# Tag Info

3

The following can be found in a table of transforms: Let $f$ have the Fourier transform $\hat{f}$. $x \mapsto f(ax)$ has Fourier transform $k \mapsto {1 \over |a|}\hat{f}({k \over a})$. $x \mapsto f(x-a)$ has Fourier transform $k \mapsto e^{-iak}\hat{f}(k)$. $x \mapsto e^{i a x} f(x)$ has Fourier transform $k \mapsto \hat{f}(k-a)$. The first two can be ...

3

Let $D=\frac{d}{dx}$ be the operator of differentiation with respect to $x$. Then $$(D-2\pi i\xi)e^{2\pi i\xi x}=0.$$ For any other $\xi'$, $$(D-2\pi i\xi')e^{2\pi i\xi x}=2\pi i(\xi-\xi')e^{2\pi i\xi x} \ne 0.$$ Therefore, \begin{align} 0 &= (D-2\pi i\xi_2)(D-2\pi i\xi_3)(\cdots)(D-2\pi ...

2

Suppose $\sigma: H\to\mathrm{GL}(W)$ is a complex irreducible representation of $H$. Call the trace $\chi$. Then character theory says that $\sum_{h\in H}\chi(h)$ equals $|H|$ if $\sigma$ is trivial and $0$ otherwise. Additionally, the sum $\sum_{h\in H}\sigma(h)$ is an interwtining operator, and Schur's lemma says the intertwining operators of a complex ...

2

Your computation of the $L^1$ norm is incorrect. Since $$\|Pf\|_1 = \int_{\mathbb R^n} |Pf(x)| \, dx = \int_{\mathbb R^n} \left| \int_{\mathbb R^m} f(x,y) \, dy \right| \, dx \le \int_{\mathbb R^n} \int_{\mathbb R^m} |f(x,y)| \, dy \, dx = \|f\|_1$$ you get $Pf \in L^1(\mathbb R^n)$ and the desired inequality. As for the Fourier transform, your formula is ...

1

HINT: Take $\phi\in \mathcal S'$ (a continuous linear functional) with the property that $$\tag{1}\langle \phi, G\rangle = 0\qquad \forall G(x)=Ae^{\frac{|x-x_0|^2}{2\sigma^2}},$$ where $x_0\in\mathbb{R}^n, A\in\mathbb{R}, \sigma >0$ are arbitrary. Claim (to be proved): $\phi=0$. Once this is proved, the exercise is done by standard duality arguments. ...

1

$L^1$ is dense in $L^2$. For example, if $f \in L^2$, then $f_R=\chi_{[-R,R]}f \in L^1\cap L^2$ where $\chi_{[-R,R]}$ is the characteristic function of the finite interval $[-R,R]$; and $\lim_{R\rightarrow\infty}\|f-f_R\|_{L^2}=0$. The operators $\tau_y$, multiplication by $e^{-2\pi j\xi y}$ and the Fourier transform are isometric on $L^2$. Therefore, $L^2$ ...

1

It's important that on the first page the paper states that, without loss of generality, all of their irreducible representations are unitary, so that $\rho(g)^\dagger=\rho(g)^{-1}$ for all $g\in G$. By the way, the beginning of section $3$ is in error. The sum $\sum_{h\in H}\rho(h)$ is not a projection, because it is not normalized correctly. Instead, ...

1

When considering a function $f(t,\mathbf x)$ on $[0,\infty)\times \mathbb{R}^n$, we can focus on one time slice at a time, fixing $t$ and dealing with a function of $\mathbf x$ only. The Fourier transform can be applied to this slice, since it's just a function on $\mathbb{R}^n$. The result can be denoted $\tilde f(t,\mathbf k)$, and is usually called the ...

1

$\sum_{1\le i,j\le d_\rho} |\hat{f}(\rho)_{i,j}|^2 = ||\hat{f}(\rho)||^2$ is just the definition of the Frobenius norm of the matrix $\hat{f}(\rho)$.

1

The function $x\cos(x)$ is an odd function on $(-\pi,\pi)$, which means that its Fourier series on $(-\pi,\pi)$ is a sin series. The function $x\cos(x)|_{x=-\pi}=\pi$ and $x\cos(x)|_{x=\pi}=-\pi$. So the Fourier series converges to $0$ at $x=\pm \pi$; otherwise it converges to $x\cos(x)$ on $(-\pi,\pi)$. Using trig identities $$\sin(nx+x) = ... 1 Take any nontrivial, nonnegative, symmetric function g \in C_c^\infty (\Bbb{R}^d). If we let h := \mathcal{F}^{-1}(g), then h is real-valued (why?) and$$ h(0) = \int g(x) \, dx > 0, $$since g \geq 0 and g \not \equiv 0. By continuity of h and by rescaling (i.e., replace g by Cg for some large C>1), we get h \geq 1 on B_{2\delta} ... 1 Plugging in n=1, the integral becomes$$ K(x) = \int_0^{\infty} t^{-\frac{1}{2}} e^{-t-\frac{x^2}{4t}} \,dt.$$Taking a Fourier transform, we have$$\mathcal{F}K(y) = \frac{1}{\sqrt{2\pi}} \int_0^{\infty} t^{-\frac{1}{2}} e^{-t}\int_{-\infty}^{\infty} e^{-ixy} e^{-\frac{x^2}{4t}}\,dx\,dt.$$The inner integral is the Fourier transform of the Gaussian ... 1 Your approach is perfectly appropriate. There was, however, a flaw in the execution of that way forward. Note that we have$$\begin{align} F(k)&=\langle H(a-|x|),e^{-ikx}\rangle\\\\ &=\frac{1}{-ik}\langle H(a-|x|),\frac{de^{-ikx}}{dx}\rangle\\\\ &=\frac{1}{-ik}\langle \text{sgn}(x)H'(a-|x|),e^{-ikx})\\\\ &=-\frac{1}{ik}\langle ...

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