7 votes
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Injectivity of a Fredholm operator

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 \...
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  • 37.4k
5 votes

Eigenfunctions and eigenvalues of an operator defined by a certain integral

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^{\...
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5 votes
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Every self-adjoint trace class operator on $L^2$ has integral kernel

Yes, this is correct. Actually, "self-adjoint trace-class" is more than you need; any Hilbert-Schmidt operator can be represented as an integral operator. The Hilbert-Schmidt operators from $L^2(X)$ ...
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5 votes
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Is the solution to the Fredholm integral equation of the first kind a continuous analogue of Cramer's rule for matrix equations?

Denote the Fredholm determinant $\Delta=\det(I+K)$, then the solution of the Fredholm equation $g=(I+K)f$ is given formally by $$f=(I+K)^{-1}g=\frac{1}{\Delta}\left(\frac{\partial\Delta}{\partial K}\...
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4 votes
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Hardy-Littlewood-Sobolev inequality using fractional sobolev norm on the RHS

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((-\...
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  • 1,163
4 votes

Asymptotic expansion of the Schrödinger kernel?

No, in general, the expansion in small times of the heat kernel $k_t(x,y)$ does not tell much about the Schrodinger semigroup. You may think about the circle case for which the kernel of $e^{t \Delta}...
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4 votes

Techniques to solve equations involving a definite integral

The problem space of symbolic computation on definite integrals is currently fairly open. The Risch algorithm answers the question if there is a closed form solution to a indefinite integral (assuming ...
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  • 223
4 votes
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Improper integral $\int^\infty_0 e^{-a x^2} \cosh (b\sqrt{1+x^2})$

Let us give only an expansion in $a,b$. Calling $I(a,b)$ the integral, we get easily $$ I(a,b)=\sum_{k\ge 0}\frac{b^{2k}}{(2k)!}\underbrace{e^{a}\int_0^{+\infty} e^{-a (x^2+1)}(1+x^2)^k dx}_{J_k(a)}. $...
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  • 13.2k
4 votes

Injectivity of an integral operator

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 ...
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  • 13.2k
4 votes
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integral kernel function for the SU(N) group

The integral kernel for ${\rm U}\,(N)$, due to Dyson, has been generalized by Katz and Sarnak to other compact groups (Random Matrices, Frobenius Eigenvalues, and Monodromy, page 121). Their result ...
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4 votes
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Solution set of integral equation/ Kernel of linear operator

I think it's easiest to work in the Hilbert space setting for this problem, i.e., to consider $F$ is a functional on the space $L^2([0,1]^2)$, where $[0,1]^2$ is endowed with the Lebesgue measure. Let ...
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4 votes

General strategy for studying the decay of eigenvalues of kernel integral operators

Birman and Solomyak have studied this question quite intensivly. The paper may not be the easiest to understand, but it does cover in a very general setup, what regularity conditions on the kernel ...
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4 votes

Calculation of an inverse Mellin transform

Because of the identity $$\frac{d}{dz}\, _1F_1(b;b-a;z)=\frac{b }{b-a}\, _1F_1(b+1;b+1-a;z)$$ the function $K(x)$ is given by $$K(x)=-\frac{(b-a)\Gamma(a)}{b\Gamma(b)}f(a,b,x)$$ with $f(a,b,x)$ the ...
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3 votes
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Inverting convolutions over finite intervals

A modification of the Wiener-Hopf method for this type of problems is described in Convolution equations on finite intervals and factorization of matrix functions and in Finite interval convolution ...
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3 votes

Improper integral $\int^\infty_0 e^{-a x^2} \cosh (b\sqrt{1+x^2})$

Although there is no closed form in terms of elementary functions, $$\int^\infty_0 e^{-a x^2} \cosh (bx)dx=\frac12\exp\bigg(\frac{b^2}{4a}\bigg)\sqrt\frac\pi a$$ should make for a decent lower limit. ...
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  • 645
2 votes
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Error estimate in the spectral theorem of compact operators on a Hilbert space

I'm following Szego's book on orthogonal polynomials. In chapter III he considers $f \in L^2\left((a,b),d\alpha \right)$, with $-\infty \leq a < b \leq \infty $. There exist a set of orthogonal ...
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  • 3,200
2 votes
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Is there a way to solve this integral equation?

There is no unique solution to the integral equation $$\int\limits_0^\infty g(\kappa, x_0) \exp{\left[-\frac{\left(\xi + \kappa \right)^2}{2\alpha\theta}\right]} d\kappa % = \exp{\left[-\frac{\left(\...
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2 votes
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When does an inverse PDE operator have a kernel (i.e. a fundamental solution?)

Following your assumptions, it seems that the mapping $L^{-1}$ sends linearly and continuously the smooth compactly supported functions into distributions and thus, from the Schwartz (Laurent) kernel ...
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  • 13.2k
2 votes

Gaussian bounds on Dirichlet heat kernel

The answer to the question is yes for the upper bound (albeit with different constants in the gaussian factor), and no for the lower bound. For the upper bound, and for any domain $\Omega \subset M$, ...
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  • 3,013
2 votes

Which utility functions are linearly transformed by normal perturbations?

$\newcommand{\al}{\alpha} \newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\varepsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\...
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2 votes
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Trace-class properties of integral operator

$k$ being compactly generated, you can as well assume that $k$ is a smooth function defined on $\mathbb{T}^2$ and $Op(k)$ acts on $L^2(\mathbb T)$ (for $\mathbb{T} = \mathbb R/\mathbb Z$ the unit ...
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2 votes
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Existence of integral kernel

If $f(x,y)=\sum_i c_i \chi_{E_i \times F_i}(x,y)=\sum_i c_i \chi_{E_i}(x) \chi_{F_i}(y)$ with $(E_i\times F_i)\cap (E_j \times F_j) =\emptyset$ for $i\neq j$, define $$\phi (f)=\sum_i c_i \int_{R^d} (...
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1 vote
Accepted

Gradient condition implies Hörmander condition

I figured out proving it by using mean value theorem (thanks to @WillieWong). Observe that \begin{align*} \int_{|x|>2|y|}|K(x-y)-K(x)|dx &\leq \int_{|x|>2|y|}|\nabla K(tx+(1-t)(x-y))||y| dx\\...
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  • 229
1 vote

Boundedness of integral operators on spaces of continuous functions

Too long for a comment. Your requirement is too stringent and it is quite likely that to get continuity from $L^\infty$ into itself, it is indeed necessary to have $$ \text{esssup}_x\int\vert k(x,y)\...
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  • 13.2k
1 vote
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Composition of a smoothing operator with an $L^2$-bounded operator, non-compact Riemannian manifold

Thanks to Jochen Wengenroth's comments, I can now give the full answer: the idea is that $C_\mathrm{c}^\infty(M) \hookrightarrow L^2(M, \mathrm{d} \mu_g) \xrightarrow{G} L^2(M, \mathrm{d} \mu_g) \...
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  • 119
1 vote

Gaussian bounds on Dirichlet heat kernel

I think you will find what you need here: http://ac.els-cdn.com/002212369090106U/1-s2.0-002212369090106U-main.pdf?_tid=c2567084-37e6-11e6-a2ed-00000aab0f6b&acdnat=...
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1 vote

Well-definedness for a singular integral

Assuming the function $f$ of class $C^1$, you find that, using Taylor's formula with integral remainder, $$ f(t)-f(s)=(t-s) f_1(t,s),\quad \text{with $f_1$ continuous},$$ so that $ (T_\alpha f)(t)=\...
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  • 13.2k
1 vote
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Various limits of the Christoffel Darboux Kernel

It seems that both questions receive some answers in this paper. In th $L^2$ norm, Section 9 seems to stipulate convergence and a limit $K$. For the uniform and pointwise convergence, we have on p. ...
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  • 3,200
1 vote

Error estimate in the spectral theorem of compact operators on a Hilbert space

Actually, the presumption that the partial sums of the kernel converge to it in $L^2$ of the product is not quite right: for example, mapping $\ell^2\to \ell^2$ by $e_n\to \lambda_n\cdot e_n$ for a ...
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  • 21.3k
1 vote
Accepted

Fredholm integral with functions constrained to [0;1]

There is not enough information for a thorough answer. An a priori bound on the solution may indeed help theoretically and practically. As usual with measured data you may not want to solve the ...
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