimate Differential operator estimate Alex A 2011-11-29T10:18:49Z 2011-11-29T10:18:49Z <p>Suppose I have an a priori estimate of the form </p> <p>$\|u\|\substack{H^1(\mathbb{R}^n)} \le C \| Pu \|_{L^2(\mathbb{R}^n)},$</p> <p>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 </p> <p>$\|u \| \substack{H^1(\Omega )}\leq C \| Pu \|_{L^2(\tilde{\Omega })}$</p> <p>for any $\tilde{\Omega }$ such that $\overline {\Omega } \subset \tilde{\Omega }$ (the bar means closure)?</p> http://mathoverflow.net/questions/81802/does-such-a-smooth-function-exist/81874#81874 Answer by Alex A for Does such a smooth function exist? Alex A 2011-11-25T12:18:09Z 2011-11-25T12:18:09Z <p>@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 </p> <p>$$2|u(t)|\frac{d|u(t)|}{dt} = 2u(t)u'(t)$$</p> <p>whence</p> <p>$$\frac{d|u(t)|}{dt} \le u'(t)\le C|u(t)|.$$ </p> <p>Now the function $f(t):=|u(t)|$ is non-negative, satisfies $f'(t)\le Cf(t)$. By Grönwall's inequality we obtain </p> <p>$$f(t)\le f(0)e^{Ct}=0 \quad \text{for }t>0$$</p> <p>and thus $u(t)=0$ for $t\ge 0$. The case $t&lt;0$ can be treated similarly. </p> http://mathoverflow.net/questions/81863/estimate-a-surface-integral Estimate a surface integral Alex A 2011-11-25T10:25:13Z 2011-11-25T10:25:13Z <p>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. </p> <p>My question:</p> <p>Is there any way of estimating the surface integral </p> <p>$$\int \limits _{\partial \Sigma } a(x)|u(x)|^2\ dS_x$$</p> <p>by some non-surface integral of these quantities that does not involve putting any derivative on $a$? One derivative on $u$ would be fine.</p> <p>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. </p> <p>What comes to my mind is some Green formula, or possibly the trace theorem, but I didn't succeed yet. </p> http://mathoverflow.net/questions/81734/non-self-adjoint-operator-estimate Non-self-adjoint operator estimate Alex A 2011-11-23T18:45:24Z 2011-11-23T18:45:24Z <p>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 </p> <p>$$D_\theta = -\frac{1}{a(x)}i\alpha \cdot \nabla + \beta +R (x,\partial _x) + V(x)$$ </p> <p>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.</p> <p>Moreover $R (x,\partial _x)$ is a first order differential operator with smooth compactly supported coefficients supported in $2&lt;|x|&lt;3$ and $V(x)$ is Hermitian and supported in $|x|&lt;1$. </p> <p>Now, my wish is to obtain a bound of the form </p> <p>$$\operatorname{Im} z \|u \|^2 \le C \|(D_\theta - z)u\| \quad \text{for }\operatorname{Im} z >0 \text{ and }\operatorname{Re}z > 1$$</p> <p>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 </p> <p>$$\operatorname{Im}\langle a(x)(D_\theta -z)u,u\rangle \le -C\operatorname{Im}z\|u\|^2$$</p> <p>Upon expanding the left hand side above and using the fact that the largest eigenvalue of $\beta$ is 1 I obtain </p> <p>$$\operatorname{Im}\langle a(x)(D_\theta -z)u,u\rangle \le (1 - \operatorname{Re}z)\langle (\operatorname{Im}a(x))u,u\rangle + \operatorname{Im}\langle a(x)Ru,u\rangle -C\operatorname{Im}z\|u\|^2$$ </p> <p>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 $2&lt;|x|&lt;|3|$ and 0 for $|x|\le 1$, then I can show</p> <p>$$\operatorname{Im}\langle a(x)Ru,u\rangle \le C\|(D_\theta - z)\chi u\| \|u\|.$$</p> <p>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 $h$. </p> <p>Any new ideas would be greatly appreciated. </p> http://mathoverflow.net/questions/81525/operators-and-eigenfunctions-is-this-correct Operators and eigenfunctions - is this correct? Alex A 2011-11-21T16:42:03Z 2011-11-21T23:36:15Z <p>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. </p> <p>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. </p> <p>Am I forgetting something? </p> http://mathoverflow.net/questions/124966/infinite-product-estimate Comment by Alex A Alex A 2013-03-20T08:26:16Z 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. http://mathoverflow.net/questions/104195/theorem-of-supports Comment by Alex A Alex A 2013-01-21T07:53:50Z 2013-01-21T07:53:50Z Thank you for your reply S&#246;nke! I know, I have seen counterexamples in other situations. I would really like to know whether anything can be said in this special case. http://mathoverflow.net/questions/108704/integral-equation Comment by Alex A Alex A 2012-10-26T14:58:48Z 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 &quot;short&quot;, which is not necessarily true in my setting. http://mathoverflow.net/questions/108704/integral-equation Comment by Alex A Alex A 2012-10-24T09:28:46Z 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. http://mathoverflow.net/questions/108704/integral-equation Comment by Alex A Alex A 2012-10-09T06:38:20Z 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. http://mathoverflow.net/questions/106026/non-analyticity-of-convolution Comment by Alex A Alex A 2012-09-03T07:33:05Z 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 $(x^2 - 1)^{1/2}x 1_{|x| &gt; 1}$ but I don't get much further after that. Actually I'm looking at this in the framework of distribution theory where $g\in \mathcal{E}'(\mathbb{R})$. http://mathoverflow.net/questions/106026/non-analyticity-of-convolution/106032#106032 Comment by Alex A Alex A 2012-09-02T15:10:11Z 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 $\varphi$. http://mathoverflow.net/questions/106026/non-analyticity-of-convolution/106032#106032 Comment by Alex A Alex A 2012-09-02T13:18:06Z 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. http://mathoverflow.net/questions/106026/non-analyticity-of-convolution/106032#106032 Comment by Alex A Alex A 2012-08-31T14:13:59Z 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 $x_0$. http://mathoverflow.net/questions/104195/theorem-of-supports Comment by Alex A Alex A 2012-08-08T11:25:30Z 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. http://mathoverflow.net/questions/96200/constructing-an-example-of-hamiltonian-flow/96328#96328 Comment by Alex A Alex A 2012-05-09T07:38:59Z 2012-05-09T07:38:59Z Thanks, maybe it's hard to give general sufficient properties of $V$. http://mathoverflow.net/questions/94628/invertible-matrix-perturbation/94651#94651 Comment by Alex A Alex A 2012-04-30T13:57:39Z 2012-04-30T13:57:39Z Of course, so obvious once you see it! Thanks! http://mathoverflow.net/questions/94628/invertible-matrix-perturbation/94651#94651 Comment by Alex A Alex A 2012-04-28T20:13:58Z 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 $x_0$? http://mathoverflow.net/questions/95151/from-microlocal-to-local/95234#95234 Comment by Alex A Alex A 2012-04-27T08:33:19Z 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 $u$? http://mathoverflow.net/questions/95151/from-microlocal-to-local/95234#95234 Comment by Alex A Alex A 2012-04-26T08:17:55Z 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 $O(h^s)$.