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A distribution is a continuous linear functional on the space $\mathcal{C}^{\infty}_c$ of smooth (indefinitely differentiable) functions with compact support. Though they appeared in formal computations in the physics and engineering literature in the late $19^{th}$ century, their formal setting was brought up by the work of S. Sobolev and L. Schwartz in the middle of the $20^{th}$ century.

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Why equality of singular supports?

$\nu$ is a distribution on the real line and the operator $P=d/dt$ is elliptic with constant coefficients. In that case we have $$ \text{singsupp $\nu$}=\text{singsupp $P\nu$} $$ for the $C^\infty$ si …
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1 vote

Proving $\int_0^\infty \sin x/x \, dx=\pi/2$ by test functions and distributions

For $t>0$, we define $ F(t)=\int_0^{+\infty}\frac{\sin x}{x}e^{-tx} dx, $ so that $$ -F'(t)=\int_0^{+\infty} e^{-tx}\sin x dx,\quad F'(t)=\Im{\int_0^{+\infty} e^{-(t+i)x}dx}=-\frac{1}{t^2+1}. $$ Since …
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Distributions more complicated than the Dirac δ and derivatives

Maybe a simple example to demonstrate that, although a distribution is of finite order on each compact, it could be of infinite order, i.e. not of finite order. Consider simply the following distribut …
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3 votes
1 answer
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A sufficient condition for a distribution to be temperate

Claim: Let $T$ be a distribution on $\mathbb R^n$ such that $\nabla T$ belongs to $L^p(\mathbb R^n)$ for some $p\in [1,+\infty]$. Then $T$ is a temperate distribution, i.e. belongs to the topological …
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2 votes

Distributions as derivatives

Maybe more a lengthy comment than an answer. Let me work with temperate distributions (dual of the Schwartz space $\mathscr S(\mathbb R^d)$). Let $u\in \mathscr S'(\mathbb R^d)$ and let us define the …
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Dirac delta composed with absolute value

Let $\kappa:\mathbb R\rightarrow\mathbb R$ be a diffeomorphism with $\kappa(0)=0$. Mimicking the change of variable formula, we would like to have $$ \int \delta(\kappa(x))\vert \kappa'(x)\vert \phi(\ …
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The fourier transform of homogeneous distribution and related topics

In the first place your presentation must be clarified. An homogeneous distribution $u$ of degree $a$ on $\mathbb R^n$ is characterized by $$\forall \lambda >0,\quad u(\lambda x)=\lambda^au(x),\qquad …
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3 votes
1 answer
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Division theorem for vector-valued distributions

The classical division theorem for scalar distributions on $\mathbb R^n$ can be formulated as follows. Let $T$ be a tempered distribution on $\mathbb R^n$ and let $P$ be a non-zero polynomial of $n$ …
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2 votes

Division theorem for vector-valued distributions

I think I have an answer to my own question : let us consider $Q$ the transposed of the comatrix of $P$. The determinant of $P$ is a polynomial and by the Lojasiewicz-Hörmander theorem, we can find a …
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Integral representation of tempered distributions

Let $\mathcal L$ be a continuous linear mapping from $\mathscr S(\mathbb R^n)$ into $\mathscr S'(\mathbb R^n)$. The Laurent Schwartz kernel theorem asserts that there exists $K\in \mathscr S'( \mathb …
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6 votes
2 answers
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A smooth function such that the second derivative of its absolute value is a distribution of...

Let $f\in C^\infty(\mathbb R;\mathbb R)$ and let us define $g(x)=\vert f(x)\vert$. It is easy to verify that $g$ is locally Lipschitz-continuous function, but I would like to find an example of a smoo …
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7 votes
1 answer
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The Schwartz space is not normable

The Schwartz space of rapidly decreasing function (as well as their derivatives) on $\mathbb R^n$ is a Fréchet space, whose (metric complete) topology is given by the usual countable family of semi-n …
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3 votes
1 answer
322 views

Characterization of convex functions

Let $\Omega$ be a convex open subset of $\mathbb R^n$ and let $f:\Omega\rightarrow \mathbb R$ be a convex function. Since $f$ is continuous, it can be considered as a distribution on $\Omega$ and then …
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1 vote

($n$-dimensional) Inverse Fourier transform of $\frac{1}{\| \mathbf{\omega} \|^{2\alpha}}$

Let's start with a simple change of notation. Let me consider first on $\mathbb R^n_x$ the function $f_{\beta}(x)=\Vert x\Vert^{\beta-n}$ for $0<\beta< n$, which is locally integrable and homogeneous …
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2 votes

Solving $x\partial_x f = 0$ over distributions

Let us work in $\mathbb R^n$. The distribution solutions of the equation $$ (x\cdot \partial _x) u=0 $$ are the distributions which are homogeneous of degree 0. Here $x\cdot \partial_x=\sum_{1\le j\le …
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