# Tag Info

4

In general, no. For example, take $f(\xi) = 1$. There is a generalization of the fourier transform, the Fourier–Stieltjes transform, and that maps a dirac delta to $f(\xi) = 1$. So, in general, this inverse could be a distribution. $L^1(\mathbb{R})$ is pretty much the largest class of functions which we can define the fourier transform for. The fourier ...

3

There are many possible conventions, but you can't get around the fact that a $2\pi$ will appear somewhere. In order to have a way of comparing different conventions easily, I wrote up restrictions on the various coefficients in the definitions and comparison between different choices.

2

Let $f_{n}(x)=e^{inx-x}$ on $[0,\infty)$ for $n=0,\pm 1,\pm 2,\pm 3,\cdots$. The Fourier transform of $f_{n}$ is $$\int_{0}^{\infty}e^{ix\zeta}f_{n}(x)\,dx =\int_{0}^{\infty}e^{inx-x}e^{ix\zeta}\,dx = \frac{1}{1-in-i\zeta}=\frac{i}{\zeta+i+n}.$$ The function $f_{n}$ is in $L^{2}[0,\infty)$ with $\|f_{n}\|=1$ for all $n$. Choose any ...

2

The fact that $f$ is in $H^1$ implies that its derivative $Df$ is in $L^2$. The Fourier coefficients of $Df$ are (up to a constant) $\{n |f_n|\}$. The Parseval relationship then implies that $\{n |f_n|\} \in \ell_2$. Since $\{n |f_n|\} \in \ell_2$ the Cauchy-Schwartz inequality implies \begin{align*} \sum_n |f_n| &= \sum_n n^{-1} n |f_n| \\ ...

2

This is an illustration of using DFT to find the product of two polynomials. It is based on the fact that Fourier transform (discrete or continuous) converts convolution into multiplication. The product of two polynomials amounts to convolution of their coefficients: $$\left(\sum_{j=0}^n a_j x^j\right)\left(\sum_{k=0}^n b_k x^k\right) = \sum_{m=0}^{2n} ... 2 Yes. x \rho_t(x) is smooth and compactly supported, and any such function f is in H^s for every s. The easiest way to see this may be this: Let k be any integer larger than s. Verify from the definition that H^k \subset H^s. (Note that (1+|\xi|^2)^s \le (1+|\xi|^2)^k). Using the fact that the Fourier transform takes differentiation to ... 2 If \widehat{\mu} is non-negative, then its square root defines a bounded Fourier multiplier operator S on L^2. In this case S is self-adjoint, so T=S^*S. Conversely, any linear and shift invariant S is a Fourier multiplier operator with an L^\infty symbol. The adjoint S^* would have a symbol that is the complex conjugate of the symbol for ... 1 As you have observed that, g should be, (f\ast f)^{\vee}. Next, observe that, g= f^{\vee}\cdot f^{\vee}= (f^{\vee})^{2}. Since f\in L^{2}(\mathbb R), we have f^{\vee}\in L^{2}(\mathbb R) (by Planchrel) and therefore (f^{\vee})^{2}\in L^{1}(\mathbb R); and so g\in L^{1}(\mathbb R). On the other hand, Given g\in L^{1}(\mathbb R), to find ... 1 A periodic function is a function that repeats its values in regular intervals or periods. We say that a function has a period of L if$$f(x) = f(x + L)$$for all x in the domain of f(x). For example, the function \sin(x) has period 2 \pi since \sin(x) = \sin(x + 2 \pi) (as you can easily verify from the graph). The function returns to the same ... 1 You just need to solve the equation$$f(x) = f(x + T)$$for T using all available means suitable for solving the equation you get. 1 Only large frequencies matter for smoothness. For every M, the part of Fourier transform with \{\xi:|\xi|\le M\} contributes a real-analytic term to the function. You know that integrability of |\xi|^\alpha \hat u(\xi) implies certain smoothness of u. So you want to show that this product is integrable for every \alpha. On every ball \{|\xi|\le ... 1 This is a factor for norming. See the wikipedia article on Fourier Transform for a list of common constants. The one I see more often uses$$\hat f(\xi) := (2\pi)^{-\frac n2} \int_{\mathbb R^n} f(x) e^{-ix\cdot \xi} \mathrm dx \\ \check f(x) = (2\pi)^{-\frac n2} \int_{\mathbb R^n} f(\xi) e^{ix\cdot \xi} \mathrm d\xi Where $\cdot$ denotes the inner product. ...

1

It's a misnomer to say "The DFT is periodic". It appears that you are learning about the DFT from a signal processing or engineering book. They teach the DFT the way you have explained, which can lead to confusion. A much more clear way to think about the DFT is to think of it as a linear transformation from a finite vector space to a finite vector space. ...

1

In fact, $\alpha\beta$ is optimal. To prove it, let $f,g:[0,1]\to\mathbb{R}$ with $f(x)=x^\alpha$ and $g(x)=x^\beta$. Note that $f\circ g\in C^{\alpha\beta}$, however, for all $\gamma>\alpha\beta$, $f\circ g$ does not belong to $C^\gamma$.

Only top voted, non community-wiki answers of a minimum length are eligible