New answers tagged

2

First, your formula $\hat{f}(n)=\sum_{n\in\mathbb{N}}a_{n}\cos(\lambda_{n}t)$ is way off. In fact $\hat f(\pm\lambda_n)=a_n/2$ and $\hat f(j)=0$ if $j\ne\lambda_n$. Note as well that assuming $f$ is $Lip_\alpha$ at one point implies $a_n=O(\lambda_n^{-\alpha})$ for $0<\alpha\le 1$, but that condition on the $a_n$ only implies that $f\in Lip_\alpha$ for $...


0

No: Recalling the quotient rule $$\frac{d}{dx}\dfrac{f(x)}{g(x)}=\dfrac{\frac{df}{dx}g(x)-f(x)\frac{dg}{dx}}{g(x)^2}=\dfrac{1}{g(x)}\dfrac{df}{dx}-\frac{f(x)g'(x)}{g(x)^2},$$ we conclude that $\dfrac{1}{g(x)}\dfrac{d}{dx}f(x) \neq \dfrac{d}{dx}\dfrac{f(x)}{g(x)}$ unless $g(x)=$ const.


1

First, note that \begin{align} \|\sin-\cos\|^2&=\|\sin(x)\|^2+2\langle\sin,\cos\rangle+\|\cos\|^2 \\ & =\|\sin\|^2+\|\cos\|^2 \\ & = \|1\|^2=\pi. \end{align} You want $$ \|f-\sin\|+\|f-\cos\| \le \frac{2}{3}\sqrt{\pi}+\frac{1}{3}\sqrt{\pi}=\sqrt{\pi} $$ If you have the above, then $$ \sqrt{\pi}=\|\sin-\cos\| \le \|\sin-f\|+\|...


0

$$ f(x)-S_n(x) = \frac{1}{2\pi}\int_{x-\pi}^{x+\pi}\left[\frac{f(x)-f(y)}{\sin{\frac{1}{2}(x-y)}}\right]\sin((n+1/2)(x-y))dy $$ Because $|f(x)-f(y)| \le M|x-y|^{\alpha}$, the expression enclosed by square brackets is absolutely integrable. So $S_n(x)\rightarrow f(x)$ for all $x$ by the Riemann-Lebesgue Lemma. The uniformity of the Lipshitz condition ...


0

You have two points ($\sin x$ and $\cos x$) in your space ($L^2$). You take a ball about each point and try to find a point in the intersection of the two balls. If there are any such points, there will be one that is a convex linear combination of sin and cos ($t\sin x +(1-t)\cos x$). This reduces it to looking for a single number between 0 and 1. Now ...


1

If $f(x)$ is real-valued, take $\dfrac{1}{3}$ times the first inequality plus $\dfrac{2}{3}$ times the second inequality to get: $$\int_{0}^{\pi}\dfrac{1}{3}(f(x)-\sin x)^2+\dfrac{2}{3}(f(x)-\cos x)^2\,dx \le \dfrac{1}{3} \cdot \dfrac{4\pi}{9} + \dfrac{2}{3}\cdot\dfrac{\pi}{9}$$ $$\int_{0}^{\pi}\left[f(x)^2-2f(x)\left(\dfrac{1}{3}\sin x - \dfrac{2}{3}\cos ...


2

Yes. By Cauchy's inequality, \begin{align*} (S+\#A)^2 = \bigg( \sum_{h=0}^{N-1} |\hat A(h)| \bigg)^2 &\le \bigg( \sum_{h=0}^{N-1} |\hat A(h)|^2 \bigg) \bigg( \sum_{h=0}^{N-1} 1^2 \bigg) \\ &= N \sum_{h=0}^{N-1} |\hat A(h)|^2 \\ &= N^2 \sum_{a\in A} 1^2 = N^2 \#A, \end{align*} where the last line follows by Plancherel's theorem applied to $x_a = 1$...


0

I believe I've found an alternative proof for this fact (using some arguably much-less-elementary facts). Given $f,g\in L^2(S^1)$, we can write their Fourier series $(\hat{f_n}),(\hat{g_n})$, defined by $$f(x)=\sum_{n\in \mathbb{Z}} \hat{f_n} e^{inx}$$ and same for $g$. Then by the convolution theorem, $h\equiv f\star g$ has Fourier coefficients $\hat{h_n}=...


0

The by Hans is good and explains under what conditions this makes sense. The OP asked me to post my comment as an answer. To justify this, I'll try to expand it with a bit more detail. There are various conventions for defining the Fourier transform. I will assume we are working with the following: $$\hat{f}(\omega) = \int_{-\infty}^{\infty}f(t) e^{-i \...


2

This is valid, so long as $|f(t)|=O(e^{b_-t})$ as $t\to\infty$ and $|f(t)|=O(e^{b_+t})$ as $t\to-\infty$ with $b_-<b_+$. Then $f(t)e^{bt}\in L^1(-\infty,\infty), \forall b\in(b_-,b_+)$ and the complex Fourier transform converges. Under this condition, the Fourier transform is analytic in $\omega=a+ib,\, a,b\in\mathbb R$ if $f(t)$ has finite number of ...


1

I saw this proof in an extract of the College Mathematics Journal. Consider the Integeral : $I$ = $\int_0^{\pi/2}ln(2cosx)dx$ From $2\cos(x)$ = $e^{ix}$ + $e^{-ix}$ , we have: $\int_0^{\pi/2}In(e^{ix}$ + $e^{-ix})dx$ = $\int_0^{\pi/2}In(e^{ix}(1 + e^{-2ix}))dx$ =$\int_0^{\pi/2}ixdx$ + $\int_0^{\pi/2}In(1 + e^{-2ix})dx$ The Taylor series expansion of ...


0

$e^{ix}$ is a vector of length 1 and argument $x$. $e^{ix}dx$ is a vector of length $dx$ and argument $x$. The sum of all these vectors is, intuitivelly speaking (or even rigorously speaking, if one is allowed to use non standard analisys), a closed polygon with infinitely many edges of lenght $dx$. This is an example figure with $n=10$ edges: $e^{ikx}...


3

A basis of convex functions is a tricky concept. Convexity is not closed under operations over vector spaces: a linear superposition of convex functions does not have to be convex (in fact, it most likely isn't). So... the entire formalism of vector spaces (Hilbert spaces actually) goes through the window and the concept of a basis doesn't make much sense. ...


2

Yes, you are correct. If let $f_r(e^{i\theta})=F(re^{i\theta})$ where $F$ is a function as you describe, then your assumptions give $$ \int_{0}^{2\pi}|f_r(e^{i\theta})|d\theta \le M, \;\;\; 0 \le r < 1. $$ If you let $\rho_r(\theta)=\int_{0}^{\theta}F(re^{i\theta})d\theta$, then $\rho_r$ is a function of bounded variation with total variation $V(\...


0

I am sorry you had to wait almost a year for someone to react! For the difference we have $$ (f\ast K_\delta)(x) - f(x) = \int_{\mathbb R^d} f(x-y) K_\delta(y)\ \text{d}y - \int_{\mathbb{R}^d} f(x) K_\delta(y) \ \text{d}y, $$ where we used property 1. Hence the difference equals $$ \int_{\mathbb{R}^d} \left(f(x-y) - f(x)\right) K_\delta(y) \ \text{d}y, $$ ...


2

Sketch of a more computational proof that the result fails: Let's take $b=1$ for simplicity. Verify that $$\int_0^1 \frac{\sin^2(\lambda x)}{x}\, dx \to \infty$$ as $\lambda \to \infty.$ This and the Riemann Lebesgue lemma allow us to inductively choose $\lambda_1 < \lambda_2 < \dots $ so that $$\tag 1 \int_0^1 \frac{\sin^2(\lambda_n x)}{x}\, dx >...


5

You can't prove this because it's false. (It's true if you assume just a tiny bit more smoothness than just continuity; see below.) First, $$\lim_{\lambda\to\infty}\int_0^b\frac{|\sin(\lambda t)|}{t}\,dt =\lim_{\lambda\to\infty}\int_0^{\lambda b}\frac{|\sin(t)|}{t}\,dt=\infty.$$ But the $L^1$ norm of $\sin(\lambda t)/t$ is the same as its norm as a linear ...


1

Claim: if $f \in C^N(\mathbb R^d)$, then $\int_{\mathbb R^d} |\check f (\xi)|^2 (1+\xi^2)^N d \xi < \infty$. Suppose for a moment that the claim is true. Then, by Holder's inequality, $$\|\check f \|_{L^1} \leq \|\check f(\xi)(1+|\xi|^2)^{N/2}\|_{L^2} \|(1+|\xi|^2)^{-N/2}\|_{L^2}.$$ Since $N>d/2$, it is easy to verify that $(1+|\xi|^2)^{-N/2} \in L^2$...


0

$f \in L^2$ $\hat{f}= \lim\limits_{u,v\mapsto \infty}\frac{1}{\sqrt{2\pi}}\int_{-u}^v fe^{-isx}dx$ multiply $f$ by a bump function $g_n$ with compact support in $[-u,v]$ such that $ \lim\limits_{n\mapsto \infty}g_nf=f$ on $[-u,v]$ let $h_n=g_nf$$\quad$ $h_n \in L^1 \bigcap L^2$ and $\hat{h}_n \in L^2 $(Plancherel theorem) Also $\lim\limits_{n\mapsto \...


1

The function $\hat{f}$ is continuous on $\mathbb{R}$ and vanishes at $\infty$ by the Riemann-Lebesgue lemma. So the definition of $f_n$ makes sense as a Riemann integral on the finite interval $[-n,n]$. The function $f_n$ converges pointwise everywhere to $f$ because $f$ is differentiable at every point of $\mathbb{R}$. You can rewrite $f_n$ in the following ...


1

Note that the range of $K$ is in a finite dimensional subspace $$\text{span} \{x, \cos x\},$$ thus $K$ is a compact operator and has only point spectrum ($0$ is a eigenvalue too). To look for eigenfunction, we need only restrict ourselve to the above subspace. Note that $$ K (ax+b \cos x) = b\pi x + a\frac{2\pi^3}{3} \cos x,$$ This implies that $$ ...


1

Very interesting paper. Your quote is on page 8; it seems likely to me that the theorem of Roth on the next page is a special case of what he means. If so then we're talking about a set $S\subset\{1,.\dots,N\}$, a "solution to a linear equation" inside $S$ is something like $n,n+a,n+2a\in S$ (which is the same as saying there exist $x,y\in S$ with $(x+y)/2\...


2

The Riemann integral doesn't make sense for general measurable $f$. Assuming $f$ is Riemann integrable, the Riemann and Lebesgue integrals coincide. Finally, $\lim_{R\rightarrow\infty}\int_{-R}^R|f|dx$ exists iff $f\in L^1$.


0

Let's do the following change of variables: $\theta = 2 \pi \left(x - {1 \over 2}\right)$. The function becomes $$f(x) = g(\theta) = {1 \over 2} + {\theta \over 2 \pi},$$ and its domain is $]-\pi, \pi[$. The geometric convenience of this is that $g$ can be thought of as a (discontinuous) function on the unit circle in the complex plane. The $k$-th ...


0

Define $e_\lambda(t)=e^{i\lambda t}$. One can easily calculate the relevant integrals explicitly, showing that $\{e_\lambda:\lambda\in\Bbb R\}$ is an orthonormal subset of $G$. Hence the completion $H$ of $G$ is a non-separable Hilbert space, since $\{B(e_\lambda,\sqrt 2/2)\}$ is an uncountable collection of pairwise disjoint nonempty open sets. Comment ...


0

For convenience, we first deduce the mean value $$a_0=\int_0^{1/2}2\,dt-\int_{1/2}^1dt=\frac12.$$ Then the function is odd, so there will be sine terms only, and by symmetry we can integrate on a half period $$b_n=2\int_0^{1/2}\frac32\sin(2\pi nt)\,dt=-\left.\frac 3{2\pi n}\cos(2\pi nt)\right|_0^{1/2}=-3\frac{\cos(\pi n)-1}{2\pi n}.$$ Only the terms for ...


0

Except for the constant term (indeed $\frac12$), this is a so-called square waveform, the transform of which is well-know. See http://mathworld.wolfram.com/FourierSeriesSquareWave.html or https://en.wikipedia.org/wiki/Square_wave.


3

Since I'm a Matlab user, here's how I would do it. A lot of professional audio suites have Fourier analysis built in, they just call it "spectrum analysis". Record the instrument using an audio recording software such as Audacity. Or, you could just record a cell phone video, then use one of many applications to extract audio from cell phone videos (do a ...


2

The trick is explained here at the end of page 1 : If $f$ is continuous and bounded by $C$, with $$A = \sup_x |(1+x^2) \phi(x)|$$ you get $$\sup_x |(1+x^2) \phi(x)f(x)| \le A C$$ i.e. $$|\phi(x)f(x)| \le \frac{AC}{1+x^2}$$ and $\displaystyle F(\phi) = \int_{-\infty}^\infty f(x) \phi(x) dx$ is a tempered distribution since $$\left|\int_{-\infty}^\infty f(x) ...


0

Let us start by proving the inequality in the hint: For fixed $x\in\mathbb{R}^{n}$, we have $$ \partial_y^\alpha(\tau_x\tilde{\psi})(y) = \partial_y^\alpha(\psi(x-y)) = (-1)^{|\alpha|}(\partial^\alpha\psi)(x-y) $$ and $$ (1+|x-z|^2)^N \leq (1+2|x|^2+2|z|^2)^N \leq 2^N(1+|x|^2+|z|^2)^N \leq 2^N (1+|x|^2)^N(1+|z|^2)^N. $$ Hence $$ \sup_{|\alpha|\leq N,y\in\...


1

This is not possible. Every $v \in W^{2,4}(\Omega)$ satisfies $\nabla v \in C(\bar\Omega)$. Hence, if $\varphi^\varepsilon \in W^{2,4}(\Omega)$ converges in $W^{1,\infty}(\Omega)$ towards $v$, we get $\nabla v \in C(\bar\Omega)$, since the derivatives converge in $L^\infty(\Omega)$. But $\nabla \varphi \in L^\infty(\Omega)$ can be discontinuous.


2

Let $g_n = \hat f_n$. We have that $g_n \to g$ in $L^1$, so by the proof that $L^1$ is a Banach space, there is a subsequence $(g_{{n_k}})$ converging to $g$ a.e. Now, $(g_{n_k})$ is a subsequence of $(g_n)$, so it must also converge to $\hat f$ a.e. as so does $(g_n)$. Thus, $g = \hat f$ a.e. By continuity, we get $g = \hat f$ on $\Bbb R$.


0

You've seen one inequality is obvious but we need the other direction. You can get this directly, constructing $f$ with $|f|\le1$ such that $\int fD_n$ is close to $||D_n||_1$. (Draw a picture of $D_n$ and imagine what $f$ should look like; you want $f=1$ on "most" of the set where $D_n>0$ and $f=-1$ on most of the set where $D_n<0$.) Or you can get ...


2

There is no "obvious proof". The proof is essentially based on the fact that Hilbert transform is $L^p-$bounded (i.e. that $$\|Hf\|_{L^p(\mathbb R)}\leq C\|f\|_{L^p(\mathbb R)}$$ for a suitable constant $C$. The proof of this is not that easy. But this boundness will allow you to show that $$T_Nf(x):=\int_{|\alpha |\leq N}\hat f(\alpha )e^{2i\pi x\alpha }\...


3

As Eric Wofsey has said, such functions need not be bounded. But I think it is important to note as well that if the function is continuous, then it will be bounded. This follows from the fact that $\mathbb{R/Z}$ is compact, and so its image under a continuous map is also compact. Since compact subsets of $\mathbb{R}^N$ are closed and bounded... Anyhow, his ...


3

There are certainly plenty of unbounded functions $f:\mathbb{R}/\mathbb{Z}\to\mathbb{R}$. For instance, let $x_0,x_1,x_2,\ldots\in\mathbb{R}/\mathbb{Z}$ be any infinite sequence of distinct points. Then you could define $f(x_n)=n$ and $f(x)=0$ if $x\neq x_n$ for all $n$. This $f$ is unbounded, since its image is all of $\mathbb{N}$. For a slightly less ...


0

You need to split the integral in to parts, since absolute value is always positive but the what's inside it goes negative when $0 < $x < $\pi$


1

The absolute value can be discarded by "folding" the integration range. It makes no difference to integrate by decomposition $$\pi-x\to \pi x-\frac{x^2}2+C$$ or by substitution $$\pi-x\to-\frac{(\pi-x)^2}2+C$$ because these two expressions just differ in a constant.


3

Both of your questions were answered in short papers by Charles Fefferman: C. Fefferman. On the convergence of multiple Fourier series. Bull. Amer. Math. Soc. 77 (1971), 744-745. C. Fefferman. On the divergence of multiple Fourier series. Bull. Amer. Math. Soc. 77 (1971), 191-195. In the second one he shows that the answer to your second question is no. ...


1

The desired equality is $$\frac1N\biggl[\frac{\sin^2(N\omega/2+\omega/2)}{\sin^2(\omega/2)}-\frac{\sin(N\omega+\omega/2)}{\sin(\omega/2)}\biggr]=\frac1N\biggl[\frac{\sin(N\omega/2)}{\sin(\omega/2)}\biggr]^2,$$ which upon multiplying both sides by $N\sin(\omega/2)^2$ and rearranging a bit simplifies to $$\sin(\omega/2)\sin(N\omega+\omega/2)=\sin^2(N\omega/2+\...


0

$$\hat G(\alpha )=\int_{-\pi/2}^{\pi/2} e^xe^{-2i\pi x\alpha }\mathrm d x=\int_{-\pi/2}^{\pi/2} e^{x(1-2i\pi \alpha ) }\mathrm d x.$$ It doesn't look that hard to compute...


0

The DFT has global self-similarity properties. That is what allows for the FFT factorizations. However each factor in the FFT spreads to affect the whole signal space. To derive a DWT is to find or build a family of functions which have local self-similarity properties to allow for a similar factorization which spreads more locally over the signal space.


1

Orthogonal wavelets allows for analysis and synthesis filters to be the same and only one filter (and it's mirrored alternating signs). For biorthogonal wavelets you need two different filters and their respective mirrored alternating signs. So you have more degrees of freedom in designing biorthogonal wavelets, although the implementation will be slightly ...


1

Why Parseval? It is way easier than that: the integral of an odd integrable function over a symmetric interval with respect to the origin is just zero.


2

Shannon wavelets have dual basis functions resembling the reconstruction functions for Fourier Transforms. If you apply a dyadic subdivision on both frequency bands what you will get is something very similar to the FFT. In this sense is rather the FFT which is a special case of DWT.


0

You look to make same confusion as your previous post about Fourier transformation. Let $S_N(f)$ the partial Fourier series. The convergence of $S_N(f)$ (if it converge to something, it converge to $f$) needs more condition that only $L^2$. In my memories, there is a something with Holder continuity. But, by a theorem, $S_n(f)$ will converge to $f$ in the $L^...


5

We have to prove that the Fourier cosine transform of $e^{-x^k}$ is non-negative and decreasing on $\mathbb{R}^+$. We have: $$ \int_{0}^{+\infty}\cos(\xi x)e^{-x}\,dx = \frac{1}{1+\xi^2},\qquad \int_{0}^{+\infty}\cos(\xi x)e^{-x^2}\,dx = \frac{\sqrt{\pi}}{2}\,e^{-\xi^2/4} \tag{1} $$ hence the claim is expected to hold by some kind of interpolation argument. ...


1

$$\begin{align} \int_{-\infty}^\infty \frac{\sin(t)}{t}\,f(t)\,dt&=\int_{-\infty}^\infty \frac{\sin(t)}{t}\,\left(\frac{1}{2\pi}\int_{-\infty}^\infty \frac{\omega}{1+\omega^4}e^{i\omega t}\,d\omega\right)\,dt\tag 1 \end{align}$$ Enforcing the substitution $\omega \to -\omega$ in the inner integral on the right-hand side of $(1)$ reveals $$\begin{align} ...


3

Hint1: what is the inverse Fourier transform of $\frac{\sin(t)}{t}$, as a function of $w$? Hint2: what is the integral over $\mathbb{R}$ of the product between the previous function and $\frac{w}{1+w^4}$? Hint3: what is the integral over $\mathbb{R}$ of an integrable odd function?


1

Since you want the sine series (if I understand you correctly), you first extend your function to an odd function: $$ \tilde f(x)= \begin{cases} -x & -2\pi<x<-\pi\\ x & -\pi\leq x\leq \pi\\ -x & \pi<x\leq 2\pi \end{cases} $$ Then calculate the Fourier coefficients of this $4\pi$-periodic function as usual: $$ a_n=\frac{2}{4\pi}\int_{-2\...



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