5
votes
On a Poincaré inequality with weight
Such an inequality cannot exist. Take $\Omega=B_1(0)\subset \mathbb{R}^n$ and assume find a constant $C>0$ independent of $\omega$, then taking a sequence $(w_k)_k \subset L^p(B_1)$ weakly ...
- 421
4
votes
Accepted
$L^1$ error between indicator function and smoothed out version
Yes, this works, and the only ingredient we need is the estimate $\int_r^{\infty} e^{-t^2}\, dt\lesssim e^{-r^2}$.
We then have (for example)
\begin{align*}
\int_r^{\infty} |f_r(x)|\, dx &=\frac{1}...
- 19.4k
4
votes
Eigenfunctions of the integral kernel $1/(x^2 + x'^2)$
I think I have a solution, however I cannot reproduce your asymptotic at $x\rightarrow\infty$ (although it agrees at $x\rightarrow 0$). Also I was unable to evaluate the original integral with my ...
- 596
4
votes
Accepted
Convergence of spectrum
$C^0$-convergence is sufficinent.
Note that $\lambda_i$ can be defined as the least lower bound on numbers $\lambda$ such that the following property holds:
There is an $i$-dimensional subspace $W$ ...
- 41.6k
4
votes
Relationship between noncommutative torus for different values of theta
Jamie Gabe has already pointed out why the answer is negative, but I wanted to point out that in fact the negative answer has nothing to do with Cstar algebras, quantum tori, or anything beyond the ...
- 25.1k
4
votes
Accepted
Is the projective limit $\mathcal{D}(\mathbb{R})$ separable?
The projective topology topology you describe is coarser than the usual inductive limit topology on $\mathscr D(\mathbb R)$ (the universal properties imply that it is enough to have continuity of the ...
3
votes
Pointwise convergence and disjoint sequences in $C(K)$
I claim that either of your properties (sequences or nets) is equivalent to having only finitely many non-isolated points.
Property $S$: Any pointwise null sequence in $C(K)$ has an almost disjoint ...
- 131
3
votes
Accepted
Weak convergence in $H^{1}$ implies different convergence in $L^{p}$?
This is true. Assume for example that $d \geq 3$. Since $(f_n)$ is bounded in $H^1$, it is bounded in $L^q$ for $2 \leq q \leq \frac{2d}{d-2}$, by Sobolev embedding. Moreover $f_n \to f$ strongly in $...
- 4,465
3
votes
optimization over moving domains
$\newcommand\R{\mathbb R}$The answer is no.
E.g., let $A=B=\R$, $B_a=[1-a^2,2]$ for $a\in A$, and $\ell(b)=b^3-3b$ for $b\in B$. Then $\ell(b)$ and $B_a$ are perfectly smooth, but $L$ is not ...
- 105k
2
votes
Solution of $\Delta f -\frac{1}{2}hf = 0$ behaves asymptotically as $f(x) = 1 - C/|x|$
For the specific question asked, just follow what Giorgio said.
Let $\phi = f-1$, then $\Delta \phi = \frac12 hf$ is a compactly supported function on $\mathbb{R}^3$.
Now let $g$ be the Newton ...
- 34.6k
1
vote
Solution of $\Delta f -\frac{1}{2}hf = 0$ behaves asymptotically as $f(x) = 1 - C/|x|$
I think you can get the result blowing-down the solution. I'll just sketch the argument since I don't want to be confusing with the details.
Set $u:=1-f$, then $u$ solves $-\Delta u = \frac{h}{2}(1-u)$...
- 421
1
vote
A possible measure-theoretic pathology
Suppose that $S$ is the graph of a continuous, strictly increasing function $\psi$. Then $S(W) = \psi(W)$, and the question asks if there is $\psi$ and $W$ such that $|W| = 1$, but $|\psi(W)| = 0$. ...
- 16.1k
1
vote
Extending Hölder functions
I think that for the $C^{k, \alpha}(A)$ space, where $A$ is bounded and the regularity of its boundary is $C^{k, \alpha}$ as well, an extension theorem holds: see for example Theorem 4 in Brian Krumme'...
- 21
1
vote
Weierstrass-type approximation of a system of the form $\{ x \mapsto f(x)^{p_n} : n \in \mathbb N\}$
This is an attempt to salvage the damaged comment of bathalf15320. We consider the case where $f$ maps $[0,1]$ onto itself. Then a sufficient condition for the completeness of the given family of ...
- 11
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