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## Hot answers tagged fourier-analysis

3

The theorem you stated is definitely related to the question. Let $s$ be a step function so that $\| p-s\|_{L^1} <\epsilon$, then $$\bigg|\int_0^{2\pi} p(x) q(nx) - \int_0^{2\pi}s(x) q(nx)\bigg|\le C\int_0^{2\pi}|p(x)-s(x)| <C\epsilon,$$ where $C$ is the bounded of $q$, that is $|q(y)|\le C$ for all $y\in [0,2\pi]$. Now if you have shown the theorem ...

2

Let $X$ be an independent Laplace random variable with $X\sim L(0,1) = \frac12 \exp{(-|x|)}$, then its characteristic function : $$\varphi_X(t)=\mathbb{E}[e^{itX}]=\frac{1}{1+t^2} \newcommand{\var}[1]{\mathrm{var}\left[#1\right]}$$ By symmetry $\mathbb{E}[X]=0$ we write (generally) : $$\varphi_X(t)=\mathbb{E}[e^{itX}]=\mathbb{E}[1+itX-t^2X^2+\cdots\,]=1-\... 2 Yes. In fact one derivative is enough. The Fourier transform of f' is itF(t) (or something like that, depending on how your definition of the Fourier transform is normalized). Since f'\in L^1 this shows that tF(t) is bounded. So$$\int_{|t|>1}|F(t)|^2\,dt\le c\int_{|t|>1}\frac{dt}{t^2}<\infty.$$We certainly have \int_{|t|\le1}|F(t)|^2\,dt&... 1 The argument establishing the inequality for \varphi\in\mathcal S would work just as well if it had been assumed only that \varphi\in L^1, but the author wanted to focus on \mathcal S for a reason. The space \mathcal S is actually closed under the Fourier transform. So \widehat\varphi, the Fourier transform of \varphi\in\mathcal S, is ... 1 Let \tau_t g denote, for a fixed t, the function x \mapsto g(x-t). We have:$$F_g f (\omega, t) = \int_{\Bbb R} f(x) \overline{\tau_t g(x)} e^{-i\omega x } dx = \mathcal F({f\overline{\tau_t g}})(\omega)$$We know that \mathcal F: L^2 \to L^2 is an isometry up to a 2\pi factor, so (wrt \omega) \|F_g f\|_{L^2(\Bbb R)}^2 = 2 \pi \|f \overline{\... 1 If s_n converges uniformly on a set of full measure then s_n converges uniformly, period. (If |s_n-s_m|<\epsilon on E then |s_n-s_m|\le\epsilon everywhere.) So at a minimum you need f=g almost everywhere, with g continuous. Of course that's far from sufficient. But the question is really about uniformly convergent Fourier series, without ... 1 This is true assuming just that f is bounded and has one-sided limits at the origin. The simplest proof is by a cheap trick: A change of variables shows that$$f*K_\epsilon(0)=\int f(\epsilon t)K_1(-t)\,dt.$$Now apply dominated convergence... 1 If you break everything into intervals, then you do have a proper orthogonal analysis. For example,$$ f(t)=\sum_{n=-\infty}^{\infty} \frac{1}{\sqrt{2\pi}}\int_{(n-1/2)\delta}^{(n+1/2)\delta}\hat{f}(\omega)e^{i\omega t}d\omega = \sum_{n=-\infty}^{\infty}f_n(t). $$In this case,$$ (f_n,f_m)=\int_{-\infty}^{\infty}f_n(t)\overline{f_m(t)}dt=0,\;...

1

Your density lemma says it suffices to prove the result for a function of the form $q(x)=1_{[a,b]}(x)$ $[a,b] \subset [0,2\pi]$. Now \begin{align*} \int _{0}^{2\pi} q(x) p(nx) d x &= \int _{0}^{2n\pi} \frac{1}{n}q(x/n)p(x) d x \\ &= \sum _{k=0}^{n-1} \int _{2(k)\pi}^{2(k+1)\pi}\frac{1}{n} q(x/n)p(x) d x \\ &= \sum _{k=0}^{n-1} \int _{0}^{...

1

If $f\in L^2(\mathbb{R})$, then $f \in L^1[-R,R]\cap L^2[-R,R]$ for any $0 < R < \infty$. Consequently $$\hat{f_{R}}(s)=\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}f(t)e^{-ist}dt \in L^2(\mathbb{R})$$ is a continuous function of $s$, and, by Parseval's identity, $$\left\|\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}f(t)e^{-ist}dt\right\|_{L^2}^2=\int_{-R}^{R}... 1 Suppose f\in L^1(\mathbb{R}). For 0 < R < \infty and f \in L^1(\mathbb{R}), the function$$ \hat{f_R}(\omega)=\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}f(t)e^{-ist}dt $$is infinitely differentiable with$$ \hat{f_{R}}^{(n)}(s)=\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}f(t)(-it)^{n}e^{-ist}dt. $$Furthermore, one has the uniform ... 1 The idea being discussed doesn't need a rigorous definition of wavelet, or even to use wavelets at all; the notion of a wavelet is, I think, mainly a clever way of generating a convenient basis in a systematic fashion starting from a basic shape. (the particular basis you listed could probably stand to be scaled so as to be orthonormal basis rather than just ... 1 by using Fourier series of f(x)=1, x\in[0,1]$$1=\sum_{n=1}^\infty\frac{4}{(2n-1)\pi}\sin (2n-1)\pi x$$integrate both sides when integration limits are x=0 \rightarrow 1$$\int_{0}^{1}1.dx=\int_{0}^{1} \sum_{n=1}^\infty\frac{4}{(2n-1)\pi}\sin (2n-1)\pi x dx1=\sum_{n=1}^\infty\frac{8}{(2n-1)^2\pi^2}\sum_{n=1}^\infty\frac{1}{(2n-1)^2}=\frac{\pi^...

1

Following proof rely on this integral identity : $$\int_{a}^{1}\frac{\arccos x}{\sqrt{x^2-a^2}}\mathrm{d}x=-\frac{\pi}{2}\ln a\qquad ;\,a\in(0,1]$$ We will prove it later on. Now, let's make a power series : \zeta(2)=\sum_{n=1}^{\infty}\frac{1}{n^2}=\int_0^1\frac{1}{x}\sum_{n=1}^{\infty}\frac{x^n}{n}\,\mathrm{d}x=-\int_0^1\frac{\ln(1-x)}{x}\,\mathrm{d}x=...

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