New answers tagged

-1

Well there are 3 conditions for a Fourier Series of a function to be exist: 1. It has to be periodic. 2. It must be single valued, continuous.it can have finite number of finite discontinuities. 3. It must have only a finite number of Maxima and minima within the period. 4. The Integral over one period of |f(X)| must converge. Each of them have Analytical ...


1

Here's a simple way to determine the form of $\hat{h}_a$: The function $h_a$ satisfies $x\cdot\nabla h_a = -a h_a$. Taking the Fourier transform gives the equation $i\nabla\cdot(ix\hat{h}_a) = -a \hat{h}_a.$ The left hand side can be rewritten as $-(\nabla\cdot x)\hat{h}_a - (x\cdot\nabla)\hat{h}_a.$ Here, $\nabla\cdot x = d$ so we have a differential ...


0

The Paley-Wiener-Schwartz theorem can be generalized quite a bit. Instead of assuming the function is compactly supported, we can assume that $u$ is a compactly-supported distribution (a tempered distribution with compact support). We often write $u\in\mathcal{E}'(\mathbb{R}^n).$ Then, the Fourier transform of $u$ (we'll write $\mathcal{F}u$) is $C^\infty,$ ...


1

Let $f$ be periodic. $\sum |\hat {f} (n)| <\infty$ implies that the Fourier series converges absolutely and uniformly. Since this condition is satisfied here (by comparison with $\sum\frac 1 {n^{2}}$) the answer is a definite YES.


5

The measure in question is $d\mu(y)=h_\lambda (y)dm(y).$ Note that $\mu$ is a positive measure on $\mathbb R$ with $\mu(\mathbb R)=1.$


1

Summability is no problem, for example if $K_n(t)$ is the Fejer kernel then $K_n(s)K_n(t)$ is an approximate identity on $\Bbb T^2$. Now, convergence for multiple Fourier series is a serious problem...


2

$$\frac{1}{\pi} \int_0^{\pi} \frac{a-b\cos\varphi}{\sqrt{a^2+b^2-2ab\cos\varphi}} \cos n \varphi\, d\varphi = \, _3\tilde{F}_2\left(\tfrac{1}{2},\tfrac{1}{2},1;1-n,n+1;\tfrac{4 a b}{(a+b)^2}\right)$$ $$\qquad\qquad\qquad\qquad-\frac{ b}{a+b} \, _3\tilde{F}_2\left(\tfrac{1}{2},\tfrac{3}{2},2;2-n,n+2;\tfrac{4 a b}{(a+b)^2}\right),$$ with $\tilde{F}$ defined ...


3

No, because a series of continuous functions cannot converge uniformly to a discontinuous function.


0

In terms of generalized functions, there is a Fourier series for $\tan(x)$ on $(-\pi/2,\pi/2)$. To see this, we proceed as follows. Let $f(x)=\tan(x)$ and let $F(x)\equiv \int_0^x f(t)\,dt$ for $0 \le x< \pi/2$ be an antiderivative of $f(x)$ on $[0,\pi/2)$. Clearly, we have $F(x)=-\log(\cos(x))$. Next, note that we can write $$\begin{align} \log(\...


1

Let $f \in L^2_{loc}( \Bbb{R}_{>0} )$ such that $f(2x) = f(x)$. Then we have the Fourier series $$f =\sum_n C_n(f) x^{2i \pi n/\log(2)} = \sum_n C_n(f) e^{2i \pi n\log_2(x) }, \quad C_n(f) = \frac{1}{\log(2)}\int_1^2 f(x) e^{-2i \pi n\log_2(x) }\frac{dx}{x}$$


3

Yes, I believe that Daubechies Wavelets of high orders satisfy this. Daubechies Wavelets have a compact support (that's why they were invented), and the higher the order, the higher the regularity. This doc relates the order of the Daubechies Wavelets (and scaling functions) with their Holder exponent. It seems that at order 7, both wavelets and scaling ...


5

This morning I had a counterexample, using this: Gap. $\sum_1^\infty\frac{\cos(nt)}{n^{1/2}}$ is not the Fourier series of a function in $L^1(\Bbb T)$. Non-proof: I swear I read yesterday that this was proved by Hardy&Littlewood. I can't find it today. But all I need is a much weaker statement: Cor. If $p>2$ there exists a sequence ...


0

Your question is not really about the Brownian bridge, but rather it is about the Fourier decomposition in the Hilbert space $L^2[0,1]$. For this reason, I will focus on a specific realization of the Brownian bridge, namely $\beta_t=\sin(\pi t)$, which although it is way too smooth to be a sample path of the Brownian bridge, it nevertheless does lie in the ...


2

Since it may be helpful to someone, here is my complete answer to this question (done, of course, with a lot of help from @FXV). Firstly, if $g(t)=f(t+x)-f(x)$, then by linearity we have that $$s_n(f;x)-s_n(g;0)=f(x).$$ In other words, if $s_n(g;0)\to 0$, then $s_n(f;x)\to f(x)$ so it suffices to consider the case $x=0$, $f(0)=0$. Now we use the ...


0

You introduced $\int u \overline{v}$ which is not a bilinear form since Bourgain's spaces $X_{\tau = h(\xi)}^{s,b}$ are complex vector spaces. Thus, you should rather consider $$\begin{cases}\mathcal{B}: &S(\mathbb{R}^{d+1})\times S(\mathbb{R}^{d+1}) &\to &\mathbb{C} \\ &(u,v) &\mapsto &\int u v .\end{cases}$$ which defines a ...


0

One issue with your argument is that you consider only the real part of the DFT (why?). You might as well consider only the imaginary part, which, for the first case ($[1,-1,1,-1]$) equals the all-zero vector ($[0,0,0,0]$), which, you could also "argue" makes no sense as you have a non-zero signal. You have to consider both the real and imaginary part of ...


0

As it was pointed out in this post, I can make use of the Plancherel Theorem. The function $f$ lies in $(L^1 \bigcap L^2) $, since it's in $L^1$ and $f(x)\le 1\ \forall x$. So it's Fourier transform also lies in $L^2$ and the inversion formula holds.


2

The aim is to show that $s_n(f,a) \rightarrow f(a) $ for all $a$, however it is easy to show that if you let $g(x)=g(x+a)-g(a)$, $$s_n(f,a)-s_n(g,0) = \sum_{-n}^n \frac1{2\pi}\int_{-\pi}^\pi (f(t)e^{ik(a-t) } - g(t)e^{-ikt})dt = f(a)$$ so if you showed that $s_n(g,0)\rightarrow 0$ you may deduce that $s_n(f,a)\rightarrow f(a)$. I think I agree with what you ...


2

As Adrian Keister pointed out in a comment, we have $$S(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2-1/4}\cos(nx).$$ In absolute value the summands are bounded above by $1/(n^2-1/4).$ Because $\sum 1/(n^2-1/4)<\infty,$ Weierstrass M tells us the series converges uniformly on $[-\pi,\pi].$ Since each summand is continuous, so is $S(x).$


5

This is a lemma in the paper Mordell's exponential sum estimate revisited by Jean Bourgain. The author proves it using Fourier analytic methods applied to finite abelian groups.


1

The estimate you want to prove is not correct in general. I will only present a counterexample for the case $p=1$ and large $n$, but I think that the claim should also fail for other $1 < p < 2$, once $n$ is large enough. Counterexample: Let $f = 1_{[-1/2, 1/2]^n}$, and define $f_k (x) := k^n \cdot f(k x)$. Note that $f, f_k \in L^1$ with $\|f_k\|_{L^...


4

Late to the party :) This is an excellent question. As you'll understand the answer, you'll realize why wavelets are such a powerful thing. It is true that scaling functions provide a way to approximate all functions in $L^2(\mathbb R^n)$. The same can be said about decomposing using Wavelets themselves. So if both scaling function and wavelet can be ...


2

Let $2\pi Rt=x$, then $$I =\lim_{R\rightarrow\infty} \int_{-2\pi R}^{2\pi R} \frac{\sin x}{x} dx= \int_{-\infty}^{\infty} \frac{\sin x}{x} dx= 2 \int_{0}^{\infty} \frac{\sin x}{x} dx=2\frac{\pi}{2}=\pi$$


0

It is not too hard to prove by induction on $k>1$ that $$\sum_{n=|j|+1}^{|j|+k}n(a_{n-1}+a_{n+1}-2a_n)(1-\frac{|j|}{n})=a_{|j|}+ka_{|j|+k+1}-(k+1)a_{|j|+k}$$ and since by the first part of the proof we have $$\lim_{n\to\infty}n(a_n-a_{n+1})=0$$ and since by hypothesis of the theorem $a_n\to 0$ as $n\to\infty$, you get the desired result by letting $k\to\...


1

A simpler case is when $f$ is $L^1$ and locally $a$-Hölder continuous, from $$\int_{-R}^{R}\hat{f}(w)e^{2i\pi x w}dw = \int_{-\infty}^\infty f(x-y)\frac{\sin(2 \pi R y)}{\pi y}dy$$ we obtain that it converges to $f$ locally uniformly. Note that $\int_{|y|> b}|\frac{f(x-y)}{\pi y} |dy< \infty$ implies $\lim_{R \to \infty}\int_{|y|> b}\frac{f(x-y)}...


0

For $\varphi \in S(R^n \times R^n), u \in S'(R^n)$ let $$T[u](x) = <u,\varphi(x,.)> \ \in R, \qquad (D_x^a x^b) T[u](x_0)=<u,(D_x^a x^b \varphi)(x_0,.)>$$ Since $u$ is a tempered distribution you know that for some $N_u,C_u$, for all $\phi \in S(R^n)$ $$|<u,\phi>| \le C_u \sum_{|r|,|s| \le N_u} \|D_y^r y^s \phi\|_\infty$$ whence $$\sup_{...


0

Since the FT of $f(x)$ as $\hat f(\omega)$ exists, we can write:$$\hat f(\omega)=\int_{-\infty}^\infty f(x)e^{-i2\pi\omega x}dx$$and by substituting we obtain$$\int_{-R}^{R}\hat{f}(w)e^{2i\pi x w}dw{=\int_{-R}^{R}\int_{-\infty}^\infty f(x_1)e^{-i2\pi\omega x_1}e^{2i\pi x w}dx_1dw\\=\int_{-\infty}^\infty f(x_1)\int_{-R}^{R} e^{-i2\pi\omega x_1}e^{2i\pi x w}...


1

Changing to hyperspherical coordinates gives $$I(R, a) = \int_{x < R} \frac {e^{i \boldsymbol k \cdot \boldsymbol x}} {x^2 + a^2} d\boldsymbol x = S_{n - 2} \int_0^R \int_0^\pi \frac {e^{i k r \cos \theta}} {r^2 + a^2} r^{n - 1} \sin^{n - 2} \theta \, d\theta dr = \\ (2 \pi)^{n/2} k^{1 - n/2} \int_0^R \frac {r^{n/2}} {r^2 + a^2} J_{n/2 - 1}(k r) dr.$$ ...


1

If $\widehat{\psi}$ represents the Fourier transform of $\psi$ then $$ \psi(x) = \frac{1}{2\pi}\int_{-\infty}^{+\infty} \widehat{\psi}(\omega) e^{j\omega x}{\rm d}\omega \tag{1} $$ Now consider the expression \begin{eqnarray} \frac{1}{2\pi}\int_{-\infty}^{+\infty} \widehat{\psi}(s\omega) e^{-j\omega a}e^{j\omega x}{\rm d}\omega &=& \frac{1}{2\pi}...


1

This is not an answer, but I would like to share my progress on the problem. If $F$ is replaced by $F+\tau$ then $F-F$ stays the same, hence we may assume $0\in F$ without loss of generality. Like any Minkowski difference, $F-F$ is symmetric with respect to zero, hence if $(F-F)\subset E$ then $(F-F)\subset(E\cap(-E))=E'$. Assuming $\alpha=\sqrt{2}$, $E'$ ...


0

To add on AD's answer, First of all there is no identity in the convolution algebra $ L^1(\Bbb{R/Z})$, to make it unital we need to add the Dirac $\delta$, it stays a Banach algebra with $N(c\delta+f) = |c| + \|f\|_{L^1(\Bbb{R/Z})}$. Let $(f_n) \in L^1(\Bbb{R/Z})$ such that $f_n$ is not algebraic over $\Bbb{C}[\delta,f_1,\ldots,f_{n-1}]$, then $\Bbb{C}[\...


1

HINT : $$ f(x)=\sum_{n=1}^\infty \frac{(-1)^n}{(n\pi)^3}\sin(n\pi x)$$ $$ f''(x)=-\sum_{n=1}^\infty \frac{(-1)^n}{n\pi}\sin(n\pi x)$$ $\sin(n\pi (x-1))=\sin(n\pi x)\cos(n\pi)=(-1)^n\sin(n\pi x)$ $$ f''(x)=-\sum_{n=1}^\infty \frac{1}{n\pi}\sin(n\pi (x-1))$$ With $X=x-1$ we get the Sawtooth wave or Sawtooth function $SW(X)$. http://mathworld.wolfram.com/...


3

Suppose $f*g=0$, then (by the multiplication theorem) $\hat{f}(n)\hat{g}(n)=0$. This is true whenever $f$ and $g$ lives on disjoint frequencies. Example (complex case) $f(t)= 1$ and $g(t)=e^{it}$, here $$f*g(t)= \int_{-\pi}^\pi f(x)g(t-x)\frac{dx}{2\pi}= \int_{-\pi}^\pi \exp(i(t-x))\frac{dx}{2\pi}=0$$ Example (real case) $f(t)= 1$ and $g(t)=\sin(t)$, then ...


0

Ok so I just found $$\int_{-1}^1\frac{x^3-x}{12}\sin(n\pi x)dx=\frac{(-1)^n}{n^3\pi^3}$$


2

Fourier transforming both sides should work fine. The trick is that you can separate the resulting double integral: \begin{align} \mathcal{F}\left[\int_{-\infty}^\infty f(v)e^{-2(x-v)^2}dv\right] =& \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty\int_{-\infty}^\infty f(v)e^{-2(x-v)^2}e^{-i\omega x}dvdx \\ =& \frac{1}{\sqrt{2\pi}}\left[\int_{-\infty}^\...


4

The point at which the over shoot is measured is not fixed. So if you fix an $x$ value your limit will be your original function (which is your bullet point #2), but if you look at the max value over a fixed small interval around your jump, that limit yields an overshoot. This Desmos graph of $f_n(x) = n(1-x)x^n$ exhibits a similar phenomenon in a much ...


1

Yes, if $f\in L^1$ then $F_N*f\to f$ almost everywhere. Katznelson refers to Zygmund for most of this. Sketch/outline: In addition to being what K calls a summability kernel, $F_N$ is even when regarded as a function on $(-\pi,\pi)$, and $F_N\ge0$ is decreasing on $(0,\pi)$. Hence Suppose $f\in L^1(\Bbb T)$. Then $F_n*f(t)\to f(t)$ for every $t$ ...


1

A simpler approach than the one I had proposed in my comment. I have taken the "other" definition of Fourier Transform (I am used to it in the framework of Signal Processing) : $$\hat{f}(k) := \int_{\Bbb R} f(x) e^{-2 i \pi k x}\,dx.$$ Let us consider the "ramp function" : $r(x)=\max(x,0)$ whose Fourier Transform is known to be : $$\hat{r}(k)=\dfrac{i}{...


0

The Fejer Kernels form an approximation to the identity: \begin{align} \int_0^1 F_N(x)dx = 1~~ \forall N \\ \sup_N \int_0^1 |F_N(x)| dx < \infty \\ \text{for all}~ $\delta>0$ ~ \int_{|x|>\delta} |F_N(x)|dx \rightarrow 0~\text{as}~ N\rightarrow 0 \\ \end{align} Additionally, in Muscalu/Schlag's Classical and Multilinear analysis I, it is a result (...


0

https://en.m.wikipedia.org/wiki/Carleson%27s_theorem It is true in $L^p $ for $p>1$ but not in $L^1$. EDIT: This is about Dirichlet kernel, not Féjer. Sorry about that mistake, I read too fast the author's question.


0

In the paper, a function with 2 real variables is defined $$ I(x,y)=\frac1{2\pi i}\int_C\frac{F(u,v)}{u+iv-(x+iy)}d(u+iv). $$ Of this function, the Fourier transform in both variables is considered, it is assumed that the integration over the (compact) curve $C$ can be moved out of the Fourier integration \begin{align} \hat I(ω_x,ω_y)&=A_{Fourier}\int_{\...


1

First, if $f\in E$ and $K\in L^1$ then $K*f$ is defined to be a vector-valued integral $$K*f=\frac1{2\pi}\int_{-\pi}^\pi K(t)f_t\,dt.$$ Lemma. $||K*f||_E\le||K||_1||f||_E$. Proof: $||K*f||_E\le\frac1{2\pi}\int_{-\pi}^\pi|K(t)|\,||f_t||_E=||f||_E\frac1{2\pi}\int_{-\pi}^\pi |K(t)|\,dt$. Now suppose $f\in E$ and $\epsilon>0$. Choose $\delta\in(0,...


1

You can use this version of the Riemann-Lebesgue lemma: If $g$ is Riemann integrable on $[a,b]$, then $$\int\limits_a^b g(x)\cos (nx)\, dx\rightarrow 0$$ as $n\rightarrow\infty$. It is clear that $g(x)=f(x)/x$ is continuous on $[0,\pi]$ if we take $g(0)=0$ (the only problem area is zero, and the condition that $|f(x)|\leq x^4$ takes care of that). Since ...


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