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0

Try to use method, which is used in the calculation of Green function of wave equation.


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There are several ways to construct $L^1(\mathbb{R})$. One way is to let $L^1(\mathbb{R})$ be the smallest Banach space containing $C_0^\infty(\mathbb{R})$ under the norm $\|\cdot\|=\|\cdot\|_{L^1}$. For other definitions of $L^1$ this is also true (because it is the same space), however then this would need a proof based on the particular construction.


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Let $f \in L¹(\mathbb{R})$. Given an $\epsilon > 0$ you can find a simple compactly supported function $\phi$ such that $\|f - \phi\|_{L^1} < \epsilon/2$. Now you just have to find a smooth and compactly supported function near $\phi$. One way to do this is by doing a convolution against a mollifier. There are other ways to prove this. Another proof ...


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Just to complement PhoemueX's answer: I decided to look for "Féjer's Conjecture" and "Féjer-Jackson-Gronwall Inequality" on Google, and turned out that these results are quite known - yet Vaaler's reference was slightly difficult to reach. For example, there is a nice paper about a generalization of some polynomial inequalities like this here, from ...


1

The whole purpose of calculating the arithmetic mean of a distribution is to minimize the expectation of the square of the error. It all makes sense if you are penalized one dollar per square of the error. So if you know that tossing 3 fair coins gives you the usual distribution of outcomes, your best guess for the number of heads to show is 1.5, even though ...


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There are two factors here. First is, as pointed out in the comments by Shalop and Gary, average value of a random variable need not correspond to an actual value assumed by that variable. There are also issues of interpretation. The well-known joke that says a man whose feet is on fire and head on ice saying he is on the average comfortable. But I would ...


1

Here is a one-page proof by Landau: http://www.digizeitschriften.de/dms/img/?PID=GDZPPN002374331 . Here is a rough translation: Fejer's conjecture $$ S_{n}\left(x\right)=\sum_{\nu=1}^{n}\frac{\sin\left(\nu x\right)}{\nu}>0\qquad\text{ for }\qquad0<x<\pi $$ was first proved independently by Mr. Jackson and Mr. Gronwall. The following proof is ...


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The Fourier transform is a special type of integral transform most commonly defined as \begin{equation*} \hat{f}(\xi)=\int^{\infty}_{-\infty}f(x)e^{-2\pi ix\cdot\xi}dx,~\xi\in\mathbb{R} \end{equation*} that takes a function from real space into Fourier space, and vice-versa. It also changes various operations, such as transforming multiplication into ...


2

Let $m=\int_{0}^{1}f(x)\,dx$ be the mean value of $f$ over $[0,1]$. By considering the Fourier series of $f$ we have: $$f(x) = m +\sum_{n\geq 1}\left( s_n \sin(2\pi n x) + c_n \cos(2\pi n x)\right)$$ and: $$ \| f'(x) \|^2 = 2\pi^2\sum_{n\geq 1} n^2 (s_n^2+c_n^2)\leq 1 $$ hence: $$ \| f(x) - m\|^2 = \frac{1}{2}\sum_{n\geq ...


1

If you know that $\sum_{k\ne 0}|{\hat{f}(n)}|^2\le {1\over 4\pi^2}$, then you only need to consider $|\hat{f}(0)|$ and you can then apply Parseval's theorem linking the sum of the Fourier Coefficients with the Integral of the function squared. So if you can modify $f$ by adding a constant to make $\hat{f}(0) = 0$, you should be good to go...


1

Examine pointwise convergence condition for a Fourier series. Note that ypur function though not continuous at 0 is Lebesgue integrable in $[- \pi, \pi]$. Let me state the point wise convergence criterion. Theorem. Suppose $f$ is Lebesgue integrable in $[-\pi, \pi]$. A necessary and sufficient condition for the Fourier series $T(x)$ of $f$ to converge ...


4

So is it implying the Fourier series doesn't converge to $f(0)$? No, it doesn't imply that, and in this case, the Fourier series of $f$ does converge to $f$ pointwise. If we look at $g(x) = f(x) - 1$, we note that $g$ is an odd function, hence in the real Fourier series, only sine terms occur, and of course $\sin (nx) = 0$ for $x = 0$, whence the ...


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If you want to work in a classical pointwise way, the inversion is a Cauchy Principle Value integral $$ f(x)=\lim_{R\rightarrow\infty}\int_{-R}^{R}\hat{f}(\xi)e^{2\pi i\xi x}d\xi . $$ If $f$ is absolutely integrable and satisfies some condition such as having left- and right-hand derivatives at $x$, or is of bounded variation on an interval $[a,b]$ with ...


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Integration is linear (that is integral of sum is sum of integrals) so yes, final result is sum. I think most od the time it's not convenient to split integrals this way though, because exponents are easier to handle than trig functions.


1

You certainly need to assume that $\hat f\in L^1$ to apply the inversion formula as written; no, that does not follow from assuming $f\in L^1$. It's been pointed out that $f.f'.f''\in L^1(\mathbb R)$ imply $\hat f\in L^1$; you might also note that $f,f'\in L^2(\mathbb R)$ imply $\hat f\in L^1$. Even if $\hat f\notin L^1$ you can still recover $f$ from $\hat ...


1

No, it doesn't imply absolute integrability. Yes, both $f$ and $\hat f$ need to be $L^1$. However, if for example $f\in C^2(\mathbb R)$ and $f',f''\in L^1$, then $$\mathcal F [f''](\xi)=(-2\pi i \xi)^2 \hat f (\xi)$$ is a bounded continuous function, and for this to be true $\hat f$ must be bounded at least by something like $$|\hat f(\xi)|\leq ...


1

$Q1.$ According to my school's EE graduate course catalog, there are a couple of classes covering Hilbert, Banach and L^p spaces; and linear functionals. Is it possible to cover these topics without a formal mathematical background? $A1.$ It is the responsibility of the school to provide a curriculum for the students that is both meaningful an doable. ...


0

The original post has $f$ defined such that $$f(x)=(e^{-ab}-1)H(x)$$ where $H$ is the Heaviside function defined as $$H(x) = \begin{cases} 1, & \text{if $x\ge 0$} \\ 0, & \text{if $x<0$} \end{cases} $$ The Fourier-Transform $\hat H(k)$ of the Heaviside function is $$\hat H(k)\equiv \int_{-\infty}^{\infty}H(x)e^{ikx}dx=\pi ...


1

It is enough to show that $\sum|\hat f(n)|<\infty$. $$ \sum|\hat f(n)|=\sum n^{-2/3}|\hat g(n)|\le\Bigl(\sum n^{-4/3}\Bigr)^{1/2}\Bigl(\sum|\hat g(n)|^2\Bigr)^{1/2}<\infty $$ since $4/3>1$ and $g$ is Riemann-integrable, and hence in $L^2$.


1

Your solution is correct, apart from the fact that you forgot the scale factors for the Dirac terms. It should be $$\hat{T}(k)=\frac{1}{\sqrt{2\pi}}\left( \frac{2i\beta -\lambda k-\mu ik^2}{k^3}\right)+\sqrt{\frac{\pi}{2}}\left( \mu\delta(k)+i\lambda\delta'(k)-\beta\delta''(k)\right)$$ It looks like Mathematica can't be trusted here. First of all, it gets ...


2

You have $g=F(f)$. The Fourier transform is a continuous linear map in many good spaces (Schwartz space, its dual, or $L^2$). The derivative of a linear map is the map itself, and so a continuous linear map is continuously differentiable (and the second derivative vanishes). That is, $\partial g/\partial f=F$. (One usually denotes this by $DF(f)=F$ or ...


0

Since orthonormal families are automatically linearly independent, the "only if" direction is trivial. On the other hand, suppose the orthonormal family is a basis, and fix $\varepsilon > 0$. Then by definition of a (Schauder) basis, there exists some positive integer $n$ and $c_i \in \mathbb{C}$ for $i = 1, \ldots n$ such that $$\left\lVert f - ...


1

I don't think so - it looks like you are missing a factor of $n$. I think it's a little simpler than this if you stick with exponentials. Write $$f(x) = \sum_{n=-\infty}^{\infty} 3^{-n^2} e^{i n x} $$ This converges, as you said, absolutely and uniformly. We may write $$f'(x) = i \sum_{n=-\infty}^{\infty} n \, 3^{-n^2} e^{i n x} $$ $$g(x) = \pi ...


2

Well, derivation under the integral sign gives you directly $$\delta(y)=\int e^{-ixy}dx$$ $$\delta'(y)=-i\int xe^{-ixy}dx$$ $$\delta''(y)=-\int x^2 e^{-ixy}dx$$ $$\delta'''(y)=i\int x^3 e^{-ixy}dx$$ Then, by integration by parts, you can verify what it does as a distribution: $$\int \delta(y)f(y)dy=f(0)$$ $$\int ...


2

To elaborate on Daniel Fischer's comment: We have Let $$\tag1 f\colon x\mapsto \sum_{n=0}^\infty a_nx^n$$ be a power series and $R=1/\limsup\sqrt[n]{|a_n|}>0$ its radius of convergence. Then the series converges absolutely for all $x$ with $|x|<R$ and we can take the derivative termwise, i.e., $f'(x)=\sum_{n=0}^\infty (n+1)a_{n+1}x^n$ where ...


1

An idea: At $\;\xi=0\;$ : $$\lim_{\xi\to0}f(\xi)\stackrel{l'Hospital}=\lim_{\xi\to0}\frac{i\lambda e^{i\lambda\xi}}i=\lambda$$ so that $\;\xi=0\;$ is a removable singularity. At any other point $\;f(\xi)\;$ is the quotient of two analytic functions and is thus analytic.


0

Consider the function \begin{eqnarray} \chi(x) = \begin{cases} e^{-\frac{1}{x}} \ & \text{ if}\ x> 0 \\ 0 & \text{ if}\ x \le 0 \end{cases} \end{eqnarray} then $f(x) = \chi(1-|x|^2)$, so it's enough to show that $\chi$ is $C^{\infty}$. We'll break the proof into several easy steps. 1.If $f$, $g$ are continuous functions on $\mathbb{R}$ and ...


0

It is enough to show $f:t\mapsto\left\{ \begin{array} [c]{c}% e^{-\frac{1}{1-t}}\text{ if }t<1\\ 0\text{ if }t\geq1 \end{array} \right. $ is infinitely differentiable at $1$. Step 1: It is easy to verify that $$ \lim_{t\uparrow1}\frac{f\left( t\right) }{\left( 1-t\right) ^{k}}=0\text{ }\forall k=0,1,2,... $$ Step 2: $f^{\left( k\right) },$ the ...


2

Consider $f(x) = e^{-\frac{1}{(1 - x^2)}}$ Then note that if $f^{n}(x) = \frac{P(x)}{Q(x)}f(x)$ then $f^{n+1}(x) = \frac{P'(x) Q(x) - P(x)Q'(x)}{Q(x)^2} + (-\frac{1}{1 - x^2})'f(x) = \frac{\hat{P}(x)}{\hat{Q}(x)}f(x)$. Where $f^n$ is the n th derivative of $f$ and $P,Q, \hat{P}, \hat{Q}$ are polinomials. Note that $Q(x) = \bigg(\frac{1}{1-x^2}\bigg)^k$ for ...


1

This answer is intended to complement David's very nice solution by showing that the partial sums of the Fourier integrals are at most be pointwise unbounded almost everywhere (a.e.), when $f\in L^{1}(\mathbb{R})$; there exists a subsequence which converges to $t$ pointwise a.e. Combining this with David's solution, we see that the partial sums of the ...


1

Define $$f(t)=\sum_{n=-\infty}^\infty \hat\mu(n)e^{int}.$$The series converges uniformly so $f$ is continuous. The uniform convergence also shows that $$\hat f(n)=\hat\mu(n).$$So uniqueness (for complex measures) shows that $\mu=f$, or more carefully $d\mu=f\,dt$. If you don't buy the uniqueness for complex measures bit: Lemma: If $\nu$ is a complex ...


0

The problem is stated to be \begin{align} g(x) = \sum\limits_{n=1}^\infty \frac{sin(nx)}{6^n sin(x)} \mbox{ for } x \not= k\pi, \mbox{ and } g(k\pi) = \lim_{x\to k\pi} g(x), \hspace{5mm} k \in \mathbb{Z} \end{align} Consider the following. \begin{align} g(x,t) &= \sum_{n=1}^{\infty} \frac{\sin(nx) \, t^{n}}{\sin(x)} \\ &= \sum_{n=1}^{\infty} ...


1

Try in the opposite direction: \begin{align} \mathcal{F}(e^{-ax}H(x)) &=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-ax}H(x)e^{-isx}dx \\ & = \frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}e^{-ax}e^{-isx}dx \\ & = \frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}e^{-x(a+is)}dx \\ & = ...


1

$g(x)=\sum\limits_{n=1}^\infty \frac{\alpha^n}{sin(x)}(sin(n-2)x cos2x + cos(n-2)x sin2x)$ $ = \sum\limits_{n=1}^\infty \alpha^n\frac{sin(n-2)x}{sin(x)} + 2\sum\limits_{n=1}^\infty \alpha^n cos(n-1)x$ $= -\alpha + \alpha^2\sum\limits_{n=1}^\infty\alpha^n \frac{sin(nx)}{sinx} + 2\sum\limits_{n=1}^\infty\alpha^n cos(n-1)x$ where we can find $g(x)$ to be: ...


9

I changed my evaluation slightly, and I was able to get the result in a very simple form. First notice that $$ \int_{-\infty}^{\infty} \text{Ei}^{2}(-|x|) e^{ikx} \, dx = 2 \int_{0}^{\infty} \text{Ei}^{2}(-x) \cos(kx) \, dx. $$ Then integrating by parts, and assuming for now that $k >0$, $$ \begin{align}\int_{-\infty}^{\infty} \text{Ei}^{2}(-|x|) ...


2

I propose another method that does not use integration. Problem is in the change of variable when completing the square : because of the imaginary unit $i$, the new variable is complex and thus the integral must be done carefully in $\mathbb{C}$. As $f$ is smooth (i.e. $f\in\mathcal{C}^\infty\left(\mathbb{R};\mathbb{R}\right)$), $\hat{f}$ is also smooth. ...


1

Ok. Note that $2\pi=1$ here... Say $f$ is Kolomogorov's example. Regard $f$ as a periodic function defined on the entire line. Let $c_n$ be the $n$-th Fourier coefficient of $f$ (we want to reserve \hat for the Fourier transform). Say $\psi\ne 0$ is a smooth function with support in $(-1/2,1/2)$. Say $\psi$ is even just so we don't have to worry about the ...


1

You forget to mention one vital thing: $f$ is periodic with periodicity $2\pi$. The fourier coefficient is as you mention: $$f_n=\frac{1}{2\pi}\int_0^{2\pi}f(x)e^{-inx}dx$$ An interesting property of this integral is that one may move the interval of integration: as $f(x+2\pi)=f(x)$ and also $e^{-in(x+2\pi)}=e^{-2n\pi i} \cdot e^{-inx}=e^{-inx}$, we may ...


0

The following proof only works for uniformly continuous functions $f$ (note that uniform continuity implies continuity). Introduce a neighbourhood $X\subset\mathbb{R}^n$ around the origin. Let $X^c=\mathbb{R}^d\setminus X$ denote the complement of $X$, and $\mathbb{R}^n=X+X^c$. The idea is to examine the convergence problem over $X$ and $X^c$, and use an ...


3

Let the Fourier transform be defined as \begin{align} f(\omega) = \frac{1}{2\pi} \, \int_{-\infty}^{\infty} f(x) \, e^{-i \omega x} \, dx \end{align} Now, for $f(x) = e^{- x^{2} + 2x}$ the following is developed. \begin{align} 2 \pi \, f(\omega) &= \int_{-\infty}^{\infty} e^{- (x^2 - (2 - i\omega) x)} \, dx \\ &= e^{-\left(1 - \frac{i ...


0

I believe that your domain of interest is incorrect. As far as I can tell, you can only get this answer if you're considering the function on the domain $x\in [0,1]$. Starting from the definition of the nth Fourier coefficient on this domain: $$f_n:=\int_{0}^{1} f(t) e^{-2\pi int}dt$$ Then, for $g(x)=f(x+c)$: $$\begin{aligned}g_n&=\int_0^{1} f(t+c) ...


0

Take the Majority function on 3-bits as an example it is not true. $\widehat{Maj}({1})\leq \widehat{Maj}({1,2,3})$.


2

Note that $$\frac{1}{\alpha+i\omega}=\frac{\alpha}{\alpha^2+\omega^2}-\frac{i\omega}{\alpha^2+\omega^2}\tag{1}$$ Taking the limit $\alpha\rightarrow 0^{+}$ of (1) gives $$\lim_{\alpha\rightarrow 0^+}\frac{1}{\alpha+i\omega}=\lim_{\alpha\rightarrow 0^+}\frac{\alpha}{\alpha^2+\omega^2}+\frac{1}{i\omega}\tag{2}$$ The real part of (2) is a scaled nascent ...


0

If the Fourier coefficients $a_k$ are periodic with period $N$, the $T$-periodic signal $x(t)$ can be written as $$\begin{align}x(t)&=\sum_{k=-\infty}^{\infty}a_ke^{j2\pi k t/T}\\&=\sum_{l=-\infty}^{\infty}\sum_{k=0}^{N-1}a_ke^{j(k+lN)2\pi t/T}\\&= \sum_{k=0}^{N-1}a_ke^{j2\pi kt/T}\sum_{l=-\infty}^{\infty}e^{j2\pi lNt/T}\tag{1}\end{align}$$ ...


2

For the equation $$u_{tt} = c^{2} \, u_{xx} -\alpha u, \, 0<x\leq L, t>0$$ $$u(0,t) = u(L,t) = 0$$ $$u(x,0) = f(x), \, u(x,L) = g(x)$$ let $u(x,t) = F(x) G(t)$ for which \begin{align} \frac{1}{c^{2}} \left( \frac{G''}{G} + \alpha \right) = - \lambda^{2} = \frac{F''}{F} \end{align} or \begin{align} F'' + \lambda^{2} \, F &= 0 \\ G'' + (\alpha + ...


2

This is related. The function $$ f(a,b,x)=\frac{\sin[b\sqrt{a^{2}+x^{2}}]}{\sqrt{a^{2}+x^{2}}} $$ has Fourier transform $$ \hat{f}(a,b,s)= \left\{\begin{array}{cc} \frac{\pi}{2}J_{0}[a\sqrt{b^{2}-s^{2}}], & 0 < s < b \\ 0 & b < s < \infty \end{array}\right. $$ The derivative of ...


0

for $\cos(x)$, this is usually done when the argument $x$ is small compared with unity. Just how small depends on how accurate you would like your results. The "quick and dirty" way to know if it's good enough, is to compare the error directly with evaluation of $\cos(x)$.


1

You can't really compute that integral analytically, since the delta function isn't really a "function," and the integral is not really well-defined as you've written it down. What you CAN do is prove that when integrated against a test-function, the integral behaves like a delta function. So let's test it. We want $\int\delta(t)f(t)dt = f(0)$ for any ...


2

Without any limiting procedures, you can use the definition of the distributional Fourier transform, which is defined by $$\langle \mathcal{F}(d),\mathcal{F}(f) \rangle = \langle d,f \rangle$$ for $d \in \mathcal{S}',f \in \mathcal{S}$. Here $\mathcal{S}$ is the space of Schwartz functions (smooth functions all of whose derivatives decay faster than any ...



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