# Tag Info

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The only criterion I know is based on a Theorem in the second book of Dunford and Schwartz, see Theorem X1.6.29 and following. If the resolvent of an Hilbert-Schimdt operator satisfies some decay estimates on some rays dividing the complex plane, then the span of the generalized eigenfuntions is dense. The theorem generalizes to operators having trace-class ...

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In the paper The Green Function for uniformly elliptic equations Manuscripta Mathematica Gruter & Widman proved - among other things - pointwise estimates about the gradient of the Green function under the Dini continuity assumption on the leading coefficients. This is in section 3 Theorem 3.3

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One way to get the claimed value of the given integral, $J$ in notation below, is by starting from the standard relation \begin{aligned} \zeta(s) &= \frac 1{\Gamma(s)}\int_0^\infty \frac {x^{s-1}}{e^x-1}\; dx\ ,\qquad \text{ so }\\ \zeta'(s) &= \frac\partial{\partial s} \left(\ \frac 1{\Gamma(s)}\int_0^\infty \frac {x^{s-1}}{e^x-1}\; dx \ \... 18 I don't have a published source for this integral, but if need be you could refer to the following derivation:\int_0^\infty \frac{x^2\ln{x}}{e^x-1}\,dx=\int_0^\infty x^2 e^{-x}\ln x\sum_{k=0}^\infty e^{-kx}=\sum_{k=0}^\infty\frac{3-2\gamma-2 \ln (k+1)}{(k+1)^3}=(3-2\gamma)\zeta(3)+2\zeta^\prime(3).$$The integral over x^2 e^{-(k+1)x}\ln x ... 0 Check Computing a family of reproducing kernels for statistical applications, 1996, by Christine Thomas-Agnan 1 Certainly not, if C is supposed to be real. For instance, suppose that \rho_1 and \rho_2 are probability densities such that \rho_1\rho_2=0. Then the left-hand side of your inequality is \infty, whereas its right-hand side is 2C. 1 Perhaps the most elementary way to see that f is not continuous at zero is to lower bound the sum at x=\pi/k. Note that summands coming from n_1=2\ell k+s and n_2=n_1+k together have non-negative contribution to the sum when s=1,\dots k (and \ell\in \mathbb{Z}_{\geq 0}), so the sum is actually lower bounded by$$ \sum_{n=1}^{2k}\frac{\sin{(n\cdot ...

10

This function is (on the real line, at least) the product of $$\exp( \mu^2 \sum_{i=1}^\infty |z_i|^2 - 2 \mu \Re(\sum_{i=1}^\infty z_i)) \quad (1)$$ and the Hadamard type product $$\prod_{i=1}^\infty E_1( 2 \mu\Re z_i - \mu^2 |z_i|^2) \quad(2)$$ where $E_1$ is the first elementary factor $$E_1(z) := (1-z) \exp(z).$$ The expression (1) is clearly entire in ...

0

This not an answer to my question (I accepted Iosif's answer) but additional information based on my most recent research, that may be helpful to the reader interested in these processes. It is too long for an update of my question. For simplicity, we use the notation $F(x)=F_0(x)$ and $F(x-k)=F_k(x)$. Theorem A Regardless of the distribution $F_k$, if $b-a$ ...

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This is not true. Indeed, suppose that $X_k=X_{s;k}=k+sZ_k$, where $s\downarrow0$ and the $Z_k$'s are any iid random variables (r.v.'s). To obtain a contradiction, suppose that, for the random Borel measure $\mu_s$ over $\mathbb R$ defined by $\mu_s(B):=\sum_{k\in\mathbb Z}1(X_{s;k}\in B)$, the distribution of the random variable (r.v.) $\mu_s(B)$ is Poisson ...

3

Hjalmar Rosengren gave a nice formula for $f_{m,n}(x)$ based on the theory of hypergeometric series. Here I provide a direct elementary proof of the same based on generating series. Let us introduce the differential operator $$D_r:=\frac{1}{r!}\frac{\partial^r}{\partial x^r}.$$ Then $$\sum_{k=0}^\infty\binom{n+k}{k}x^k=D^n\left(\frac{1}{1-x}\right)=\frac{1}{(... 3 With proofs abound, I would like to record yet another instance of application due to the Wilf-Zeilberger techniques. The aim is to prove the identity shown above (from GH from MO, Hjalmar Rosengren):$$\sum_r\binom{m+n+k-r}{k-r}\binom{m}r\binom{n}r\binom{m+k}k^{-1}\binom{n+k}k^{-1}=1.$$The mechanical process begins with letting (suppressing other variables)... 23 See Meyer, Walter; Kay, David C., A convexity structure admits but one real linearization of dimension greater than one, J. Lond. Math. Soc., II. Ser. 7, 124-130 (1973). ZBL0271.52002. Theorem 4. If V and W are real vector spaces with \dim V>1, and f:V\to W is a one-to-one mapping which preserves convexity, then f is either linear (if f(0)=0) ... 6 This is Gauss' hypergeometric function F(n+1,m+1,1;x). You can then apply the huge theory of hypergeometric functions to derive further expressions. For instance, Euler's transformation formula gives the alternative expression$$\frac 1{(1-x)^{m+n+1}}\,F(-m,-n,1;x)=\frac 1{(1-x)^{m+n+1}}\sum_{k=0}^{\min(m,n)}\binom mk\binom nk x^k$$for the same series as ... 1 Come on, for the first property just use the fact that$$ \int_{-n}^n K(x) dx = \sum_{j = -n+1}^{n-1} \phi(j) + \tfrac12(\phi(-n)+\phi(n)) $$has a finite limit as n \to \infty, together with convergence of$$ \biggl| \int_{-a}^a K(x) dx - \int_{-\lfloor a\rfloor}^{\lfloor a\rfloor} K(x) dx \biggr| \leqslant |\phi(-\lfloor a\rfloor-1)| + |\phi(-\lfloor a\...

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Rohrlich has conjectured that the multiplicative relations in $\mathbb{C}^\times / \overline{\mathbb{Q}}^\times$ between values of $\Gamma$ at rational numbers are generated by the multiplication formula and the reflection formula. In conceptual terms, Lang says that $\Gamma$ is an odd punctured distribution on $\mathbb{Q}/\mathbb{Z}$, and that conjecturally,...

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No, we cannot. Formally, $\varphi$ is the eigenfunction of the unbounded operator $L_s$ on $L^2(\Omega)$, defined initially by $$L_s u(x) = (-\Delta)^s u(x) \qquad \text{for } x \in \Omega ,$$ where $u \in C_c^\infty(\Omega)$ (and it is understood that $u(x) = 0$ for $x \notin \Omega$), and then extended to an appropriate domain (e.g. by means of ...

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As $g(t)=\log{t}$ is a strictly increasing function, what you really needs is just the maximum of $\frac{A}{B}.$ Also, after some experiments, it seems that, for pure imaginary numbers only, the maximum occurs when the imaginary part is equal to $\lambda$. That is (in your notation), when $a=0$ the maximum is obtained with $b=\lambda$. So "divide and ...

3

I find it useful to represent the Legendre function in terms of a hypergeometric function, using a formula from Wikipedia, $$f(\theta)=\frac{ (1+\cos \theta)^{\mu/2} \, _2F_1\left(-\nu,\nu+1;1-\mu;\frac{1}{2} (1-\cos \theta)\right)}{\Gamma (1-\mu)\sin^\mu\theta(1-\cos \theta)^{\mu/2}}.$$ Then Mathematica gives me the small-$\theta$ expansion f(\theta)=\...

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I think it is not possible to do much better than $\frac{1}{\sqrt n}$. More precisely, I believe the best $\alpha$ is $\frac{1}{\sqrt n}$ whenever $n$ is a power of $2$, and therefore (since $\alpha$ is non-decreasing in $n$) at most $\frac{\sqrt{2}}{\sqrt n}$ in general. Indeed, consider $f(x)=\frac 1 2 \|A^{-1} x\|_2^2$ for a linear invertible map $A$ on $\... 12 The answer is yes indeed. It is a special case of Fox-H function, a variation of the confluent Fox-Wright$_{1}\Psi_{1}$function (a generalization of the confluent hypergeometric function$_{1}F_{1}$) providing the inverse function. See a previous answer here for details and references. For this particular case solution is (Setting$\alpha = a$), for$\...

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If $a$ is rational, then the root, say $x_*$, of your equation is algebraic, and (say) Mathematica will find for you with any degree of accuracy. Otherwise, one can approximate $a$ by rational numbers. Another way to get bounds on $x_*$ is to use a combination of the Newton and secant methods to bracket the root, as follows. For $a\in(0,1)$, using the ...

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