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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 ...


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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{\...


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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,\;...


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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}...


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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 $x=0 \rightarrow 1$ $$\int_{0}^{1}1.dx=\int_{0}^{1} \sum_{n=1}^\infty\frac{4}{(2n-1)\pi}\sin (2n-1)\pi x dx$$ $$1=\sum_{n=1}^\infty\frac{8}{(2n-1)^2\pi^2}$$ $$\sum_{n=1}^\infty\frac{1}{(2n-1)^2}=\frac{\pi^...


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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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