# Tag Info

1

In general, a conformal map preserves angles (but not necessarily length). In the case of an inner product space, since the angle between two vectors $v,w$ is defined to be the quantity $\frac{\langle v,w\rangle}{|v||w|}$, a conformal map $c:V\to W$ is one that preserves this quantity. In particular, since $|v| = \sqrt{\langle v,v\rangle}$, if $c$ ...

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Yes, you need to prove compactness. This is an instance of Rellich–Kondrachov theorem, but presumably the point of the problem is to show compactness directly. To get uniform convergence, it suffices to have Fourier coefficients converging in $\ell^1(\mathbb{Z})$, so let's look at those. Let $g_n^{(k)}$ be the $k$th coefficient of $g_n$. Then $g_n''$ has ...

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Yes, for decimation by an integer factor $M$, the DTFT of the decimated signal is given by $$Y(e^{j\theta})=\frac{1}{M}\sum_{m=0}^{M-1}X\left(e^{j(\theta-2\pi m)/M}\right)$$ which shows that the resulting spectrum is an expanded and aliased version of the original spectrum.

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If you have an orthonormal basis $\{ e_{k} \}_{k=1}^{N}$ on a finite-dimensional space, such as what you would obtain with Gram-Schmidt, then every vector $x$ is expressed as $$x = \sum_{k} (x,e_{k})e_{k}.$$ This extends to $L^{2}[0,2\pi]$ using $e_{k} =\frac{1}{\sqrt{2\pi}}e^{ikx}$: $$f = \sum_{k}(f,e_{k})e_{k}$$ The ...

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You can imagine derivative as a matrix such as this, as the corresponding values of each element approach zero (limit definition): Or integration as like this(riemann sum): Moreover, fourier transform already has a matrix representation for discrete case https://en.wikipedia.org/wiki/DFT_matrix You need extend this matrix to infinity and shrink the ...

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Let $$f_a(x)=\frac{1}{\sqrt{|x|}} \chi_{[-a,a]}$$ Observe that $f_a \in L^1$ but not in $L^2$. Now just compute the fourier transform and play with $a$. \begin{align} \hat{f_a}(y) &= \int_{-a}^a \frac{1}{\sqrt{|x|}}e^{-iyx} \, dx \\ &=\int_0^a\frac{1}{\sqrt{x}}e^{-iyx}+\frac{1}{\sqrt{x}}e^{iyx} \,dx \\ & =2 \int_0^{\sqrt{a}}e^{-iys^2} + ...

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I don't want to write $\Delta$ so I will use $\Delta x=a$ and $\Delta t-i\epsilon=t$. The expression becomes: $$\frac 1 {a} \int_0^\infty dk \, e^{-ikt} \sin k\,a=\frac 1 {t^2 -a^2}$$ This integral does not appear to converge at infinity so I will do a trick. I will multiply it by $e^{-bk}$ assuming $b>0$, and then insert $b\rightarrow 0$ at the very ...

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I would start with writing the sin as sum of 2 exponential functions. Then the integrand becomes the sum of exponential functions, so is easy.

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For the second part, it follows from the fact that you have inversion formula if $f$ is in Schwartz, i.e. $$f(x)=\int\hat f(\xi)e^{2\pi ix\xi}d\xi$$

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This is a consequence of the Poisson summation formula, specifically equation 3 of the article. The idea of Poisson summation is that the periodization of a function $F$, obtained by summing its translates, has a Fourier series whose coefficients are found by sampling $\hat F$. If $\hat F$ is compactly supported, this Fourier series becomes a trigonometric ...

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For 1, write $\mathbb{Z}$ as the union of two disjoint infinite subsets, $\mathbb{Z} = A \cup B$ and look at the subspaces spanned by $\{ e_k \}_{k\in A}$ and $\{ e_k \}_{k\in B}$. For 2, think some more about 1.

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At least it does not mean that $\hat{f} \in L^2(\mathbb{R})$, because then by the Plancherel theorem $f$ itself would have to be $L^2$. But this can fail; for instance $f(x)=x^{-2/3} \chi_{(0,1]}(x)$ is $L^1$ but not $L^2$. I would imagine that $\hat{f}(n)$ is not summable on $\mathbb{Z}$ either (at least with an example like mine which is quite far from ...

2

It is just a change of variable: $$\int_{-\infty}^\infty h(-\tau)\,e^{i\tau t}\,d\tau=\int_{-\infty}^\infty h(\tau)\,e^{-i\tau t}\,d\tau=\int_{-\infty}^\infty h(\tau)\,\overline{e^{i\tau t}}\,d\tau=\overline{\int_{-\infty}^\infty h(\tau)\,e^{i\tau t}\,d\tau}$$ because $h$ is real valued.

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It's better to write the formula as $$\hat{f}(w) = \frac13 \frac{\exp(\pmb{i}w)}{((2/3)+\pmb{i}w)}$$ because now $e^{-at}u(t)$ (Heaviside function is also needed here) with $a=2/3$ gets you pretty close. The factor of $1/3$ can be easily attached: the transform is linear. To get the factor of $\exp(\pmb{i}w)$, use the time-shifting property of the Fourier ...

3

From Parseval's Theorem, we have $$\frac{2}{\pi}\int_0^{\pi}f(x)^2dx=\sum_{k=1}^{\infty}\left(\frac{1}{k+1}\right)^2$$ Since the series converges, then $f$ is in $\mathscr{L}^2$ We can show this equality as follows \begin{align} \int_0^{\pi}f^2(x) &=\int_0^{\pi}f(x)\sum_{k=1}^{\infty}\frac{\sin kx}{k+1}dx \tag 1\\\\ ... 0 Well, I think this is a solution to the problem: 1 - If there was convergence, then, by the Uniform Boundedness Principle, as \sup_R \|S_R f \|_1 < \infty, then the operators S_R should be bounded in L^1. It can be shown that, conversely, it is enough that the latter condition holds for the convergence to hold. 2 - As the multiplier for S_R is ... 2 Just a sketch of a possible approach, quite close to one of the usual proofs of the Fejer-Jackson inequality. By periodicity and symmetry, we just need to prove the inequality over (0,\pi). We have:\sum_{n=1}^{N}\frac{\sin(nx)}{n}=\int_{0}^{x}\sum_{n=1}^{N}\cos(nt)\,dt=-\frac{x}{2}+\int_{0}^{x/2}\frac{\sin((2N+1)u)}{\sin(u)}\,du \tag{1}$$and we may ... 0 If f is continuous and has compact support then xf and f are in L^1. Then x\mapsto f(y)e^{-ixy} is differentiable for each y\in\mathbb R and |\frac{\partial}{\partial x} f(y)e^{-ixy}| = |-iyf(y)e^{-ixy}| \leq |yf(y)|\in L^1. Now we have a theorem that states that x\mapsto \hat f(x) = \int f(y)e^{-ixy}dy is differentiable in this case with ... 0 If f\in C_c(\Bbb R), then you can show that \hat f is differentiable by definition and dominated convergence theorem. If we want to build f\in C(\Bbb R) such that \hat f is not differentiable, we can use the fact that for even functions f = \hat{\hat f} (up to a multiplicative constant). Now take, for example, \hat f = \mathbf 1_{[-1,1]}(x). ... 0 Assuming that A and f_0 are nonzero constants, \int_{-\infty}^{\infty} \left|2Af_0 \frac{\sin(2\pi f_0 t)}{2\pi f_0 t}\right|\,dt diverges. This integral would be a constant multiple of \int_{-\infty}^{\infty} \left|\frac{\sin(t)}{ t}\right|\,dt if it converged, so it’s enough to show that \int_{-\infty}^{\infty} \left|\frac{\sin(2\pi t)}{ ... 1 There may be a misunderstanding here; I don't see how one would use Jensen's inequality to control L^p norm by L^1 norm. Anyway, the proof of$$ \lim_{n\to\infty}\|f-h_n*f\|_p =0 $$is essentially the same for 1<p<\infty as it is for p=1. For continuous functions g we have h_n*g\to g uniformly: just split the integral in ... 5 First, as pointed out in the comments above, the right hand side needs a factor of (2\pi)^{-1}. With that in mind, here we go. Let's pick our favorite 2\pi periodic functions. Namely, for j=0,1,2,\dots let f_j(x)=e^{ijx} (where i is the imaginary constant). Now let's see if the equality holds for these functions. For j=0, it is easy to see that ... 1 Let us start with a general claim: Let (g_n),\quad n=1,2,\ldots, be a sequence of functions which converges uniformly to the function g. Let h be a square-integrable function. Then the sequence (g_nh) converges by integral to gh. To prove the claim, we may assume g=0. By the Cauchy-Schwarz inequality we have$$\left(\int g_nhdx\right)^2\leq\int ...

2

No. The function $$f(x)=\begin{cases}0 & |x|>1\\ (1-\log(1-x^2))^{-1} & |x|\le1\end{cases}$$ is an explicit example. For more comments and a proof (due to Terry Tao) see this post.

2

No, there exist continuous functions on $S^1$ whose Fourier coefficients are not in $l^{\beta}$ for every $\beta <2$ ( while certainly being in $l^2$). The first example seems to be given by T. Carleman in 1916. This should be discusses in some books on classical Fourier series. I learned about it from this paper of Hausdorff.

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If you consider a minimal state space representation of $G$ $$\dot{x}=Ax+Bu\\y=Cx+Du$$ then $A$ must be a stable (Hurwitz) matrix for stable $G$. If $u\in\mathcal{L}_2$ and $A$ is stable then $x\in\mathcal{L}_2$ Proof: If $A$ is stable there exists $P=P^T>0$ such that $$PA+A^TP=-I$$ Define now the nonnegative function $V=x^TPx$. Then ...

2

It helps to recall Which functions are tempered distributions? — exactly the distributional derivatives of continuous functions of polynomial growth. The full statement of this theorem isn't needed here, but it gives an idea where to look for weird tempered distributions that are represented by a locally integrable function: take the derivative of a ...

0

First note that $$x\phi''(x)+\phi'(x)+\frac{\lambda}{x}\phi(x) = 0$$ can be written as $$-(x\phi'(x))' =\frac{\lambda}{x}\phi(x)$$ Now suppose that $\phi$ is a solution of the above for which $\phi(1)=\phi(2)=0$. Then \lambda\int_{1}^{2}\phi(x)^{2}\frac{1}{x}dx = -\int_{1}^{2}(x\phi'(x))'\phi(x)dx \\ = \left\{ ... 0 You can use the Fourier sine and cosine transforms instead, and these only depend on real functions. This is how Fourier originally worked. \begin{align} f & \sim \frac{1}{\pi}\int_{-\infty}^{\infty}\cos(sx)\int_{-\infty}^{\infty}f(y)\cos(sy)dy \\ & +\frac{1}{\pi}\int_{-\infty}^{\infty}\sin(sx)\int_{-\infty}^{\infty}f(y)\sin(sy)dy. ... 2 The change u(t,x)=e^{ax+bt}\,v(t,x) $$for appropriate choice of a,b\in\mathbb{R} will transform the equation into$$ v_t=v_{xx},\quad v(0,x)=e^{-ax}\,g(x). $$0 I found my answer, so maybe useful for someone else as well. I need to apply conjugate symmetry condition for 2D fourier transform, both in magnitude values and phase components. I applied it as below for a n*m matrix: "H" is the magnitude: H(1,1) = 0; H(1,m/2+1) = 0; H(n/2+1,m/2+1) = 0; H(n/2+1,1) = 0; H(2:end,2:m/2) = rot90(H(2:end,m/2+2:end),2); ... 0 Assuming suitable conditions on f,$$ \widehat{f''}+2i\widehat{f'}-2\widehat{f} = \widehat{g} \\ -s^{2}\widehat{f}-2s\widehat{f}-2\widehat{f}=\widehat{g}\\ \{s^{2}+2s+1\}\widehat{f}=-\widehat{g} \\ \widehat{f}=-\frac{\widehat{g}}{s^{2}+2s+1} $$Therefore,$$ f= ...

6

No, this doesn't follow. Fix a smooth bump function $\varphi$ supported in $[-1,1]$ and let $$f(x) = \sum_{n=1}^\infty \phi(2^n x-n)$$ The $n$th term is supported in $[n-2^{-n}, n+2^{-n}]$; these supports are disjoint. The series converges in every $L^p$ space. The function $f$ does not tend to zero at infinity, through. On the Fourier side, $$\hat ... 1 The Riemann-Lebesgue Lemma states that if f is \mathscr{L}^1 integrable, then$$\lim_{\omega \to \infty}\int_{-\infty}^{\infty}f(t)e^{i\omega t}dt=0$$. Obviously, for real-valued f, we have$$\lim_{\omega \to \infty}\int_{-\infty}^{\infty}f(t)e^{i\omega t}dt=\lim_{\omega \to \infty}\int_{-\infty}^{\infty}f(t)\cos(\omega t)dt+i\lim_{\omega \to ...

0

If $f$ is odd and integrable, then $$g(x) = \int_{0}^{x}f(y)dy$$ is even and bounded with limits at $\pm\infty$. Therefore, \begin{align} \int_{-R}^{R}e^{-isx}f(x)dx & = e^{-isx}g(x)|_{x=-R}^{R}+is\int_{-R}^{R}e^{-isx}g(x)dx \\ & = -2i\sin(sR)g(R)+is\int_{-R}^{R}e^{-isx}g(x)dx. \\ ...

0

If $f(t)$ is an odd function, then $$|\int_a^A{\frac{d \alpha}{\alpha}\int_R{f(t)e^{i\alpha t}dt}}| = |\int_a^A{\frac{d \alpha}{\alpha}\int_{R_+}{2f(t)sin(\alpha t)dt}}| \le M_{L_1(f)} \int_a^A{\frac{sin(\alpha t)d\alpha t}{\alpha t}} \lt M_{L_1(f)} \frac{\pi}{2}$$ Where $M_{L_1(f)}$ is $L_1$ measure of $f(t)$

3

Now that you ask the question, I have to admit that I had to think about it for a little bit. It seems that most courses/books focus more on the magnitude portion of the Fourier Transform, and tend to neglect the phase spectrum despite its overwhelming importance. Most of my examples are drawn from electrical engineering and signal processing, so hopefully ...

0

Use contour integration. Case 1: $t>0$ Observe the integral over $C$, where $C$ is comprised of the real line plus an infinite semi-circle, $C_{\infty}$, enclosing the upper-half plane of the complex $\omega$ plane. $$\frac{1}{2\pi}\oint_C\frac{e^{i\omega t}}{a+i\omega}d\omega=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{e^{i\omega ... 0 Hint: Use the fact that the Fourier transform is an L^2 isometry, i.e.$$\langle \mathcal{F}f,\mathcal{F}f\rangle = \langle f,f\rangle,where in your case f(x) = \begin{cases} 1 & |x|\le 1 \\ 0 & |x|>1\end{cases}. 0 As @Daniel Fischer pointed out it does cancel in contrary to what I thought then the Fourier coefficients of f are given by f_j = \int_{-\pi}^\pi e^{ij\lambda} f(\lambda) d \lambda. To this end by integration by parts \begin{align*} \int_{-\pi}^\pi e^{ij\lambda} f'(\lambda) d \lambda &= \left[ e^{ij\lambda} f(\lambda) \right]_{-\pi}^\pi - ... 0 You shoud subsitute in formla then show that every term is equal or grater than zero 1 Here's a summary of, and extending remarks and explanations to, the comments. Note that my description of u(t,x) is based on experimental observations, not formal proofs. I plotted the sum for u(t,x) for a few values of t: within a moderate range for x, it would converge, although I did need extended accuracy (used Maple for this). My understanding ... 1 The two dimensional Fourier Transform of a function f(x,y) is\mathscr{F}\{f\}(k_x,k_y)=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x,y)e^{ik_xx}e^{ik_yy}dxdy$$For f(x,y)=\text{rect}_{L_x}(x)\text{rect}_{L_y}(y), the Fourier Transform is ... 3 Symmetry is your friend. Since \psi is radially symmetric, so is \phi. So wlog we can assume {\bf p} = p \bf k, and {\bf p} . {\bf r} = p r \cos(\theta). Then$$\eqalign{\phi(p {\bf k}) &= \dfrac{1}{(2 a \hbar)^{3/2} \pi^2} \int_0^{2\pi} d\varphi \int_{0}^\infty r^2 \; dr \int_{0}^{\pi} \sin(\theta)\; d\theta\; e^{ip r \cos(\theta)} ...

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Hint: $\dfrac{1}{|s|^\alpha+1}=\dfrac{1}{|s|^\alpha\left(1+\dfrac{1}{|s|^\alpha}\right)}=\dfrac{1}{|s|^\alpha}\sum\limits_{n=0}^\infty(-1)^n|s|^{-\alpha n}=\sum\limits_{n=0}^\infty(-1)^n|s|^{-\alpha(n+1)}$

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There is no ambiguity on the function definition. It is defined as $x-\pi$ on $(0,\pi]$ and even with period $2\pi$, so that it is $-x-\pi$ on $(-\pi,0]$. Anyway, the Fourier series is to be computed on the range $(-2\pi,2\pi]$, there will be no $2\pi$ harmonic.

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Before deleting it, can you please tell me why this would be the graph ?

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The Fourier series converges to a certain extension of the given function. cosine series on $[0,L]$: even periodic extension sine series on $[0,L]$: odd periodic extension full Fourier series (sines and cosines on $[-L,L]$): periodic extension. To obtain the required extension, Define $f(x)=f(-x)$ on $[-L,0)$ and proceed to 3. Define $f(x)=-f(-x)$ ...

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The solution is $\hat{u}(\xi, t)$= $\hat{f}(\xi)\cos\Big(\sqrt{-a^2+c^2\xi^2}\Big)$ After that you shoud convert $cos at$ to $(e^{ait }+e^{-ait })/2$

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