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As taken from my upcoming textbook: There is yet another solution to the Basel problem as proposed by Ritelli (2013). His approach is similar to the one by Apostol (1983), where he arrives at $$\sum_{n\geq1}\frac{1}{n^2}=\frac{\pi^2}{6}\tag1$$ by evaluating the double integral $$\int_0^1\int_0^1\dfrac{\mathrm{d}x\,\mathrm{d}y}{1-xy}.\tag2$$ Ritelli ...


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First note that Theorem 4.1 holds for $\psi\in L^2$ by a simple approximation argument. Now assume $f\in L^2$, $||f||_2=1$. Then, since $|x|\le L_1/2$ for $x\in I_1$, $$\left(\frac{L_1}{2}\right)^2\ge \left(\frac{L_1}{2}\right)^2\int_{I_1}|f|^2\ge\int_{I_1}x^2|f(x)|^2\ge\frac12\int_{\Bbb R}x^2|f(x)|^2.$$Similarly ...


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Recall that the sine function can be represented as the infinite product $$\sin x=x\prod_{n=1}^{\infty}\left(1-\frac{x^2}{n^2\pi^2}\right) \tag 1$$ Taking the logarithmic derivative of $(1)$, we obtian $$\begin{align} \cot x&=\frac1x+2x\sum_{n=1}^{\infty}\frac{1}{x^2-n^2\pi^2}\\\\ &=\sum_{n=-\infty}^{\infty}\frac{x}{x^2-n^2\pi^2}\\\\ ...


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Here's a partial answer, too long for a comment. I'm not sure about necessary and sufficient conditions for continuity or differentiability, but it's not too hard to find sufficient conditions. Certainly, if $\sum |F(n)| < \infty$ then the series $\displaystyle \sum F(n) e^{iun}$ converges absolutely and uniformly, by the Weierstrass M-test, so $f$ is ...


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As pointed out in my earlier comment, we need to assume $s\in (0,1)$. Out of habit, I'm going using the convention $\mathbb{T}\cong[0,1)$. You can modified everything to your own convention. The expression $$[f]_{W^{2,s}(\mathbb{T})}=\left(\iint_{\mathbb{T}\times\mathbb{T}}\dfrac{\left|f(x)-f(y)\right|^{2}}{\left|x-y\right|^{1+2s}}dxdy\right)^{1/2}$$ ...


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I assume $-ikx$ on the right should be $ikx$. You can use what you have instead, but it is not standard. \begin{align} \int_{-R}^{R}f(x)e^{-ixl}dx & = A\int_{-\infty}^{\infty}\hat{f}(k)\int_{-R}^{R}e^{ix(k-l)}dxdk \\ & = 2A\int_{-\infty}^{\infty}\hat{f}(k)\frac{\sin(R(k-l))}{k-l}dk \end{align} The sinc function $$ ...


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I'm aware of one necessary and sufficient condition. It comes out of the Spectral Theorem for selfadjoint operators. The domain of $L=\frac{1}{i}\frac{d}{dx}$ on $L^{2}[-\pi,\pi]$ consists of all periodic absolutely continuous functions on $[-\pi,\pi]$ with $f' \in L^{2}[-\pi,\pi]$. $L$ is selfadjoint on this domain. From the Spectral Theorem: $$ ...


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We note here that all of the ensuing analysis uses notation that is interpreted in the sense of Distributions or Generalized Functions. With that note, let's begin by breaking the problem down into components, each of which is hopefully elementary. STEP 1: First, we know that the Fourier Transform of the Dirac Delta $\delta$ is $$\begin{align} ...


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In the link I shared in the comments section you can see that the following holds: If $f$ is absolutely continuous, $$ \left| \widehat f(n) \right| \le {K \over |n|}. $$ If $f$ is a $BV$ function, $$ \left|\widehat f(n)\right|\le {\|f\|_{BV}\over 2\pi|n|}. $$ If $f \in C^p$, $$ \left|\widehat{f}(n)\right|\le {\| f^{(p)}\|_1\over |n|^p}. $$ If $f\in C^p$ ...


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$\underline{Hint}$: To show that $\{ e_n\mid n\in\mathbb{Z}\}$ is an orthonormal basis (ONB) for $L^2(\mathbb{T})$, we must show that $\operatorname{span}\{e_n\mid n\in\mathbb{Z}\}$ is dense in $^2(\mathbb{T})$, which is equivalent to stating that $\exists g\in \operatorname{span}\{ e_n\mid n\in\mathbb{Z}\}$ such that $$ \|f-g\|_{L^2(\mathbb{T})}=\left( ...


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Yes you can simply abandon the integrals because the Fourier transform is unique: if $\hat{f}=\hat{g}$ then $f=g$ (almost everywhere). To see this more clearly write your last equation as $$\frac{1}{\sqrt{2\pi}}\int[\rho(k) + k^2f(k)]e^{ikx} = 0$$ Now the Fourier transform of $0$ is simply $0$ so by uniqueness $\rho(k) + k^2f(k) = 0$. A slightly simpler ...


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Of course there are many functions with this property. An interesting way to get some: Say $\phi\in C^2_c(\Bbb R)$ (this is stronger than necessary) and the support of $\phi$ is contained in $[-\pi,\pi]$. Let $$f(x)=\frac1{\sqrt{2\pi}}\int e^{-itx}\phi(t)\,dt\quad(*).$$Then $$\int f(x)\,dx=\sum_{n\in\Bbb Z}f(n).$$First the inversion formula for the Fourier ...


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For illustration let us consider the case where the ambient space is $\mathbb{R}^3$ and $M$ is the $(x,y)$-plane (which is a surface of 0 Gaussian curvature). Let $g$ be a function of compact support in $\mathbb{R}^3$ whose support include the origin. Let $g_{\nu}(x,y,z) = g(x, y, z/\nu)$. Its inverse Fourier transform scales like $$ ...


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Multiply both sides by $\exp(-i\mathbf{p}' \cdot \mathbf{x})$ and integrate over $\mathbb{R}^3$. Swap order of integration. The dirac delta result is can be found many places. See rule 303 here. Also, you shouldn't use the same name for a function and its Fourier transform. It's confusing; use $\hat{\phi}$ or something. Also also, you mean ...


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When you solve the equation for $X$ for $\lambda =\alpha^2>0$, this gives you $$ X=A\sin(\sqrt{\alpha}x)+B\cos(\sqrt{\alpha}x). $$ Applying the initial conditions gives $$ 0=A\sin(\sqrt{\alpha}L) $$ which implies that either $A=0$ or $\sin(\sqrt{\alpha}L)=0$. Note that $A=0$ gives a trivial result so we must have $\sin(\sqrt{\alpha}L)=0,$ ie, ...


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From the conditions on $\psi$ we have $$ |\psi(\xi)|=\Bigl|\int_\mathbb{R}\psi(x)(e^{-ix\xi}-1)\,dx\Bigr|\le\Bigl(\int_\mathbb{R}|\psi(x)|(1+|x|)^\alpha\,dx\Bigr)\sup_{x}\frac{|e^{-ix\xi}-1|}{(1+|x|)^\alpha}. $$ The first estimate follows from the inequality $$ |e^{-ix\xi}-1|\le\min(2,|x|\,|\xi|). $$ As for the second estimate, are you sure that there is no ...



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