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246 views

Riemann-Liouville integral of $f$ is zero implies $f =0$ a.e

The Riemann-Liouville integral is defined by $$ I^\alpha f(x)=\frac{1}{\Gamma(\alpha)} \int_a^x f(t)(x-t)^{\alpha-1} d t $$ where $\Gamma$ is the gamma function and $a$ is an arbitrary but fixed base ...
Grandes Jorasses's user avatar
0 votes
1 answer
140 views

Singular integral bounded by Dirichlet form?

We define for some fixed $L$ $$\Omega:=\{(x_1,x_2) \in ([-L,L]^2 \times [-L,L]^2) \setminus \{x_1=x_2\}\},$$ in particular $x_1,x_2 \in \mathbb R^2.$ Let $f \in C_c^{\infty}(\Omega)$, then I am ...
António Borges Santos's user avatar
0 votes
1 answer
127 views

asymptotic of ratio between two summations (l1 / l2 norm)

Let $B$ as a $n\times n$ matrix where $$B_{ij}(\pmb{\theta})=(\theta_i-\theta_j)\sin(\theta_i-\theta_j), 1\leq i<j\leq n$$ and other entries equals to $0$, and $$\theta=[\theta_1,\cdots,\theta_n]\...
tony's user avatar
  • 405
0 votes
1 answer
110 views

Functions for which $|f^{(k)}|_{C^{0,\alpha}(0,1)} \le \Vert f \Vert_{L^1(0,1)}$

Let $f \in C^k(0,1)$ and assume that the $k$-th derivative is $\alpha$-Hölder continuous. Assume that $f(x) = 0$ in a fixed interval $(a,b) \subset (0,1)$. Can we characterize (or at least find some ...
Hiro's user avatar
  • 131
0 votes
1 answer
175 views

Asymptotic of ratio between l1 / l2 norm of a structured vector

As suggested in this discussion, I would like to inquire about the following question: Consider a matrix B of size $n\times n$ defined as: $$B_{ij}(\pmb{\theta})=(\theta_i-\theta_j)\sin(\theta_i-\...
tony's user avatar
  • 405
0 votes
0 answers
161 views

Superharmonic extension

We know the following classic result. If $K\subset\mathbb{R}^m$ ($m>1$) is a compact set and $u$ is superharmonic on a neighborhood of $K$, then we can extend $u$ to a superharmonic function $\...
M. Rahmat's user avatar
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-1 votes
1 answer
208 views

Does this function belong to $L^2(\mathbb{D})$?

Edit: After the answer of Prof. Eremenko to the previous version, I realized that a weaker assumption works for the main motivation of this post. so I revise the question. The unit ...
Ali Taghavi's user avatar
-1 votes
1 answer
519 views

Poisson kernel is the Cauchy distribution, reference?

Let $d = 2$, and consider the domain $D = \mathbb{H}$, the upper half-plane. Can someone give me a reference to a proof that the Poisson kernel is the Cauchy distribution?
Roger Smyth's user avatar
-1 votes
1 answer
236 views

Natural candidates for sub-half-exponential which limit to half-exponential function from below

There are no closed form candidates for half-exponential functions "Closed-form" functions with half-exponential growth. However sub-half-exponentials (functions whose composition grows ...
VS.'s user avatar
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1 answer
204 views

Cauchy reduction formula with measure (a variation)

The Cauchy reduction formula conveniently compresses $n$ integrations of a function $F(x)$ into a single integral. Here I am interested in reducing the following "curved-space" ...
Math2024's user avatar
  • 141
-1 votes
1 answer
369 views

Would this go to 0 [closed]

Let $t_{m}$ be the sup of the sum of the pairwise distances between any $2m$ points in the unit disk. Does $t_{m}/m^{2}$ go to $0$ as $m\rightarrow\infty$?
u51245's user avatar
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-2 votes
1 answer
314 views

Series representation for $\log(|\zeta(\frac{1}{2}+it)|)$

(Question is short and straight-forward. ) What is/are "nice and non-trivial" series representation/s of $\log(|\zeta(\frac{1}{2}+it)|)$ ?? By "nice and non-trivial" I mean contains no ...
bambi's user avatar
  • 375
-6 votes
2 answers
2k views

Is there a transformation or a proof for these integrals?

Here are certain weighted Gaussian integrals I have encountered for which numerical computation reassures equality. Question. Is this true? If so, is there an underlying transformation or just a ...
T. Amdeberhan's user avatar
-6 votes
1 answer
614 views

Proof of formula for $\pi$ [closed]

The number $\pi$ can be expressed as $\pi=\lim_{n\to\infty} \frac{n\sqrt[n]{-1}-n}{\sqrt{-1}}$ or more poetically $\pi=\frac{\infty\sqrt[\infty]{-1}-\infty}{\sqrt{-1}}$. Here we choose the principal ...
Mikhail Katz's user avatar
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