# Tag Info

3

By definition we can write for any test function $\phi$ $$\langle x\delta_0,\phi(x)\rangle = \langle \delta_0, x\phi(x)\rangle = 0\cdot \phi(0)=0=\langle 0,\phi(x)\rangle,$$ which means that $x\delta_0=0$ in the sense of distributions. In the same spirit, we write $$\langle x\delta'_0,\phi(x)\rangle = \langle \delta_0', x\phi(x)\rangle =- \langle ... 2 The expression " f(x)\delta(x-x_0) = f(x_0)\delta(x-x_0) " is a shorthand. It doesn't mean (in any ordinary sense) that If g and h are functions g(x) = f(x)\delta(x-x_0), h(x) = f(x_0)\delta(x-x_0) then g = h in that g(x) = h(x) for all x. To start, neither g and h are functions \mathbb R \to \mathbb R, because the delta ... 2 A function in \mathcal{S} decays faster than the reciprocal of any polynomial at infinity and is bounded. (This is usually built into the definition.) So its magnitude is bounded by some M_1 on the ball of some radius R centered at the origin, and by M_2/|x|^N outside this ball. Choose N sufficiently large (depending on the dimension and p) to ... 1 For your first question, the dirac delta function \delta(x) is undefined at x=0, you can think of it as being 'infinite' there, but this isn't strictly true. Thus f(x) is undefined at x=\frac{1}{2}. Your intuition in terms of 'blips' is correct however. For your second question, note that:$$\int_{-\infty}^{+\infty}\delta(x)dx=1,$$but ... 1 You have$$\int_{\Bbb R^d} u(x)\phi(x)dx = \int_{\Bbb R^d} \hat u(\xi)\bar{\hat \phi}(\xi)d\xi =\int_{\Bbb R^d} \hat u(\xi) (1+|\xi|^2)^{-s/2} \bar{\hat \phi}(\xi)(1+|\xi|^2)^{s/2}d\xi. $$Now you have \hat u(\xi) (1+|\xi|^2)^{-s/2} is a L^2 function. Take, for example, arbitrary \phi\in S, therefore, functions of the form \bar{\hat ... 1 First note that the integral you've written (that the author assumes is zero) is precisely the inner product \langle \phi, u\rangle_{H^s}. Now pick an element u in the orthogonal complement of D in H^s. Our goal is to prove that u is zero, and hence \overline D = H^s. (This implies the result by the standard fact that for a Hilbert space, ... 1 I am going to gloss over the subtleties of the dirac-delta function which isn't quite a function and get straight to the essential problem: Theorem: if \int_a^b f(x)\,dx = \int_a^b g(x)\,dx, then f and g are equal (almost everywhere). This is what I want to convince you is generally true. In order to do so, I'm going to begin with the following ... 1 You can show that$$\int_a^b f(x)\delta(x-x_0) dx = \int_a^b f(x_0)\delta(x-x_0) dx$$For all real a and b. If a< x_0 <b you get f(x_0) at each side. Otherwise 0 1 The intuitive way to understand this is with the "informal" definition of the dirac delta: It's a "function" (Really a generalized function/distribution) with the property that it is 0 everywhere except at 0, and a sort of "infinity" at 0. The specifics of the definition of the infinity is such that$$\int _a ^b \delta (x)dx=1$$if and only if$$o\in ...

1

For a distribution $u$ and polynomials $P,Q$, we define the distribution $P(M_{{\rm id}_{\mathbb{R}}}) Q(\partial)u$ as $$\tag{1} \left(P(M_{{\rm id}_{\mathbb{R}}}) Q(\partial)u\right) [\varphi]~:=~ u\left[ Q(-\partial)P(M_{{\rm id}_{\mathbb{R}}})\varphi\right],$$ where $\varphi$ is a test function; where $M_f(g):=fg$ denotes the multiplication operator ...

1

Consider, for fixed $s$, the function $$g_s(\tau) = \exp(\tau + s)\sinh \tau$$ we have $$g_s'(\tau) = \exp(\tau +s)(\sinh\tau + \cosh\tau) = \exp(2\tau + s)$$ Hence, $g_s$ is monotone, therefore we have by substitution for $\phi \in C_c^\infty(\def\R{\mathbf R}\R)$: \begin{align*} \int_\R f_1(s,\tau)\phi(\tau)\, d\tau &= \int_\R ...

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