13

A classical connection is the Atiyah-Jänich Theorem, see
Klaus Jänich: Vektorraumbündel und der Raum der Fredholm-Operatoren. Math. Ann. 161 (1965) 129–142.
Let ${\mathcal{F}}$ be the space of Fredholm operators with the operator norm, and $X$ any compact space. To a map $F\colon X\to {\mathcal{F}}$ one can associate the virtual vector bundle $(ker(F(x)))...

at.algebraic-topology fa.functional-analysis oa.operator-algebras kt.k-theory-and-homology integral-operators

8

Yes it is possible. Let $K\in S'(\mathbb{R}^2)$ be the distribution acting on test functions $f(x,y)$ by
$$
K(f)=\int_{|x-y|\ge 1} \frac{f(x,y)}{|x-y|}\ dx\ dy
\ + \int_{|x-y|< 1} \frac{f(x,y)-f(x,x)}{|x-y|}\ dx\ dy\ .
$$
What is impossible is to do this extension to the diagonal while preserving the degree -1 homogeneity. The only homogeneous ...

answered Jun 29 '17 at 16:49

Abdelmalek Abdesselam

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7

Surprisingly (to me), the statement is false. My counterexample is a little messy, but the idea is fairly simple.
Take $T = 1$ and set $a_n = \frac{1}{n}$ and $b_n = 1 - \frac{1}{n}$ for $n \in \mathbb{N}$. Define $f$ by setting $f(t) = 1$ on the intervals $[a_{2n+2}, a_{2n+1})$ and $f(t) = -1$ on the intervals $[a_{2n+1},a_{2n})$. The point of using $\frac{...

7

Such an integral over a simple (non-closed) curve is called the Cauchy type
integral.
The curve $\gamma$ is oriented, so for a function $F$ defined in $C\backslash\gamma$ and $z\in\gamma$ different from an endpoint,
we can talk of the right limit $F^+(z)$ and left limit $F^-(z)$.
Then we have Sokhotski formula
$$F^-(z)-F^+(z)=f(z),\quad z\in\Gamma$$
As in ...

answered Dec 4 '18 at 15:58

Alexandre Eremenko

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6

We show below a slightly more general claim (to simplify notation, I'll write only in terms of matrices).
$\newcommand{\reals}{\mathbb{R}}$
Def. We say a kernel $\psi: X \times X \to \reals$ is negative definite (nd) if $\sum_{ij}c_ic_j\psi(x_,x_j) \le 0$ for all $c$ such that $\sum_i c_i = 0$.
Lemma. Let $\psi$ be cpd. Fix some $x_0 \in X$. Then, $$k(x,...

6

Nemo's representation of $F(\eta)$ in terms of a hypergeometric function can be evaluated without difficulty for large $\eta$:
$$F(\eta)=\frac{\sqrt{3} {\eta}^2 \Gamma \left(\frac{2}{3}\right) \; _1F_2\left(\frac{2}{3};\frac{4}{3},\frac{5}{3};-\frac{{\eta}^3}{27}\right)}{6\pi }-\frac{12 {\eta} \; _1F_2\left(\frac{1}{3};\frac{2}{3},\frac{4}{3};-\frac{{\eta}^3}...

answered Feb 10 at 20:03

Carlo Beenakker

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5

(Edit: see below... I misread the question...) One particular way to regularize that integral, relating to meaningful things, is by viewing it as Hilbert transform applied to one Schwartz function, producing (at worst) a tempered distribution, which is then applied to the second Schwartz function.
Part of the point is that the Hilbert transform, or already ...

5

For the first question, the answer is not necessarily.
Very rough idea: The rank-nullity theorem doesn't always hold on infinite dimensional spaces.
Rough idea: Let the operator $A$ be defined on $L^2(M)$ be a injective mapping such that its range does not include the constant function. More precisely, since $M$ is compact we can enumerate its ...

5

You said in comments that you are OK with using orthogonality of matrix coefficients, but let's suppose you changed your mind. Let $\langle\cdot, \cdot\rangle$ be a $G$-invariant pairing on the space $V$ of $\pi$, and let $\mathrm dg$ be the Haar measure on $G$ normalised to give it total mass $1$. What follows is all very standard, so I assume it's not ...

4

One has, for $f,\,g\in \dot{H}^{-1/4}(\mathbb R)$,
$$
\begin{aligned}
\left|\int_{-\infty}^\infty\int_{-\infty}^\infty |x-y|^{-1/2}f(x)
\,\overline{g(y)}\,dx dy\,\right| &= C_0 \left|\left((-\Delta)^{-1/4}f,g\right)
\right| \\
& \leq C_0\|(-\Delta)^{-1/4}f\|_{\dot{H}^{1/4}}\|g\|_{\dot{H}^{-1/4}}
= C_0\|f\|_{\dot{H}^{-1/4}} \|g\|_{\dot{H}...

4

Do you mean the Airy transform?
A more recent reference is The Airy transform and the associated polynomials (2010).
In your notation the function $F_3$ is the Airy transform of the Fourier transform $\hat{f}$ of $f$:
$$F_3(x)=\int_{-\infty}^\infty dt\,\exp\left(\tfrac{1}{3}it^3+itx\right)f(t)$$
$$\qquad=\frac{1}{2\pi}\int_{-\infty}^\infty d\xi\,\int_{-\...

answered Apr 20 '17 at 8:28

Carlo Beenakker

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4

Willie Wong answered the general case, and I'd like to give a counterexample for the symplectic case:
Let $M=S^2$, the standard 2-sphere embedded in $\mathbb{R}^3$ as the unit sphere, with the standard symplectic form corresponding to the Euclidean metric on $\mathbb{R}^3$. Take on $(M \times M, \Omega = \omega \oplus - \omega)$ the functions
\begin{align}
...

4

I am going to elaborate on Nemo's comment. You first need to note that $\frac{d^2}{dx^2}e^{|x|} = 2\delta_0(x)\cdot e^{|x|} + e^{|x|}$ in the sense of distributions, where $\delta_0(x)$ represents the dirac delta function at $0$. This follows from a simple computation and noting that $\frac{d}{dx}\mathrm{sign}(x) = 2\delta_0(x)$.
Using the above fact, we ...

4

If $\lambda=1$ is an eigenvalue, then $\int_0^1 K(s,x)u(s) ds=0$ for the corresponding eigenfunction. Hence a condition to rule this out is that $\{K(s,x),x \in (0,1)\}$ spans a dense subspace of $L^2$.

4

Your operator $K$ is a Hilbert-Schmidt operator since its kernel belongs to $L^2$. As a result this is a compact operator whose spectrum contains a sequence of eigenvalues $\\{\lambda_k\not=0\\}$ with finite multiplicities such that $\lim_k\vert \lambda_k\vert=0$. To deal with the self-adjoint case, you can find an orthonormal set $\\{\mathbf e_k\\}$ such ...

3

Let's first make the interval equal to $(-1,1)$ by the change of variable
$$
x = \frac{a+b}{2} + \frac{b-a}{2}\, t ;
$$
then the equation becomes
$$
\int_{-1}^1 u(s)[c+\log |s-t|]\, ds = g(t) , \quad\quad\quad\quad (1)
$$
with $c=\log (b-a)/2$ and $g=2f/(b-a)$ (and I use the sloppy but convenient notation where $u(s)$ really means $u(x(s))$, and similarly ...

answered Sep 4 '16 at 4:01

Christian Remling

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3

This looks like a convex optimization problem. Fix a convex set $D \subseteq \mathbb{R}^N$. Fix parameters $t_1<t_2$. Let $x=(x_1, \ldots, x_N)$. Define the set $\mathcal{A}$ and function $g:\mathcal{A}\rightarrow \mathbb{R}$ by:
\begin{align}
\mathcal{A} &= \left\{x \in D : t - \sum_{i=1}^N x_i f_i(t)>0 \quad \forall t \in [t_1, t_2]\right\} ...

3

Well, the answer, if there is any, must be subtle. Take the following example. Let $H$ be the Hilbert transform. Then
$$\frac{d}{dx}Hu=f$$
is an integral equation, where the kernel is the Fourier transform of $\xi\mapsto|\xi|$. It cannot be rewritten as a set of differential equations (ODEs) within $\mathbb R$. Nevertheless, it can be recast in terms of PDEs,...

answered Jan 10 '17 at 19:58

Denis Serre

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3

Or: more generally, if $\phi$ is a continuous convex function on a Banach space $E$, and $f : [0,T] \to E$ is continuous, then we have
$$
\phi\left(\frac{1}{T}\int_0^T f(t)\;dt\right) \le \frac{1}{T}\int_0^T \phi(f(t))\;dt
$$
And $[0,T]$ can be replaced by a more abstract finite measure space, continuity of $f$ by appropriate measurability.

3

I am not sure that it is a proper question, but one has to excercise in integration once in a while, so:
First of all, choose coordinates so that vector $v$ is directed along $x$-axis. Then define $\Omega\subset\mathbb{R}^2$ to be the disc $||r||< a$, and function $f(r)=\cos(2v\cdot r)$.
$$
I=\int_\Omega \cos^2(v\cdot r)d^2r=\frac{\pi a^2}{2}+\frac{1}{2}\...

3

Your operator $T$ is a Fourier multiplier with symbol $m = \mathcal{F}[Y^m_k/r^n]$; that is, $\mathcal{F}[Tu]=m \mathcal{F}[u]$.
It is a relatively simple exercise to show that the norm of $T$ on $L^2$ is equal to $\|m\|_\infty$, the essential supremum of $|m|$. In other words, $C(n)=1$. Therefore, the supremum norm of $m$ cannot be change to any other norm....

answered Sep 10 '17 at 19:34

Mateusz Kwaśnicki

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3

The book by Melrose was almost exactly written to address issues like the ones you mentioned above, where the very first example is the one in the question. The construction fo the parametrix is the focus point of the book. However the book is supposed to be very dense and difficult to read, and it also has a fair amount of typos. Now there are more reader ...

3

By complementing Deane Yang's strategy with Jochen Wengenroth's observations in the related question Composition of a smoothing operator with an $L^2$-bounded operator, non-compact Riemannian manifold, I can now write a fully detailed answer to my question.
Theorem. Let $M$ be a smooth manifold equipped with a smooth measure $\mathrm{d}\mu$ (arising from ...

3

What you want to do is the pull-back of distributions. And there is a theorem (cf. Hörmander 1, Theorem 8.2.4) that if the set $\{(\phi(x),\eta) \colon \phi'(x) \eta = 0\}$ and $\operatorname{WF}(\delta) = \{ (0,\eta) \colon \eta \not = 0\}$ have empty intersection, then the pull-back is well-defined.
If you are only interested in the delta-Distribution, ...

3

You can immediately translate this integral equation into an easy second order linear ODE on $[0,1]$ with boundary conditions $x(0)=0$, $x(1)=1$. (I'm not going to do it for you). Just note that your integrand $ts-\min(t,s)$ is the Green function of the Laplacian in dimension $1$ (i.e. the second derivative) with Dirichlet boundary condition: one has
$$\...

3

$\newcommand{\la}{\lambda}
$
Let us formalize the question as follows:
Take any complex number $\la$. Let $K(s,t):=s\wedge t-st$ for $s,t$ in $[0,1]$. Find $x\in L^2[0,1]$ such that
\begin{equation*}
x(t)-\la \int_0^1 K(s,t)x(s)\,ds=t \tag{1}
\end{equation*}
for almost all $t\in[0,1]$.
Note that the kernel $K$ is the covariance function of the ...

3

The inverse operator is $L^{-1} = 1/2 \cdot (x-1)^{-1} d^2/dx^2 - 1/2 \cdot (x-1)^{-2} d/dx$. Its eigenfunctions are derivatives of Airy functions, $Ai^{\prime } ((2\lambda )^{1/3} (x-1))$, $Bi^{\prime } ((2\lambda )^{1/3} (x-1))$, where $\lambda $ are the corresponding eigenvalues. Without further specification of the space of functions $f$, i.e., boundary ...

2

I used Mathematica, doing first the theta integration. The hypergeometric function that arises is actually a Bessel function, but Mathematica does not find that on its own. The final result I get is
$${\pi a^2\over 2}(1+{J_1(2a\sqrt{v_x^2+v_y^2})\over a\sqrt{v_x^2+v_y^2}}).$$

2

It holds more generally replacing $L^p(\Omega)$ by a normed space $(B,\lVert \cdot \rVert_B)$, assuming $M\colon t\mapsto X_t(\cdot)$ goes from $[0,T]$ to $B$ is continuous.
We have for each integer $N$ that
$$\left\lVert\frac TN\sum_{j=1}^NX_{TkN^{-1}}(\cdot)\right\rVert_B\leqslant\frac TN\sum_{j=1}^N\left\lVert X_{TkN^{-1}}(\cdot)\right\rVert_B.$$
As $...

2

I will give a partial answer. If you don't like it I can delete it. I might also be restating what fedja is saying, in which case I apologize. It isn't clear to me that this is what he/she had in mind though.
For $|\lambda| > 1$ the answer is no. We can implement a fixed point argument to show this.
First, let $X = \{f\in L^2([0,1] \; : \; f(1-x) = -f(x)...

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