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Banach spaces, function spaces, real functions, integral transforms, theory of distributions, measure theory.
41
votes
Motivation for and history of pseudo-differential operators
Let us follow the history of classical PDE with a few (hopefully) well-chosen examples illustrating the rôle of pseudodifferential operators in classical analysis.
(1) Before 1950. Prehistory. A lon …
22
votes
Pullback measures
A simple-minded answer. The push forward of a measure is a triviality: take a measure space $(X,\mathcal M, \mu)$ and a mapping
$f:X\rightarrow Y$. Then defining $\mathcal N=${$B\subset Y, f^{-1}(B)\ …
11
votes
2
answers
701
views
Poincaré lemma for distributions
Let us consider a current on $\mathbb R^n$, that is a differential form whose coefficients are distributions. For simplicity, let us check the case of a $1$-form
$$
u=\sum_{1\le j\le n} u_j dx_j,\quad …
9
votes
3
answers
677
views
Lipschitz-free spaces of $\mathbb R^n$
We define
$$
\text{Lip}_0(\mathbb R^n)=\{f:\mathbb R^n\rightarrow \mathbb R, \text{such that $f(0)=0$ and }
\sup_{x\not=y}\frac{\vert f(x)-f(y)\vert}{\vert x-y\vert}<+\infty.
\}
$$
It is well-known th …
8
votes
Accepted
Gagliardo-Nirenberg inequality
The mother of all Gagliardo-Nirenberg inequalities is
$$
\Vert u\Vert_{L^{\frac{n}{n-1}}(\mathbb R^n)}\le c_n \Vert \nabla u\Vert_{L^{1}(\mathbb R^n)},
\tag {GN}
$$
where $c_n$ depends only on $n$ and …
8
votes
Why is multiplication on the space of smooth functions with compact support continuous?
The spaces $C^\infty_c(\mathbb R^d)$ and $C^\infty_c(\mathbb R^d)\times C^\infty_c(\mathbb R^d)$ are $LF$ spaces (inductive limit of Frechet spaces) and their standard topologies are not metrizable. W …
8
votes
PDEs and algebraic varieties
A most important result is missing in the previous answers, namely the characterization by Lars Hörmander of hypoellipticity in his seminal paper,
On the theory of general partial differential operato …
7
votes
when a pseudo-differential operators to be compact?
First a simple remark: in the formulation of the question $\mu$ should be replaced by $\mu/2$ to get
$S(m,g)=S^\mu_{1,0}$.
Next the "if and only if" is correct but misleading since it is a conditio …
7
votes
Accepted
Integral representation of tempered distributions
Let $\mathcal L$ be a continuous linear mapping from $\mathscr S(\mathbb R^n)$ into
$\mathscr S'(\mathbb R^n)$. The Laurent Schwartz kernel theorem asserts that there exists $K\in \mathscr S'( \mathb …
7
votes
2
answers
467
views
Eigenstates of Fourier transformation
Let $\gamma$ be defined on $\mathbb R^n$ by $\gamma (x)=e^{-π x^2}$. With $\mathcal F$ standing for the Fourier transformation defined on the Schwartz space by
$$
(\mathcal F u)(\xi)=\int e^{-2iπ x\cd …
6
votes
0
answers
159
views
Real interpolation space between the Wiener algebra and $L^2$
The Wiener algebra $W_n$ is the image by the Fourier transform of $L^1(\mathbb R^n)$. What is the (complex) interpolation space between $W_n$ and $L^2(\mathbb R^n)$? It is probably not true that for $ …
6
votes
Accepted
Hormander's bracket condition for the adjoint of an operator
The hypoellipticity result is more precise:
you have
$$
Lu \in H^s_{loc}\Longrightarrow u\in H^{s+2-\delta}_{loc}\quad\text{ for some $\delta\in [0,2)$,}
$$
and that $\delta$ is linked to the number o …
6
votes
Accepted
When is the adjoint of a hypoelliptic operator also hypoelliptic?
Hormander's operator $L=X_0+\sum_{1\le j\le k} X_j^2$, where the $X_j$ are real smooth vector fields with the Lie algebra of $\{(X_j)\}_{0\le j\le k}$ generating the tangent space is hypoelliptic as …
6
votes
Are Besov spaces $B^{s}_{p,q}$ invariant under Fourier transform?
Let me note $\phi_k(D)$ the Fourier multiplier $\phi_k(\xi)$, i.e.
$
\text{Fourier}\bigl(\phi_k(D)u\bigr)(\xi)=\phi_k(\xi)\hat u(\xi).
$
$\bullet$ The answer to (1) is yes since
$$
\Vert{u}\Vert_{L^1 …
6
votes
0
answers
369
views
Paving conjecture for Toeplitz matrices
Let me first recall what is the so-called paving conjecture:
for any $\epsilon >0$, there exists $r\in \mathbb N$ such that
for any bounded operator $A$ on $\ell^2(\mathbb Z)$, there exists a partitio …