# Tag Info

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The answer to the first question is simply the following: take $f(z)=z^4$ in the spectral mapping theorem and use $\mathfrak{F}^4=1$.

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The spectrum is not as stated. The spectrum of the Hamiltonian for the non-relativistic Hydrogen atom has eigenvalues corresponding to the bound states of Hydrogen, and these are negative. The positive spectrum corresponds to unbound states, and is a continuous spectrum. In spherical coordinates, for a function which depends on the radius $r$ only, one has ...

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This is only a partial answer and would be far too long for a comment, but I must quote which is limited in comments. How are x and z related? Are the $J_k$ the Bessel functions of the first kind? For them it is correct that $$\frac{d}{dx}(x^kJ_k(x))=x^kJ_{k-1}(x)$$ see http://dlmf.nist.gov/10.6#ii or Abramowitz and Stegun 9.1.30. So I guess you just ...

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I am not sure if I get your question right, but I'll try to explain what I think is going on. What the author seems to try to grasp is the 'roughness' of the curve and this can be done using Fourier methods. Generally energy in higher Fourier coefficients corresponds to high frequency content ('rough' curves), while lower coefficients carry low frequency ...

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A straightforward way you can show that $\hat{f}$ is an entire function is write $$\hat{f}(z) = \int_{-R}^{R}f(t)e^{2\pi izt}\,dt = \int_{-R}^{R}f(t)\sum_{n=0}^{\infty}\frac{(2\pi izt)^{n}}{n!}\,dt = \sum_{n=0}^{\infty}\frac{(2\pi iz)^{n}}{n!}\int_{-R}^{R}t^{n}f(t)\,dt,$$ where $f(t)=0$ outside $[-R,R]$. ...

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I do not know if this solution is what you are looking for! First notice that $$(t/\lambda-x)^2 \geq 0 \implies -(t/\lambda-x)^2 \leq 0 \implies e^{-(t/\lambda-x)^2}\leq 1.$$ and $$|e^{-2\pi i w t}| =1.$$ So we have $$|F_{\lambda}(x, w)| \leq \int_{\mathbb{R}}e^{-t^2}dt < \infty.$$

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The function ${\displaystyle {x \over (x^2 + 4)^2}}$ is ${\displaystyle -{1 \over 2}}$ times the derivative of ${\displaystyle {1 \over x^2 + 4}}$, so its Fourier transform will be ${\displaystyle -{i \xi \over 2}}$ times the fourier transform of ${\displaystyle {1 \over x^2 + 4}}$. This one is standard and is given by ${\displaystyle {\pi \over 2} ... 3 We can use residue calculus here to find the integral value: $$\int_{-\infty}^{\infty}e^{-i \xi x}f(x)dx = \int_\gamma e^{-i \xi z} \cdot \frac{z}{(z^2 + 4)^2}dz$$ which have double poles at$z = \pm 2i \Rightarrow z_1 = (z-2i)^{-2}, z_2 = (z + 2i)^{-2}. We need to find the residues for e^{-i \xi z}f(z) = (z - 2i)^{-2}ze^{-i \xi z}(z + 2i)^{-2} = ... 2 One can rather easily compute the integral by residues. For example, for \xi\geq0 we can write \begin{align*} \mathfrak{F}(\xi)=-2\pi i\cdot \operatorname{res}_{z=-2i}\frac{ze^{-i\xi z}}{(z-2i)^2(z+2i)^2} =-2\pi i \cdot\left(\frac{ze^{-i\xi z}}{(z-2i)^2}\right)'_{z=-2i}=-\frac{\pi i}{4}\xi\, e^{-2\xi}. \end{align*} But since the answer should be an odd ... 1 Think of it as a spinning wheel, where a positive frequency means that the wheel is spinning counter clockwise (positive in a coordinate system) while a negative frequency means that the wheel is spinning the other way. the wheel can be described as e^{i \theta} and if \theta increases, we move counterclockwise but if \theta decreases or if we write ... 0 It depends on what you are after: what do you want to do with the transform? What you described is a reasonable thing to do; it amounts to extending f by zero outside of the interval [-\pi,\pi] and applying the usual formula on the line. A possible drawback: if f is not zero at \pm \pi , the steep drop-off will result in low accuracy near the ... 1 No, for x\neq \omega you have:\int_{-\pi}^{+\pi} e^{ixt} e^{-i \omega t} dt=2i\frac{e^{i\pi(x-\omega)}-e^{-i \pi(x-\omega)}}{2i(x-\omega)} =2i\frac{\sin(\pi (x-\omega))}{x-\omega} \neq 2 \pi \delta(x- \omega)$$1$$ \color{#00f}{\large\mbox{No}}.\quad\mbox{The last one is equal to}\quad \left\lbrace\begin{array}{lcl} 2\,{\sin\left(\left[x - \omega\right]\pi\right) \over x - \omega} & \mbox{if} & x \not= \omega \\[3mm] 2\pi & \mbox{if} & x = \omega \end{array}\right. $$2 Yes, the convolution of an integrable function f with compact support, and a Schwartz class function g belongs to the Schwartz space again. Since all derivatives of Schwartz class functions belong to the Schwartz space, in particular are bounded, the convolution$$(f\ast g)(x) = \int f(y)g(x-y)\,dy$$is smooth, since the dominated convergence theorem ... 0 I do not think that it is what Tao means. If you have this inequality for the unit sphere, you also have it for the sphere that is centered at 0 of radius 1+\varepsilon. Just use the result for the function obtained from f by the dilation that transforms this sphere into the unit sphere. Next you integrate in \varepsilon in the interval (-1/R, ... 4 No, this cannot be done. If \phi\in C_c^\infty(R), then its Fourier transform extends to be an entire analytic function (cf. Paley--Wiener theorem). Should it be constant for (real) x in some neighborhood of x_0, it would be identically constant by the uniqueness theorem for analytic functions, and hence zero, because the Fourier transform of a smooth ... 1 Assume g has support in [-M,M]. Then, for x\in[-N,N],$$ h(x)=\int f(x-y)g(y)dy=\int_{[-M,M]} f(x-y)g(y)dy=\int \tilde f(x-y)g(y)dy $$where \tilde f=1_{[-N-M,N+M]}f, 1_K denoting the indicator function of the set K. Now show that \tilde f\in L^1 and conclude that h\in C^k. 3 This is not a function in the usual sense but generalized function or distribution.$$\sum_{n=-\infty}^\infty (-1)^ne^{-iny}=\sum_{n=-\infty}^\infty e^{in(-y+(2m+1)\pi)}=\delta(-y+(2m+1)\pi),\ \forall m\in{\mathbf Z},$$for different branch cuts. So$$\hat S = \frac{i}{y}(e^{-iy}-1)\delta(-y+(2m+1)\pi),\ \forall m\in{\mathbf Z}$$for different branch cuts, ... 1 There are some constants missing somewhere. However,$$ \frac{1}{1+y^{2}} = \frac{1}{2}\int_{-\infty}^{\infty}e^{-|x|}e^{-ixy}\,dx. $$Use the scaling property of the Fourier transform for a > 0:$$ \begin{align} \frac{1}{a^{2}+y^{2}} & =\frac{1}{a^{2}}\frac{1}{1+(y/a)^{2}} \\ & = ... 0 Hint: Note that|y| = y$when$y\geq 0$, and$|y| = -y$otherwise. Similarly,$|x-y| = x-y$when$x-y \geq 0 \iff y \leq x$and$|x-y| = -(x-y)$otherwise. First, consider$x \geq 0$. Then, consider evaluating the integral$\int_{-\infty}^{\infty} e^{-|y|} e^{-|x-y|}\mathrm{d}y$via the three parts$\int_{-\infty}^{\infty} = \int_{-\infty}^{0} + ...

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Using the inverse Fourier transform, you may write (everything in distributional sense) $f(x)=\frac{1}{2\pi}\int_\mathbb{R}e^{ixt}\int_\mathbb{R} e^{-i\omega t} dt d\omega =\frac{1}{2\pi}\int_\mathbb{R}(\int_\mathbb{R}e^{ixt} e^{-i\omega t}dt) d\omega=\frac{1}{2\pi}\int_\mathbb{R}\delta(x-\omega)f(\omega)d\omega$ And therefore your Integral equals $2\pi ... 5 $$\int_{-\infty}^{+\infty} e^{ixt} e^{-i \omega t} dt=\int_{-\infty}^{+\infty} e^{ixt-i \omega t} dt=\int_{-\infty}^{+\infty} e^{i(x- \omega )t} dt=2 \pi \delta(x- \omega)$$ EDIT: It is known that: $$\delta(x-a)= \frac{1}{2 \pi}\int_{-\infty}^{\infty} e^{ip(x-a)} dp$$ 0 First we shall assume that the basis are orthonormal, i.e. $$\left\langle {{x_i},{x_j}} \right\rangle = \left\{ \begin{array}{l}1,i = j\\0,i \ne j\end{array} \right.$$ now, merely expanding the arbitrary vector in terms of these basis yields $$x = \sum\limits_{i = 1}^\infty {\left\langle {x,{x_i}} \right\rangle {x_i}}$$ Now, the "energy" of the vector ... 1 Consider$h_n = f \cdot \chi_{\Bbb{R}^n \setminus B_{1/n}(\gamma_0)}$and use dominated convergence. This implies that$h_n \to f$in$L^p$and each$h_n$is constant on the ball$B_{1/n}(\gamma_0)$with radius$1/n$around$\gamma_0$. EDIT: Ok, to get a solution to the precise formulation of your problem, take $$h(x) = (f(x) - f(\gamma_0)) \cdot ... 0 This is called index shift. Introduce m=n+N. Since n runs from -N to N, the new index m runs from 0 to 2N. So,$$\sum_{n=-N}^{n=N} e^{ 2 \pi i n t} = \sum_{m=0}^{m=2N} e^{ 2 \pi i (m-N) t} = \sum_{m=0}^{m=2N} e^{ 2 \pi i m t}e^{ -2 \pi i N t}$$Here we see the common factor e^{ -2 \pi i N t} and move it outside to get$$e^{ -2 \pi i N ... 0 Actually, you can change it a little,$\chi_{(-t,t)} (y)=\chi_{(\max(-y,y), +\infty)} (t)=\chi_{(|y|,+\infty)} (t),$so we change the integral to$J=\int_{a-b}^{a+b} \chi_{(|y|, +\infty)} (t) dt= \mu((|y|, +\infty)\cap (a-b, a+b ))$. When$|y| \leq a-b, J=2b.$When$|y| \geq a+b, J=0.$When$a-b<|y|<a+b, J=a+b-|y|$. Then we consider the sign of ... 1 I'm assuming$\chi_{(-t,t)}$is the characteristic function of$(-t,t)$, i.e. $$\chi_{(-t,t)}(y) = \begin{cases} 1, & y \in (-t,t), \\ 0, & y \notin (-t,t). \end{cases}$$ The way you explain the cases isn't clear, though, as you state that they depend on$t$? But$t$is the variable of integration, it ranges from ... 2 The integral with$e^{-ixy}$works for functions that are integrable on the real line (are in$L^1(\mathbb R$), but the domain of the Fourier transform extends way beyond that. How do we compute it when the integral diverges? Either by some approximating process, or by using general properties of the transform such as its interplay with derivatives. I'll use ... 1 Hint: You can use the sifting property of the delta function $$\int_{-\infty}^{\infty} \delta(x-x_0) f(x) dx = f(x_0) .$$ Added: Following my last comment we advance as $$\hat{f}(y)= \int_{-\infty}^{\infty} e^{-ixy} = \delta(y) \implies \frac{d\hat{f}(y)}{dy}= \int_{-\infty}^{\infty}(-ix) e^{-ixy} = \delta'(y) \longrightarrow (*).$$ So by ... 2 Fourier analysis requires a function to be defined on a (locally compact, abelian) group. The set of real numbers is a group under addition; this is the setting of Fourier transform. So is the circle$\mathbb R/(a\mathbb Z)$for some$a>0$; this is the setting of Fourier series. So is the cyclic group$\mathbb Z/(n\mathbb Z)$, which is the setting of the ... 1 First, your definition of$||f||_s$is wrong: the square root on the right-hand side is missing. Now, if we use the correct definition $$||f||_s=\sqrt{\sum_m(1+m^2)^s|\hat f(m)|^2},$$ then everything is pretty straightforward: $$\sup_x|f(x)|=\sup_x\left|\sum_m\hat f(m)e^{imx}\right|\le\sum_m|\hat f(m)|=\sum_m |\hat f(m)|(1+m^2)^{s/2}(1+m^2)^{-s/2}$$ (by ... 0 Expand$\exp(itx)$using Euler's and note the symmetry of the bounds. Explicitly we have$ \begin{align} I &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\exp\left(\frac{-x^{2} - 2itx}{2}\right)\,\mathrm{d}x \\ &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\exp\left(-\frac{x^{2}}{2}\right)\big(\cos(tx) - i\sin(tx)\big)\,\mathrm{d}x \\ &= ...

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Consider $L^{2}(\mathbb{R})$. The Fourier transform and its inverse implement the Spectral Theorem for the selfadjoint operator $Af = \frac{1}{i}\frac{d}{dx}f$ on the domain $\mathcal{D}(A)$ consisting of absolutely continuous $f \in L^{2}(\mathbb{R})$ for which $f'\in L^{2}(\mathbb{R})$. The spectral measure $E$ is $$E[a,b]f = ... 1 When a function f is defined on \mathbb R^n, its Fourier transform \hat f is also defined on \mathbb R^n. But it would be counterproductive to think of both f and \hat f as coinhabiting the same space. Forming expressions like f+\hat f would be nonsensical. The functions are really defined on different spaces. To emphasize this distinction, ... 0 You have e^{-x^2/2} e^{-itx}. The exponent is$$ \begin{align} -\frac{x^2}{2} - itx & = -\frac 1 2 (x^2 + 2itx) \\[10pt] & = -\frac{1}2 \left((x^2+2itx+ (it)^2) - (it)^2\right) \tag{completing the square} \\[10pt] & = -\frac 1 2 \left( (x+it)^2 - t^2\right) \end{align} $$So you're integrating$$ e^{-(1/2)(x+it)^2} \cdot \underbrace{{}\ \ ...

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One way your identity $1_{A}B^{s}f = B^{s}(1_{A}f)$ can fail: $1_{A}f$ may not be in the domain of $B^{s}$ even if $f$ is differentiable. For example, let $f$ be piecewise linear with $f(-2)=0$ and derivative $$f'=\chi_{[-2,-1]}-\chi_{[1,2]}.$$ Then $f \in \mathcal{D}(B^{p})$ for $0 \le \rho \le 1$. However, ...

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The Fourier transform turns differentiation into multiplication and multiplication into differentiation, which is definitely related to the ODE $$\frac{d}{dx}\left(e^{-ax^{2}}\right) = -ax\left(e^{-ax^{2}}\right),\\ \left(e^{-ax^{2}}\right)|_{x=0} = 1.$$ So the Fourier transform of $e^{-ax^{2}}$ must also be a Gaussian ...

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Hint: $$e^{-x^2/2} e^{-itx} e^{t^2/2} = e^{ \frac{(ix-t)^2}{2} }$$

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Well, using @martini 's idea I wrote a proof: Let $\varepsilon>0$ be given and fix $\alpha, \beta\in\mathbb N_0^n$. If $p\in\mathbb N$ then using the multinomial theorem: \begin{align} \displaystyle |x|^{2p} |\partial^\beta f(x)|^2&=(x_1^2+\ldots+x_n^2)^p |\partial^\beta f(x)|^2\\ &=\sum_{|\upsilon|=p} ... 1 Since it's bounded for any multiindex add 1 to \alpha_i to show that |x^{\alpha} \partial^\beta f(x)|\leq \frac{M_i}{|x_i|} for every i. Selecting the largest M_i to be M we get |x^{\alpha} \partial^\beta f(x)|\leq \min_i\frac{M_i}{|x_i|}\leq\frac{M}{\max_i|x_i|} for all x\neq0. But if |x|\to\infty then \max_i|x_i|\to\infty. 1 Hint. The point for (\Leftarrow) is, that you have boundedness for all \alpha. Let \alpha, \beta be given, we have |x^\alpha \partial^\beta f(x)| \le |x|^{|\alpha|}|\partial^\beta f(x)| = \frac{|x|^{|\alpha| + 1}|\partial^\beta f(x)|}{|x|} $$If we can bound the numerator, we are done. 0 Your plots help rule out any obvious bugs like incorrect signs, thank you for posting them. That is simply roundoff error. One way to check is to compute the same thing with higher precision. Also note that for n terms the error is about O(n\epsilon_\mathrm{mach}). In general, the error will be the sum of discretization error that goes as ... 1 Here's another explanation of the failure of inequality$$\tag{1} \|f\ast g\|_{L^p(\mathbb{R}^N)}\le C \|f\|_{L^p(\mathbb{R}^N)}\|g\|_{L^p(\mathbb{R}^N)},\qquad f, g\in \mathcal{S}for p>1. (The OP is about N=1 but there is no added difficulty in considering the general case). It is a routine application of the so-called scaling argument. Assume ... 2 The inclusion stated in the title follows from the fact that \mathcal{S}\subset L^1. But the inequality would involve the L^1 norm of f, not its L^p norm. (Namely, Young's inequality for convolution.) The point is, smoothness is irrelevant to L^p norm estimates of this sort. To see why you can't have \|f\|_{L^p}, consider f=\chi_{[0,M]} ... 3 If I remember things correctly, the interpretation would be that the temperature curve can be described as 4 functions: T1 = 30*sin(2*pi * t) T2 = 15*sin(pi * t) T3 = 60*sin(2*pi/3 * t) T4 = 15*sin(pi/4 * t) and the total temperature is T = T1 + T2 + T3 + T4 this, however, does not mean that the temperature reaches 60 degrees, it could be canceled out by ... 1 I am guessing that you would like to know \lim_{s \to 0+} f(s) = \lim_{s \to 0+} \int_0^\infty a(t) e^{-st}dt. Let (s_n) be any sequence of positive numbers converging to 0 and set f_n(t) = a(t)e^{-s_nt} \to a(t) as n \to \infty. Moreover, |f_n(t)| \leq |a(t)| so if a(t) is Lebesgue integrable, then the limit equals \int_0^\infty a(t) dt ... 0 If 1<p<2, it holds. Note that f \in C(R) implies f \in L^{\infty}. Since f is in L^p \cap L^\infty, we get f \in L^2. Thus we get the desired result. For f \in L^p with  2<p<\infty, Fourier transform of f is not defined unless f \in L^q for some 1 \leq q \leq 2. 1 But your function m is L^\infty(\mathbb R), note that \begin{align*} \lim_{k \to 0} \frac{e^{-ik} - 1}k &= \lim_{k\to 0} \left(\frac{\cos k - 1}{k}-i\frac{\sin k}k\right)\\ &= 0 -i = -i \end{align*} So there is some neighbourhood of 0, where m is bounded, and m is (trivially) bounded on the complement of every neighbourhood of 0, ... 0 This is not an answer, but it's too long for a comment. Classical Fourier transform can be seen as an isomorphism between L^2 functions on S^1 or \mathbb{R}^n and l^2 sequences given by a particular choice of basis functions. These functions are actually group characters. Now group characters encode the irreducible representations of these groups. ... 0 There are manipulations where it is desired that the inverse corresponds to a complex-valued signal, such as the computation of the analytic signal. If, however, the resulting signal after manipulation is desired to be real-valued then the conjugate symmetry condition of the Fourier transform of a real-valued signal must remain satisfied:X_k=X^*_{N-k} ...

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