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1

$f(w)e^{-π \delta w^2}\xrightarrow[\delta\to0]{}f(w)$ almost everywhere in $w$ and we have the dominating function $f$. By Lebesgue dominated convergence theorem, the result follows.

2

Since $f\in\mathcal{S}(\mathbb{R})$ we have that both $x\cdot f(x)$ and $|x|^{3/2}\cdot f(x)$ belong to $L^2(\mathbb{R})$, so: $$\left|\int_{\mathbb{R}} f(x)\left(1-e^{-\pi\delta x^2}\right)\,dx\right| \leq \|x\cdot f(x)\|_2 \cdot \sqrt{2\pi(2-\sqrt{2})\sqrt{\delta}}=\color{red}{C\cdot \delta^{1/4}} \tag{1}$$ or: $$\left|\int_{\mathbb{R}} ... 4 Yesterday I said that this was Tauber's Theorem, the original tauberian theorem. It's not; Tauber's Theorem is the analogous result for Abel summability. This is Someone's Theorem. I'm going to give a proof of ST organized in what I feel is the "right" manner, deriving convergence from a sort of "maximal inequality"; then as a bonus we will see that one can ... 1 Note that the \mathcal{C}^\infty_c functions are dense in \mathcal{L}^1. This theorem is true for a smooth function; you can see this by integrating by parts. Now invoke the density to see it holds for any f\in\mathcal{L}^1. 1 Your function f satisfies f(-\pi)=f(\pi) and has a natural 2\pi periodic extension f_p to \mathbb{R}. The Fourier series for f on [-\pi,\pi] is periodic with period 2\pi. The Fourier series converges pointwise and uniformly on every closed interval [a,b]\subseteq \mathbb{R} for which f_p is continuous on a slightly larger interval ... 2 To derive a recurrence relation for such coefficients is the same as finding the structure of the continued fraction of e^2 (explanation at the bottom of this answer): that can be done by exploiting the Gauss continued fraction for the hyperbolic cotangent, but it is easier to compute such coefficients directly. If we start from$$ ...

3

If it's continuous, it's bounded (image of the compact set $[-\pi,\pi]$ is a compact hence bounded set), so $0\le |f(x)|^2\le M$. Then by Lebesgue's DCT or Fatou's lemma or whatever you want, $$\int_{-\pi}^\pi |f(x)|^2\,dx\le \int_{-\pi}^\pi M\,dx = 2\pi M < \infty$$ which is exactly what it means to be in $L^2([-\pi,\pi])$

1

Notice that for all $t \in \Bbb R$, $$\left| \left ( \frac{(-1)^{n-1}}{4n^2-1} \right )\cos(2nt) \right| \le \left| \frac{(-1)^{n-1}}{4n^2-1} \right| \underbrace {|\cos (2nt)|} _{\le 1} \le \left| \frac{(-1)^{n-1}}{4n^2-1} \right| = \frac 1 {4n^2-1}$$ and $\sum \frac 1 {4n^2-1}$ is convergent (use the limit comparison test and compare it with \sum \frac 1 ... 1 As noted in the comments there are some mistakes in the question, and I don't understand what the Fourier series have to do with the problem. Anyway, we have: $$-\pi<x<0 \Rightarrow \sin x<0 \Rightarrow -i\sin x =i |\sin x| \Rightarrow \mbox{arg}(i |\sin x|)=\frac{\pi}{2}$$ x=0 \Rightarrow \sin x=0 \Rightarrow -i\sin x =0 \Rightarrow ... 1 This is true \text{Arg}(-i\sin(x))=\frac{\pi}{2}\cdot \text{sgn}(x), x\in(-\pi,\pi) Recall that the principal argument \text{Arg}(z) lies in the domain -\pi<\text{Arg}(z)\le \pi. Setting y=\sin x, then y \in [-1,1]. Note that z=-iy is a pure imaginary complex number, and the sign of the imaginary part is \pm (liying on the y-axis) ... 1 You have noted that \begin{align} a_n & = \frac{2n+1}{2}\int_{-1}^{1}e^{-x}\frac{1}{2^n n!}\frac{d^{n}}{dx^{n}}(x^2-1)^{n}dx \\ & = \frac{2n+1}{2^{n+1}n!}\int_{-1}^{1}e^{-x}\frac{d^{n}}{dx^{n}}(x^2-1)^ndx \end{align} And, \begin{align} \int fg^{(n)}dx&=fg^{(n-1)}-f^{(1)}f^{(n-2)}+f^{(2)}{g^{(n-3)}}-\cdots+(-1)^{n}\int f^{(n)}gdx ... 1 To answer your question, yes. For a proof, first require a few prerequisites. From F. G. Friedlander and M. Joshi's Introduction to the Theory of Distributions, we have Theorem 1.2.1. Let f \in C_c^k(\mathbb R), let \rho \in C_c^\infty(\mathbb R) be such that \rho \geq 0 and \int \rho = 1, let \rho_\varepsilon = \varepsilon^{-1} \rho( x / ... 1 Fourier's Treatise on heat conduction was first submitted in 1807. In that work he introduced the method of separation of variables, and various expansions in orthogonal functions. Fourier died in 1830. Gauss looked at Potential Theory at roughly the same time. By 1830, Gauss was heavily involved in the study of Electromagnetism. He gave a minimization ... 0 I read the proof with your notation in "Introduction to Distribution Thoery" by F.G. Friedlander. I try to explain. We assume initially u=\phi \in \mathcal{S}(\mathbb{R}^n) \subset L^1(\mathbb{R}^n), then \displaystyle \widehat{\phi}(\xi):=\int e^{i x \cdot \xi} \phi(x) dx and by inversion formula is known that \displaystyle \phi = \mathcal{F}^{-1} ... 1 Let's call \nu=\frac{2\pi}{T} so we have the coefficient of the Fourier series of the triangual function is a_k = \color{blue}{\frac{1}{T} \int_{0}^{T/2} 2 \frac{t}{T} \mathrm e^{-i \nu kt} \mathrm dt} + \color{red}{\frac{1}{T} \int_{T/2}^{T} 2 \left(1-\frac{t}{T} \right)\mathrm e^{-i \nu kt} \mathrm dt} $$Observe that integrating by parts$$ \int ... 1 You want the Fourier coefficients of $$f(t) = \left\{\begin{array}{cc} \frac{2t}{T}, & 0 \le t \le \frac{T}{2} \\ \frac{2(T-t)}{T}, & \frac{T}{2} < t \le T \end{array} \right.$$ Fork \ne 0, these can be written as \begin{align} c_k & = \frac{1}{T}\int_{0}^{T}f(t)e^{-2\pi ikt/T}dt \\ ... 1 Using that\mathcal{F}\circ\mathcal{F}(f)(x)=f(-x)$and in the comments you mentioned that you know that$\hat\delta =1$, we get $$\hat 1 = \hat{\hat \delta} =\delta(-\cdot) =\delta$$ with some normalizing coefficient depending on your Fourier transform. 2 By definition we have$\displaystyle \langle \varphi, \widehat{\delta} \rangle= \langle \widehat{\varphi}, \delta \rangle = \widehat{\varphi}(0)=(2\pi)^{-n/2} \int_{\mathbb{R}^n} e^{-i x \cdot 0} \varphi(x)dx = (2\pi)^{-n/2} \langle \varphi , 1 \rangle$by arbitrariness of$\varphi$get the identity. Similary$\displaystyle \langle \varphi, \widehat{1} ...

0

The distributional derivative of the Dirac delta distribution is the distribution $\delta′$ defined on compactly supported smooth test functions $\varphi$ by $$\delta'[\varphi] = -\delta[\varphi']=-\varphi'(0).$$ The first equality here is a kind of integration by parts, for if $\delta$ were a true function then $$\int_{-\infty}^\infty ... 2 Home-made proof of that H(f) is bounded: Suppose f is in the Schwartz space. Let \epsilon \in (0,1), x \in \mathbb {R}. Then$$\tag 1\int_\epsilon ^\infty \frac{f(x-t)-f(x+t)}{t}\,dt = \int_\epsilon^1 \frac{f(x-t)-f(x+t)}{t}\,dt + \int_1^\infty \frac{f(x-t)-f(x+t)}{t}\,dt.$$In the first integral we can use the MVT to see the integrand is bounded by ... 2 Presumably |H(\xi)| \leq |f(\xi)| was a typo for |Hf(\xi)| \leq |f(\xi)|, but I don't see why you think that's so. In fact since H is a Fourier multiplier operator it's a theorem that it's bounded on L^p if and only if it's bounded on L^{p'}. So on the plus side, having shown that it's bounded on L^p for p\in(1,2] it follows that it's ... 0 Well first, what you say about C^\infty([-\pi,\pi]) is not true. For example, f\in C^\infty([-\pi,\pi]) if f(t)=t, but the coefficients of that function are not O(1/k^2). Let's talk about 2\pi-periodic functions - that's what you meant, or what you should have meant. I'd just say the coefficients were rapidly decreasing, and specify what that ... 0 The periodic extension of f(x) is$$ f_p(x)=\sum_{n\in\Bbb Z} f(x-2n\pi)=\arcsin(\cos x) The function is even and then a_0=0 and b_n=0. \begin{align} a_n&=\frac{1}{\pi}\int_{-\pi}^\pi f(x)\cos(nx)\,\mathrm dx\\ &=\frac{1}{\pi}\int_{-\pi}^0 \left(\frac{\pi}{2}+x\right)\cos(nx)\,\mathrm dx+\frac{1}{\pi}\int_{0}^\pi ... 0 I am not completely sure what your question is, so I'm sorry if I misunderstand you. You have to integrate over the entire period, i.e on the interval [-\pi, \pi] (the function is 2\pi periodic and therefore the interval should have length 2\pi). Hence you have that \begin{align} a_n &= \frac1\pi\int_{-\pi}^\pi f(x)\cos (n x)\,\mathrm{d}x\\ b_n ... 0 Well, someone asked for so there is a compact solution (I can provide more details of course) : First, since f is one to one we deduce that f'(z) \neq 0 for all z \in \bar D. Then we set D_r = \{ z, \; |z| \leqslant r\}. Then if A_r denotes the area of f(D_r) we have :A_r= \iint_{f(D_r)} \mathrm{d}x \, \mathrm{d}y =\iint_{D_r} |f'(x+iy)| ...

0

Integrating we find \begin{align} a_n&= \frac{1}{\pi}\int_{0}^{\pi}\sin x \cos (nx) \,\mathrm dx\\ &=\frac{1}{2\pi}\int_{0}^{\pi}\left[ \sin ((n+1)x) - \sin ((n-1)x)\right]\,\mathrm dx\\ &=\frac{1}{2\pi}\left[\frac{\cos((n+1)x)}{n+1}-\frac{\cos((n-1)x)}{n-1} \right]_0^\pi\\ &=\frac{1}{2\pi}\left[\frac{-2 (n \sin x \sin(n x)+\cos x \cos(n ...

0

Probably it is easier to write $f(x)$ as $$f(x)=\frac{1}{2}\left(\sin(x)+\left|\sin(x)\right|\right)$$ then recall that: $$\left|\sin(x)\right| = \frac{2}{\pi}-\frac{4}{\pi}\sum_{n\geq 1}\frac{\cos(2nx)}{4n^2-1}.$$

0

If we define $$\widehat{f}(s)=\mathcal{F}(f)(s) = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}f(x)e^{isx}\,dx,\qquad \mathcal{F}^{-1}(f)(s) = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}f(x)e^{-isx}\,dx$$ we have: $$\mathcal{F}^{-1}\left(\frac{1}{1+x^2}\right) = \sqrt{\frac{\pi}{2}}\,e^{-|s|} \tag{1}$$ $$... 2$$ q\cdot(sgn(1-k) + sgn(1+k)) =\begin{cases} 0 \ \ k > 1 \\ 2q \ \ -1<k<1 \\ 0 \ \ k < - 1 \end{cases} $$Which yields$$ f(t) = \frac{1}{2\pi} \int_{-1}^{1} \frac{2e^{itw}}{1+w^2} \frac{\pi}{4} dw $$1 I don't fully understand the question: if you agree that \left\{\frac{1}{\sqrt{2\pi}}, \frac{1}{\sqrt{\pi}}\cos kx, \frac{1}{\sqrt{\pi}}\sin kx\right\} are a basis for the function space of 2\pi-periodic functions, it follows immediately that any function in this space can be expressed as a linear combination of these basis functions (and the a_i and ... 1 Suppose the support of f is contained in [-1,1], and \hat f (x) = 0 for |x|>N \in \mathbb N. Applying a standard Fourier series argument on [-\pi,\pi] then shows$$f(x) = \sum_{-N}^{N}\hat f (n) e^{inx}, x \in [-\pi,\pi].$$Thus f is a trigonometric polynomial that vanishes on [1,\pi]. But a trigonometric polynomial on [-\pi,\pi] that ... 0 I'm not sure what you mean by moderate decrease. I suspect you use an argument such as the following, assuming f(t)t is absolutely integrable, for example:$$ \hat{f}(\omega+h)-\hat{f}(\omega)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)\{e^{-i\omega t}-e^{-i(\omega+h)t}\}dt \\ ...

1

By the Schwartz's Paley-Wiener theorem the Fourier transform of a compact-supported function is an entire function, and non-zero entire functions cannot be compact supported by Liouville's theorem.

1

If $f$ is compactly supported, then $\hat{f}$ is holomorphic. You can see that because the Fourier transform integral extends into the complex plane and is differentiation in the complex variable $s$: $$\hat{f}(s) = \frac{1}{\sqrt{2\pi}}\int_{-R}^{R}e^{-ist}f(t)dt \\ \hat{f}'(s) = \frac{1}{\sqrt{2\pi}}\int_{-R}^{R}(-it)e^{-ist}f(t)dt.$$ ...

0

Here the author probably meant that $T\mu$ is the restriction map $$T\mu(\xi_1,\cdots, \xi_{n})=\mu(\xi_{1},\cdots, \xi_{n-1},0)$$ Now writing out explicitly we have $$\widehat{T\mu}(\xi')=\int\mu(\xi,0)e^{-i \xi\cdot \xi'}d\xi$$ But recall that formally we have (for 1D functions) $$f(0)=\frac{1}{2\pi}\int ... 0 for a energy signal i.e physically realizable, exist a uncertainty principle between duration of signal and her bandwidth. If Bq is her quadratic band and Dq is her quadratic duration then BqDq >= (8*pi)^-1 . Therefore a signal can't have both duration and band zero. But if duration tend to zero then band tend to infinity and vice versa. To notice that the ... 0$$F(x) = \frac{a-b}{2}\ |x| + \frac{a+b}{2} x$$for -\pi<x<\pi. The first part is an even function that corresponds to the a_n's. Without looking at tables, the formula for a_n, n> 0 is$$\begin{align*} a_n &= \frac2{2\pi}\int_{-\pi}^\pi\frac{a-b}{2}\ |x|\cos nx\ dx\\ &= \frac{a-b}{2\pi}\int_{-\pi}^\pi|x|\cos nx\ dx\\ &= ...

1

The reason why the result equals zero is that Wolfram Alpha uses the unilateral $\mathcal{Z}$-transform defined by $$\mathcal{Z}\{f(n)\}=\sum_{n=0}^{\infty}f(n)z^{-n}\tag{1}$$ as detailed here. Since your sequence equals zero for $n\ge 0$, its unilateral $Z$-transform is obviously zero. The bilateral $\mathcal{Z}$-transform of the given sequence exists ...

0

Actually it was (as you can check in the lecture notes you've referenced) $$\frac 1 {2\pi} \int_{\mathbb R} \int_{\mathbb R} e^{-ixt} g(x)\ dt\ dx.$$ If you integrate by $t$ first then it can be rewritten as (I'll assume below that integration interval is always $\mathbb R$) $$\int dx\ g(x) \left(\frac 1 {\sqrt{2\pi}} \int \frac{1}{\sqrt{2\pi}} ... 0 Take g outside the inner integral on the left and you have the fourier transform of 1. This is the dirac delta, a distribution that integrates to 1 and is infinity at x. In this sense only the value of g where the dirac delta is non-zero becomes relevant. The point is then moving with x in your case. 2 The Parseval identity (a.k.a. Plancherel's Theorem) for the Fourier transform is$$ \int_{-\infty}^{\infty}|f(t)|^2dt = \int_{-\infty}^{\infty}|\hat{f}(t)|^2 dt,\;\;\; f \in L^2(\mathbb{R}). $$I assume you're familiar with this theorem. If you have a function f\in L^2(\mathbb{R}) that is supported in [0,\infty) only, then the Fourier transform ... 1 If I am reading your problem correctly, f \in L^2, and \mathcal{F}f is band limited, which puts \mathcal{F}f \in L^1\cap L^2, and gives$$ \mathcal{F}^{-1}\mathcal{F}f = \frac{1}{\sqrt{2\pi}}\int_{-\lambda}^{\lambda}\mathcal{F}f(s)e^{isx}ds. $$And, it is also true that$$ f = ...

0

We define $$\hat{f}(k)=\mathcal F\{f(x)\}=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty f(x) e^{-i k x}\,\mathrm dx$$ and $$f(x)=\frac{1}{x^2+a^2}$$ It's easy to find that $$\mathcal F\{\mathrm e^{-a|x|}\}=\sqrt{\frac{2}{\pi}}\frac{a}{k^2+a^2}$$ and using the duality property $\mathcal F\{\hat f(x)\}=f(-k)$, we have $$\mathcal ... 3 If the partial sums S_N(x) converge to S(x), then the Cesaro sums$$ \sigma_N(x)=\frac1N\sum_{k=1}^NS_N(x) $$also converge to S(x). But the Cesaro sums of a continuous function f converge uniformly to f, so that S(x)=f(x) for all x. 1 The problem whether the Fourier series of any continuous function converges almost everywhere was posed by Nikolai Lusin in the 1920s, resolved positively in 1966 by Lennart Carleson in L^2 and generalized by Richard Hunt to L^p for any p > 1. This result is known as the Carleson–Hunt theorem. This article may help. Andrey Kolmogorov, as a student ... 0 I'm not aware of any relation to standard trigonometric series. There are trigonometric series expansions that do not have evenly spaced eigenvalues, and those definitely are not going to be related. Fourier considered such an example in his original Treatise on Heat Conduction from 1807:$$ X''(x)+\lambda^2 X(x) = 0,\;\;\;\; 0 \le x \le r.\\ ...

0

$$f(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} F(\omega) \, \mathrm e^{-i \omega x} \,\mathrm d\omega$$ so we have f'(x) = \frac{\mathrm d}{\mathrm dx}\!\left( \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} F(\omega) \,\mathrm e^{-i \omega x} \,\mathrm d\omega \right)= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} (-i \omega) \, F(\omega) ...

0

You're confusing yourself with $x$ and $y$. You want to find the transform of $g(x) = f(x+h)$, not $g(y) = f(x+h)$. Putting $g(x) = f(x+h)$, We see \begin{align*} \hat g(x) = \int_\mathbb R g(x) e^{-2\pi i x w} dx = \int_{\mathbb R} f(x+h) e^{-2\pi i xw} dx &= \int_\mathbb R f(x) e^{-2\pi i(x-h)w} dx \\ &= e^{2\pi i h w}\int_{\mathbb R}f(x) e^{-2\pi ...

2

Let's just take $f_n=\varphi_n *f$, with $\varphi_n(x)=n\varphi(nx)$, and $\varphi\in C_0^{\infty}$, $\int\varphi =1$. We're assuming that $f,xf,\xi\widehat{f}\in L^2$. Then $\int_{-L}^L |f_n-f|^2\to 0$ by a standard approximation argument, and if we choose $L>0$ large enough, then both $\int_{|x|>L}x^2|f|^2$ and $\int_{|x|>L}x^2|f_n|^2$ are small. ...

0

It is true : By the containment of $L^p$ spaces see David M. Bressoud's book (A Radical approach to Lebesgue's theory of Integration) on page 260 : $(\int_a^b |H_n|^p dx)^{\frac {1}{p}} \le (b-a)^{\frac {(q-p)}{pq}} (\int_a^b |H_n|^q dx)^{\frac {1}{q}}$ you can take p=1 and q=2 By Bessel's Inequality $\int_a^b |H_n|^2 dx \le \int_a^b |f|^2 dx$ This ...

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