User user17240 - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T23:40:29Z http://mathoverflow.net/feeds/user/2274 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/100739/smoothness-of-the-convolution-of-a-singular-measure-with-itself Smoothness of the convolution of a singular measure with itself user17240 2012-06-27T03:25:57Z 2012-07-04T23:59:06Z <p>Let $\gamma:I\subset\mathbb{R}\rightarrow\mathbb{R}^2, s\mapsto \gamma(s)$, denote the arclength parametrization of a smooth, convex curve $\Gamma:=\gamma(I)\subset\mathbb{R}^2$. Equip $\Gamma$ with arclength measure $\sigma$.</p> <p>Here is my question: what extra conditions on $\Gamma$ will ensure that the triple convolution $\sigma\ast\sigma\ast\sigma$ is a <em>smooth</em> function of $x\in\mathbb{R}^2$ inside its support (i.e. the set $\Gamma+\Gamma+\Gamma\subset\mathbb{R}^2$), up to the boundary? I am also interested in partial regularity results, and counterexamples.</p> <p>Here is a possible approach: for any $x\in\mathbb{R}^2$ and $\epsilon>0$, <code>$$\int_{|y-x|&lt;\epsilon} \sigma\ast\sigma\ast\sigma(y)dy=|\{(s,s',s'')\in I^3: |\gamma(s)+\gamma(s')+\gamma(s'')-x|&lt;\epsilon\}|.$$</code></p> <p>Understanding how these sets look like will probably help. I am familiar with results in this spirit for <em>double</em> convolutions (e.g. Fefferman 1970), but not for triple ones. </p> <p>References would be much appreciated.</p> <p>Thank you.</p> http://mathoverflow.net/questions/78067/l1-norm-of-the-fourier-transform-of-a-truncated-gaussian $L^1$ norm of the Fourier transform of a truncated Gaussian user17240 2011-10-13T21:28:23Z 2012-02-22T00:06:10Z <p>I asked this question on Math StackExchange recently but the only useful comment I got was that this could be a good question for Math Overflow. Here it goes:</p> <p>Consider the Gaussian $G(x):=e^{-x^2}$ on the real line, and localize it to the region $|x|\sim 2^k$ by multiplying it by an appropriate smooth cut-off. More precisely, take $\phi\in C_0^\infty(\mathbb{R})$ supported in the region $${x\in\mathbb{R}: \frac{1}{2}&lt;|x|\leq2}$$ such that $0\leq\phi\leq 1,$ and let $\phi_k(x):=\phi(2^{-k}x)$. Consider: $$G_k(x):=\phi_k(x)G(x).$$ It is straightforward to check that $\|G_k\|_{L^1}\lesssim 2^ke^{-4^k}$, which tends (very quickly) to $0$ as $k\rightarrow\infty$. Also, using Young's convolution inequality one can easily show that $\|\widehat{G_k}\|_{L^1}\leq \|\phi\|_1\|G\|_1$, but this gives no decay in terms of $k$. </p> <p>My question is: does $\|\widehat{G_k}\|_{L^1}$ decay as $k\rightarrow\infty$? If so, how fast? Can you prove sharp bounds (in $k$)?</p> <p>Thank you.</p> http://mathoverflow.net/questions/66140/geometric-meaning-of-derivatives-of-the-curvature Geometric meaning of derivatives of the curvature user17240 2011-05-27T01:41:21Z 2011-05-27T23:03:56Z <p>Let $\Gamma$ be a sufficiently smooth, closed, convex curve in the plane parametrized by arclength $s$. Further assume that $\kappa>0$ and that the second derivative of $\kappa$ with respect to $s$ (denoted $\kappa''$) is such that $$(\kappa''\cdot \kappa^{-3})(p)&lt; C$$ for every point $p\in\Gamma$ at which $\kappa$ attains a global minimum (for some absolute constant $C&lt;\infty$). </p> <p>What can be said about the curve $\Gamma$? In particular, is the quantity $\kappa''\cdot \kappa^{-3}$ of geometric significance and has it been considered before? Can it for instance be expressed in terms of the (centro-)affine curvature of $\Gamma$? References would be most appreciated! </p> <p>Motivation comes from trying to reduce several local problems involving the restriction of the Fourier transform to $\Gamma$ to the corresponding (more tractable) problem for an appropriate osculating conic.</p> <p>Thank you.</p> http://mathoverflow.net/questions/64693/phase-perturbations-in-oscillatory-integrals Phase perturbations in oscillatory integrals user17240 2011-05-11T21:02:07Z 2011-05-12T22:07:32Z <p>I am interested in learning about quantitative refinements of the method of stationary phase which allow to treat small perturbations in the phase of an oscillatory integral (of the first kind, in Stein's terminology).</p> <p>I have the following concrete problem, which originated from studying the restriction problem for the Fourier transform on the 2-sphere. Let $f\in C_0^\infty (\mathbb{R}^2)$ be supported in $B(0,R_1)$, and consider the phase $\phi_\epsilon (y)=\frac{|y|^2}{2}+\epsilon\frac{1}{8}|y|^4.$ Look at the integrals</p> <p>$$I_\epsilon(x,t)=\int_{\mathbb{R}^2}e^{-ix\cdot y}e^{-it \phi_\epsilon(y)}f(y)dy$$ and </p> <p>$$J_\epsilon(x,t)=\int_{\mathbb{R}^2}e^{-ix\cdot y}e^{-it (\phi_\epsilon(y)+\epsilon^2 |y|^6)}f(y)dy.$$</p> <p>Let $R_2>0$ be sufficiently large. Is it true that, for $|(x,t)|\geq R_2$,<br> $$|I_\epsilon(x,t)|^4-|J_\epsilon(x,t)|^4=C\epsilon^2 |(x,t)|^{-4} \;\;\textrm{as }\;\; \epsilon\rightarrow 0^+,$$ where the constant $C$ is allowed to depend on $R_1$ and $R_2$ (and possibly on some appropriate norm of $f$ and its derivatives), but on nothing else?</p> <p>As a consequence one would have that <code>$\|J_\epsilon\|_{L^4_{x,t}(\mathbb{R}^3)}^4=\|I_\epsilon\|_{L^4_{x,t}(\mathbb{R}^3)}^4+O(\epsilon^2)$</code> (which is what I am really interested in). Thank you!</p> http://mathoverflow.net/questions/47197/a-tricky-integral A tricky integral user17240 2010-11-24T04:55:07Z 2010-11-25T00:45:20Z <p>Let $\alpha>0$ and $\beta\in\mathbb{R}$. I am looking for an explicit formula for the integral</p> <p>$$\int_{-\infty}^{\infty} (1+x^2)^{-1/2}e^{-\alpha x^2}e^{-i \beta x}dx.$$</p> <p>I tried several changes of variables, and contour integration doesn't seem to work. </p> <p>Motivation comes from the following closely related kernel $$K(s,t)=e^{-\frac{(s-t)^2}{4}}\int_{-\infty}^{\infty}(1+x^2)^{-1/2}e^{-\frac{(s-t)^2}{4} x^2}e^{-i (s^2-t^2) x}dx,$$ which provides an example of a compact integral operator on $L^2(\mathbb{R})$ that is not Hilbert-Schmidt. I would like to check the details. </p> <p>Thank you!</p> http://mathoverflow.net/questions/8085/range-of-the-fourier-transform-on-l1 Range of the Fourier transform on L^1 user17240 2009-12-07T07:22:49Z 2010-02-01T06:36:23Z <p>It is well known that the Fourier transform $\mathcal{F}$ maps $L^1(\mathbb{R}^d)$ into, but not onto, $\overline{C_0^0}(\mathbb{R}^d)$, where the closure is taken in the $L^\infty$ norm. This is a consequence of the open mapping theorem, for instance. </p> <p>My question is: what's an explicit example of a function in $\overline{C_0^0}(\mathbb{R}^d)$ which is not in the image of $L^1(\mathbb{R}^d)$ under the Fourier transform? </p> <p>I would also like to know whether there is a useful characterization of $\mathcal{F}(L^1(\mathbb{R}^d))$. </p> <p>Remark: it is easy to see that the Banach space $\overline{C_0^0}(\mathbb{R}^d)$ consists of all continuous functions $f$ on $\mathbb{R}^d$ such that $f(\xi)\rightarrow 0$ as $|\xi|\rightarrow\infty$.</p> <p>Thank you!</p> http://mathoverflow.net/questions/7645/decomposition-of-holder-continuous-functions Decomposition of Hölder continuous functions user17240 2009-12-03T07:23:12Z 2009-12-04T17:16:02Z <p>Let $\alpha\in(0,1)$ and $\eta\in\Lambda_0^\alpha(\mathbb{R})$ be a compactly supported Hölder continuous function of order $\alpha$. I would like to show that, for any $n\in\mathbb{N}$, it is possible to decompose $$\eta=f+g$$ in such a way that $f\in C^n(\mathbb{R})$ and <code>$||f||_{C^n}=O(R^C)$</code>, and $g\in L^\infty(\mathbb{R})$ with <code>$\|g\|_{L^\infty}=O(R^{-1})$</code>. </p> <p>Here $C$ is a universal constant. On the other hand, the real parameter $R$ can be chosen as large as we want (at the expense of increasing $\|f\|_{C^n}$).</p> <p>Thank you!</p> http://mathoverflow.net/questions/2340/what-is-the-first-interesting-theorem-in-insert-subject-here/7760#7760 Answer by user17240 for What is the first interesting theorem in (insert subject here)? user17240 2009-12-04T07:06:45Z 2009-12-04T07:06:45Z <p>Harmonic Analysis: Plancherel's theorem</p> http://mathoverflow.net/questions/78067/l1-norm-of-the-fourier-transform-of-a-truncated-gaussian/79061#79061 Comment by user17240 user17240 2011-10-27T00:06:53Z 2011-10-27T00:06:53Z Well, &quot;decay to 0&quot; is equivalent to <i>some</i> kind of bound. My second sentence can be rephrased as &quot;how will this imply that the $L^1$ norm of $\widehat{G_k}$ tends to $0$ as $k\rightarrow\infty$?&quot;. $L^1$ and $L^\infty$ norms are in general not comparable... http://mathoverflow.net/questions/78067/l1-norm-of-the-fourier-transform-of-a-truncated-gaussian/79061#79061 Comment by user17240 user17240 2011-10-26T01:04:51Z 2011-10-26T01:04:51Z I agree that the $L^\infty$ norm of $\widehat{G_k}$ is small (it is bounded by $\|G_k\|_{L^1}$). But how will this imply bounds on the $L^1$ norm of $\widehat{G_k}$? I don't think dominated convergence will do it... http://mathoverflow.net/questions/78067/l1-norm-of-the-fourier-transform-of-a-truncated-gaussian Comment by user17240 user17240 2011-10-13T22:27:45Z 2011-10-13T22:27:45Z @Yemon: Yes, I should have mentioned that. Thanks! http://mathoverflow.net/questions/47197/a-tricky-integral Comment by user17240 user17240 2010-11-24T18:35:09Z 2010-11-24T18:35:09Z @Willie: edited, thanks! http://mathoverflow.net/questions/7645/decomposition-of-holder-continuous-functions/7671#7671 Comment by user17240 user17240 2009-12-04T08:31:29Z 2009-12-04T08:31:29Z By the way, I meant to get C independent of f but not necessarily n. I should have been more clear about that. http://mathoverflow.net/questions/7645/decomposition-of-holder-continuous-functions/7671#7671 Comment by user17240 user17240 2009-12-04T08:30:37Z 2009-12-04T08:30:37Z @Juli&#225;n: Thank you for your answer. I understand the bound on the remainder since $\eta\ast\phi_R\rightarrow\eta$ uniformly as $R\rightarrow\infty$, but how do you get the bound on '$\|f\|_{C^n}$'?