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Both examples are $L^1_{loc}$ functions which are bounded at infinity, thus are tempered distributions, that is continuous linear forms on the Schwartz space $\mathscr S$ of rapidly decreasing functio …

The Schrödinger equation is $ \frac1{i}\frac{\partial}{\partial{t}}-\Delta_{x}. $ It looks similar to the heat equation, but is in fact drastically different. We define a distribution $E$ by the follo …

The function $$ k(x,y)=\frac{H(x)H(y)}{\pi(x+y)},\quad \text{with}\quad H=\mathbf 1_{\mathbb R_+} $$ is the kernel of the Hardy operator which is bounded on $L^2(\mathbb R)$ with operator norm 1.

The question is not meaningful since $\ln x$ is not defined for $x\le 0$. You may define, with derivatives in the distribution sense $$ f(x)=\ln\vert x\vert\text{ (even)},\quad f'(x)=pv\frac1{x} \text …

A version of the uncertainty principle says that a function and its Fourier transform cannot be both with compact support: it is not difficult to prove since a compactly supported distribution has an …

To make sense of your Fourier multiplier, you need only to assume that $p\mapsto G(\vert p\vert^2)$ is a temperate distribution on $\mathbb R^d$. It is true whenever $G$ is a continuous function incre …

A very basic answer: think about the classical Hamiltonian, $$ a(x,\xi)=\vert \xi\vert^2-\frac{\kappa}{\vert x\vert},\quad \text{$\kappa>0$ parameter}. $$ The classical motion is described by the inte …

A strong argument is given above on the heat equation; let me be more specific. The heat equation, one of the most basic in PDE and mathematical physics, already known to Fourier, is $$ L=\frac{\parti …

Consider the most classical example, $D=\frac{d}{dx}-x$, which is the creation operator. Note that $D$ is injective since, with $L^2$ norms and dot-products and say $u$ in the Schwartz class, $$ \Vert …

Let $C$ be a convex subset of $\mathbb R^{2n}$ and $\mathbf 1_C$ be the characteristic function of $C$. Is it true that $$\forall u\in\mathscr S(\mathbb R^n),\quad \langle\mathbf 1_C^{Weyl}u,u\rangle\ …

To solve an ODE of type $ y''=f(y) $ you start by multiplying both sides by $2y'$ so that you get $$ 2y'y''=2 y' f(y)\quad\text{which gives}\quad y'^2 =F(y), \ \text{where}\quad F'=2f, $$ then assumin …

As a rather recent revisitation of Wigner's article, one may also quote James Glimm, who wrote in the article "Mathematical perspectives" (Bull. Amer. Math. Soc. (N.S.) 47 (2009), no. 1, 127–136), In …