User alext87 - MathOverflow

Bounding a recursively defined sequence

2013-04-28T07:07:32Z

I have a sequence $\lambda_0,\lambda_1,\ldots,$ which is defined recursively as

$$ \lambda_0 = \frac{1}{2},$$

and

$$\lambda_{k+1} = \max_{\lambda\in [1,b]} \left(\frac{1}{2\lambda}\prod_{0\leq j\leq k}\left(\frac{\lambda-\lambda_j}{\lambda+\lambda_j}\right)^2\right), \qquad k\geq 0,$$

where $b>1$. I would like to find a relatively sharp upper bound for $\lambda_k$. Numerically, it seems that $\lambda_k$ is bounded by something like,

$$\lambda_k \leq \mathcal{O}\left(e^{-8k/\log(b)}\right). $$

For instance, the following very simple argument gives an upper bound that is very weak and I'm seeking techniques to do better:

Note that we have, for $k\geq 0$,

$$ \lambda_{k+1} \leq \max_{\lambda\in[1,b]} \left(\frac{1}{2\lambda}\prod_{0\leq j\leq k-1}\left(\frac{\lambda-\lambda_j}{\lambda+\lambda_j}\right)^2\right)\max_{\lambda\in[1,b]}\left(\frac{\lambda - \lambda_k}{\lambda+\lambda_k}\right)^2\leq \lambda_k \left(\frac{b-1}{b+1}\right)^2$$

and hence,

$$\lambda_k \leq \left(\frac{b-1}{b+1}\right)^{2k}\lambda_0 =\frac{1}{2}\left(\frac{b-1}{b+1}\right)^{2k},\qquad k\geq0.$$

I've already asked this on the Math Stack exchange, but it didn't receive any response:

http://math.stackexchange.com/questions/372661/bounding-an-implicitly-defined-sequence

Thank you in advance.

Perturbation analysis for three term recurrences

2013-02-09T13:07:33Z

Jacobi polynomials, denoted by $J^{(\alpha,\beta)}_n$, on $[-1,1]$ satisfy a three term recurrence

$$ J_{n+1}^{(\alpha,\beta)}(x) = (A_n+B_nx)J^{(\alpha,\beta)}_n + C_nJ_{n-1}^{(\alpha,\beta)}(x), \qquad J^{(\alpha,\beta)}_0 = \gamma_0,\quad J_1^{(\alpha,\beta)}(x) = \gamma_1x+\alpha_1.$$

I'm interested in the stability of the three term recurrence under small perturbations of its coefficients. For instance, if $\tilde{X}$ denotes the quantity $X$ perturbed by $\epsilon|X|$ where $|\epsilon|\ll 1$, and $\tilde{J}^{(\alpha,\beta)}_n$ satisfies

$$ \tilde{J}_{n+1}^{(\alpha,\beta)}(x) = (\tilde{A}_n+\tilde{B}_n x)\tilde{J}^{(\alpha,\beta)}_n + \tilde{C}_n\tilde{J}^{(\alpha,\beta)}(x), \qquad \tilde{J}^{(\alpha,\beta)}_0 = \tilde{\gamma}_0,\quad \tilde{J}_1^{(\alpha,\beta)}(x) = \tilde{\gamma}_1x+\tilde{\alpha}_1,$$

then is anything known about the magnitude of

$$ | J_{n}^{(\alpha,\beta)}(x) - \tilde{J}_{n}^{(\alpha,\beta)}(x)|$$

$$ || J_{n}^{(\alpha,\beta)}(x) - \tilde{J}_{n}^{(\alpha,\beta)}(x) ||_2?$$

Numerical observations on the Chebyshev polynomial $T_k(x) = \cos(k\cos^{-1}(x))$, $x\in[-1,1]$ suggests that

$$||J_{n}^{(\alpha,\beta)}(x) - \tilde{J}_{n}^{(\alpha,\beta)}(x)||_2 \leq \epsilon\mathcal{O}( n^2).$$

Decay rate of the singular values of functions

2012-11-24T18:30:10Z

Suppose a function $f:[-1,1]^2\rightarrow \mathbb{C}$ has a singular value decomposition:

$$ f(x,y) = \sum_{k=1}^\infty \sigma_k u_k(y) v_k(x), \qquad \sum_{k=1}^\infty \sigma_k^2 <\infty, $$

where $\sigma_1,\sigma_2,\ldots,$ is a nonincreasing sequence of real numbers, and $\{u_1(y),u_2(y),\ldots,\}$ and $\{v_1(x),v_2(x),\ldots,\}$ are two sets of orthonormal functions.

I would like to understand how the smoothness of $f(x,y)$ effects the decay of the singular values $\sigma_1,\sigma_2,\ldots,$ and, in particular, find proofs of statements similar to the following:

If $f(x,y)$ is an analytic function such that for every $a\in[-1,1]$ the univariate functions $f(a,y)$ and $f(x,a)$ are analytically continuable to bounded functions on a neighbourhood containing $[-1,1]$ then the singular values decay, at least, exponentially.
If $f$ is $k$ times continuously differentiable, $f\in\mathcal{C}^k$, then the singular values decay, at least, algebraically with order $k$.
If $f$ is $k-1$ times continuous differentiable and $f^{(k)}$ is Lipschitz smooth then the singular values decay, at least, algebraically with order $k+1$.

Thanks in advance.

Computing a determinantal representation of a bivariate polynomial

2012-03-19T15:46:52Z

Let $p(x,y,z)$ be a homogeneous irreducible polynomial of degree $d$, with real coefficients. From Dickson in 1920 we know that there exists $A$, $B$ and $C$ such that

$ det(Ax + By + Cz) = c p(x,y,z)$,

where $c$ is some constant.

Vinnokov in 1988 was able to describe all the non-equivalent determinantal representations as points on the Jacobian variety that are not on the exceptional sub variety. The theoretical work in this paper is relatively constructive, but is still a long way from a numerically stable constructive algorithm for $A$, $B$ and $C$.

Given any polynomial $p(x,y,z)$, can one triple $(A, B,C)$ be constructed in a numerically stable way?

Thanks in advance.

Quantifying the failure of the Cholesky factorization test for indefinite matrices

2012-10-03T11:20:08Z

The Cholesky factorization is the classic test to check if a matrix is positive definite. In infinite precision it is also an exact test: A matrix has a Cholesky factorization iff it is positive definite. However, in floating point arithmetic the Cholesky test is not perfect and two types of errors can (probably) be made:

A false positive error: The Cholesky test passes when the matrix is indefinite.
A false negative error: The Cholesky test fails when the matrix is positive definite.

The conditions for false negative errors is pretty well understood. For example, if a matrix $A$ is positive definite then defining $D = \hbox{diag}(A)^{1/2}$ and $A = DHD$, a false negative error is made if $\lambda_{\min}(H) \leq -n \gamma_{n+1}/(1-\gamma_{n+1})$, $\gamma_{n} = \mathcal{O}(n u)$ with $u$ being unit round off.

This is a satisfactory result because it can also be shown that the test makes no error on positive definite matrices if $\lambda_{\min}(H) > n \gamma_{n+1}/(1-\gamma_{n+1})$. For more details see theorem 10.7, page 200 of Accuracy and Stability of Numerical Algorithms By Nicholas J. Higham.

The situation seems more difficult for quantifying false positive error. On indefinite matrices the Cholesky factorization is numerically unstable so one would expect examples where an indefinite matrix (which is far from positive definite) still passes the Cholesky test. Does anyone know of an example? Can the Cholesky factorization be used in finite precision as a test without quantifying the false positive error? Is it just a highly improbable event that an indefinite matrix causes a false positive error?

When is Sobolev space a subset of the continuous functions?

2010-05-21T11:35:45Z

If we let $\Omega\subset\mathbb{R}^d$ with $d=1,2,3$ and define $\mathcal{H}^1(\Omega)=(w\in L_2(\Omega): \frac{\partial w}{\partial x_i}\in L_2(\Omega), i=1,...,d)$. My tutor has repeated several times:

If $d=1$ then $\mathcal{H}^1(\Omega)\subset\mathcal{C}^0(\Omega)$.
If $d=2$ then $\mathcal{H}^2(\Omega)\subset\mathcal{C}^0(\Omega)$ but $\mathcal{H}^1(\Omega)\not\subset\mathcal{C}^0(\Omega)$.
If $d=3$ then $\mathcal{H}^3(\Omega)\subset\mathcal{C}^0(\Omega)$ but $\mathcal{H}^2(\Omega)\not\subset\mathcal{C}^0(\Omega)$.

I was interested in trying to show these relationships. Does anyone know any references that would be useful.

Thanks in advance.

Sobolev-Slobodeckij spaces for p=infinity

2011-04-11T13:23:38Z

For $1\leq p<\infty$ an approach to define fractional Sobolev spaces is by Sobolev-Slobodeckij spaces a generalisation of Hölder continuity. For example letting $U\subset\mathbb{R}^n$ then,

$ \left\|u\right\|^p_{W^\mu_p(U)} = \left\|u\right\|^p_{W^{\lfloor\mu\rfloor}_p(U)} + \sum \int_U \int_U \frac{|D^\alpha u(x)-D^\alpha u(y)|^p}{|x-y|^{n+p[\mu]}}dxdy $

where $[\mu]=\mu-\lfloor\mu\rfloor$ and the sum is taken over all multi-indices $\alpha$ with $|\alpha|=\lfloor\mu\rfloor$

This is from Chapter 14 of The mathematical theory of finite element methods By Susanne C. Brenner, L. Ridgway Scott.

Does the above hold for $p=\infty$? For example, for $p=\infty$ do we have (or something similar),

$ \left\|u\right\|_{W^\mu_p(U)} = \left\|u\right\|_{W^{\lfloor\mu\rfloor}_p(U)} + \sup \sup_U \sup_U \frac{|D^\alpha u(x)-D^\alpha u(y)|}{|x-y|^{[\mu]}} $

where the $\sup$ is taken over all multi-indices $\alpha$ with $|\alpha|=\lfloor\mu\rfloor$. Can this be shown by considering the limit of the case $p<\infty$ as $p\rightarrow\infty$?

Thanks in advance.

Algebraic curve cannot suddenly end

2011-11-08T09:23:33Z

This is a literature request for (hopefully) an English version to a rigorous proof that a complex algebraic curve cannot abruptly end.

That is, if the algebraic curve enters a closed region it must also leave it.

This has a historic significance because Gauss's proof in his Phd thesis assumed this property holds. From looking around it seems that A Ostrowski rigorously proved the result around the 1920's. Is this correct? I am unable to find the title of the paper.

Is there also a proof that a real algebraic curve does not end abruptly?

I don't regard this property as obvious, but it doesn't seem to be well commented in the literature. Maybe, I'm wrong.

Thanks in advance.

Abruptly end: Given an irreducible polynomial $p$, we define $V(p)$ to be the complex algebraic curve associated to $p$. I say that $V(p)$ does not abruptly end at $(x,y)\in\mathbb{C}^2$ with $p(x,y)=0$ if there is a disc small enough so that the boundary contains exactly two points in $V(p)$.

(This is a first attempt and maybe needs some corrections.)

Eigenvalue distribution of positive-definite analytic function

2011-09-12T14:30:53Z

Let $f:[0,1]^2\rightarrow \mathbb{R}$ be a real-valued symmetric, positive definite function. Let $\{(x_i,y_i)\}_{1\leq i\leq N}\subset[0,1]^2$ be a finite, distinct set of coordinates. The point-wise evaluation matrix $A\in\mathbb{R}^{N\times N}$ is defined by,

$ A_{jk} = f(x_j,x_k). $

What can be said about the distribution of the eigenvalues of the matrix $A$ assuming (i) $f$ is analytic or (ii) $f\in H^k([0,1]^2)$ for some $k\geq1$?

Note that any change of order of the coordinates produces a similar matrix to $A$ and hence the same eigenvalues.

Thanks in advance.

Probability of a black path on a random chess board

2011-08-17T12:45:35Z

Take a $2n$ by $2n$ chess board (oriented so the grid lines are vertical and horizontal). Usually there are $2n^2$ squares coloured black and $2n^2$ squares coloured white so that a black square is only adjacent to white squares. (Here, two squares are adjacent if they have a common edge.)

Suppose instead we start with a blank $2n$ by $2n$ chess board. We pick $2n^2$ squares at random and assign them black. The other half of the squares are assigned white.

What is the probability the resulting chessboard has a monotonic black path? (Here, a monotonic black path is one which starts in the South-West corner and finishes in the North-East corner, and consists entirely of black squares adjacent along their North or East edge.
What is the probability that the resulting chessboard has a black path from the South-West corner to the North-East corner? (Here, a black path is a sequence of adjacent black squares)

Answer by alext87 for Eigenvalues of certain positive matrices

2011-08-18T13:33:11Z

Suppose we are given $s_1\geq s_2\geq \ldots \geq s_n>0$ and let $Q\in GL_n(\mathbb{C})$ satisfying the two conditions. Then

$Q\overline{Q}=\lambda I_n$

for some $\lambda\in\mathbb{C}$ and hence, by transposing, $Q^*Q^T=\lambda I_n$. Pick $v_k\in\mathbb{C}^n$ such that

$Q^*Qv_k = s_kv_k$

Then, $Q^*Q^T(Q^{-T}Qv_k) = Q^*Qv_k = s_kv_k = \lambda(Q^{-T}Qv_k)$ that is

$\lambda Qv_k=s_k Q^Tv_k$

Multiplying by $\overline{Q}$ on both sides and conjugate we obtain,

$\overline{\lambda} Q\overline{Q}\overline{v}_k = s_k Q Q^*\overline{v}_k$

Since, $Q\overline{Q}=\lambda I_n$ and $s_k\neq0$ we have,

$Q Q^*\overline{v}_k = (|\lambda|^2/s_k)\overline{v}_k$

Moreover, $Q^* Q$ and $Q Q^*$ have the same eigenvalues and the monotonicity conditions on $s_1,\ldots ,s_n$ ensure we have,

$\frac{|\lambda|^2}{s_n}=s_1,\text{ } \frac{|\lambda|^2}{s_{n-1}}=s_2,\text{ }\ldots , \text{ } \frac{|\lambda|^2}{s_1}=s_n$

This shows the choice $s_n=s_{n-1}$, $s_1\neq s_2$ allows no such $Q$.

Bessel Potential Space inequality

2011-07-25T13:02:19Z

The Bessel Potential Space is defined for $s\in\mathbb{R}$ as,

$H^s(\mathbb{R}^d) = \{f\in L_2(\mathbb{R}^n) : (1+|\cdot|)^{s/2}\hat{f}(\cdot)\in L_2(\mathbb{R}^n)\}. $

This defines a Hilbert space such that for any $f,g\in H^s(\mathbb{R}^n)$,

$ \langle f, g\rangle = \int_{\mathbb{R}^n} \hat{f}(\omega)\overline{\hat{g}(\omega)} (1+|\omega|)^{s}d\omega. $

For any open set $\Omega\subset\mathbb{R}^n$ we have $H^s(\Omega)$ being the set of restrictions with norm,

$ \left\|f\right\|_{H^s(\Omega)} = $

$ \inf_{g\in H^s(\mathbb{R}^n)}\{\left\|g\right\|_{H^s(\mathbb{R}^n)} : g|\Omega=f \} $

Does this definition of the norm ensure we have the following: Given an open set $\Omega\subset \mathbb{R}^n$ and open sets $\Omega_1, \Omega_2\subset \mathbb{R}^n$ such that $\Omega = \Omega_1\cup \Omega_2$ and $f\in H^s(\Omega)$,

$ \left\|f\right\|^2_{H^s(\Omega)}\leq \left\|f\right\|^2_{H^s(\Omega_1)} + \left\|f\right\|^2_{H^s(\Omega_2)}. $

Maximising the volume of a parallelotope in a cone with a fixed vertex

2011-07-14T13:47:35Z

Let $C\subset\mathbb{R}^n$ be a cone with vertex at the origin, aperture $\theta$ and height $h$. Since a cone is a convex region in $\mathbb{R}^n$ we know there is a parallelotope $P$, completely contained in $C$ such that

$Vol_n(P)\geq n^{-n}Vol_n(C),$

where $Vol_n(A)$ is the $n$-dimensional volume of $A$. This result is Lemma 8 of A compactness theorem for Affine equivalence-classes of convex regions by Macbeath. (note the statement of the Lemma has a typo - the inequality is stated the wrong way round!)

I was wondering what happens if we fix a vertex of the parallelotope to the origin. That is if $\mathcal{P}$ is the family of parallelotopes completely contained in $C$ with a vertex at the origin.

What are lower bounds on

$\sup_{P\in\mathcal{P}} Vol_n(P)/Vol_n(C)?$

Domain Intersection and recification

2011-07-11T16:18:43Z

Let $\Omega\subset\mathbb{R}^d$ be a Lipschitz domain that also satisfies an interior cone condition with parameters $r$ and $\theta$. Suppose I take an annulus, $h>3t$:

$ A(x,t,h) = B(x,th)\setminus B(x,(t-1)h) $

with $x\in\Omega$.

The open set $A(x,t,h)\cap \Omega$ may not even be connected but I would like to rectify each component back to a Lipschitz domain that also satisfies an interior cone condition with parameter $\approx h$ and $\tilde{\theta}\approx \theta$. The rectified domain should be contained in $A(x,t,3h)\cap\Omega$.

Here is my attempt so far which rectifies each component to an interior cone domain:

Take one component of $A(x,t,h)\cap \Omega$, let's call it $A_1$. Let $C_y(r)$ be the cone centred at $y\in\Omega$ that is completely contained in $\Omega$ (this is guaranteed by the interior cone condition). Define,

$ A_1 = \bigcup_{y\in A(x,t,h)\cap \Omega} C_y(h) $

Now since $C_y(h)\subset \Omega$ and not farther than $h$ from $A(x,t,h)$ we have,

$ A_1\subset A(x,t,3h)\cap\Omega $

Moreover, a cone $C_y(h)$ is an interior cone domain with parameter $\tilde{r}\approx h$ and $\tilde{\theta}=\min\{{\pi/5,\theta\}}$. The union of interior cone domains is also an interior cone domain with the same parameters. This shows that $A_1$ is an interior cone domain but not necessarily Lipschitz. Due to the smoothness of the boundary of $\Omega$ can I conclude that $A_1$ is Lipschitz? Any better constructions for $A_1$?

Thanks in advance.

Are there ill-conditioned problems in infinite precision arithmetric?

2011-06-01T18:40:12Z

It is hoped that in the future with the advent of quantum computing that fundamental operations on a computer will have arbitrarily high precision. Moreover, that even with such high precision, computation times with be realistic. An important concept in Numerical Analysis is ill-conditioning.

For instance it is common to call the following problem, an ill-conditioned problem:

Finding the roots of a quadratic polynomial. The example is from Datta, BN Numerical Linear Algebra and Applications SIAM, Second edition (2010)

$ z^2-2z+1=0 \rightarrow z^2-2.0001z+1=0 $

since a relatively small change in the polynomial coefficients can cause a much larger perturbation in the polynomial roots.

However the ill-conditioning doesn't seem to be part of the problem because in arbitrarily high precision we have the quadratic formula allowing us to compute the roots. The ill-conditioning seems to be caused by working in finite precision which is not inherent to the problem.

Researchers often talk about the difference between an ill-conditioned problem and an ill-conditioned method of solving. This distinction seems to be entirely blurred to me.

To help clarify the situation I have two closely related questions:

Is ill-conditioning an issue in arbitrarily high precision?
Is there an ill-conditioned problem which remains ill-conditioned even in infinite precision computing?

Answer by alext87 for Showing block diagonal structure of matrix by reordering

2011-06-17T12:23:25Z

Given a symmetric matrix $M\in\mathbb{R}^{n\times n}$ you are able to visualize the matrix as an adjacency matrix of a graph. Where node $i$ and $j$ have an edge between them if the $(i,j)$ entry of $M$ is non-zero. When the matrix is not symmetric then you end up looking at a directed graph.

A $k$ clustering algorithm on the associated graph should do a good job. Where $k$ is the number of blocks you want. This should also work when the matrix is not completely block diagonal.

Extension operator for Lipschitz domain for fractional Sobolev spaces

2011-06-15T17:25:25Z

Let $\Omega\subset\mathbb{R}^n$ be a bounded domain, with Lipschitz smooth boundary. Then a well known result by Stein gives that there exists an extension operator $E: H^k(\Omega)\rightarrow H^k(\mathbb{R}^n)$ such that

$Eu(x)=u(x)$ for all $x\in\Omega$
$\left\|Eu\right\|_{H^k(\mathbb{R}^n)}\leq C\left\|u\right\| _ {H^k(\Omega)} $

where the constant $C$ depends on $k$, $n$ and Lipschitz constant of $\Omega$. See the Thesis: A Degree-Independent Sobolev Extension Operator by Luke Rogers for a really nice summary of extension operators in the integer case.

DeVore and Sharpley in Besov spaces on bound domains of $\mathbb{R}^n$ extended this to the fractional case by real interpolation. There result is:

There exists an extension operator $E: H^\tau(\Omega)\rightarrow H^\tau(\mathbb{R}^n)$ such that:

$Eu(x)=u(x)$ for all $x\in\Omega$
$\left\|Eu\right\|_{H^\tau(\mathbb{R}^n)}\leq C\left\|u\right\| _ {H^\tau(\Omega)}$

but this time the constant depends on $\tau$, $n$ and $\Omega$. So the dependence is on $\Omega$ and not on the Lipschitz condition it satisfies.

This to me seems a highly unsatisfactory situation. Does there exist an extension operator so that the bound in the fractional Sobolev setting depends on $\Omega$ only through its Lipschitz condition? Does anyone know a reference which has this result?

Extension theory with bump function

2011-06-06T11:36:32Z

Let $B_t(0)$ denote the $n$ dimensional ball of radius $t$ centered at the origin. Does there exist a $\phi\in C(\mathbb{R}^n)$ function with the properties:

$ \phi (x) = \begin{cases} 1&x\in B_r(0) \\ 0&x\not\in B_{r+3}(0) \end{cases} $

and for any real-valued function $f\in \mathcal{H}^\tau(\mathbb{R^n})$ ($\tau\in\mathbb{R}$, $\tau>d/2$) we have

$ \left\|\phi(\cdot) f(\cdot)\right\|_{\mathcal{H}^\tau(A_1)}\leq C\left\|f\right\|_{\mathcal{H}^\tau(A_2)} $

where $C$ is a constant independent of $f$, $A_1 = B_{r+2}(0)\setminus B_{r+1}(0)$ and $A_2 = B_{r+3}(0)\setminus B_{r}(0)$.

This has a similar feel to extension theory results if we think of $\phi$ as an extension operator which preserves $f$ on $B_r(0)$.

Finite Element Method Inverse Estimate

2011-02-23T14:41:37Z

Looking for a proof in the literature of the following lemma:

Let $K\subset\mathbb{R}^d$ be a bounded domain. Let $P_X$ be a finite dimensional subspace of $\mathcal{H}^k(K)$ then

$$\left\|v\right\|_{\mathcal{H}^k(K)} \leq C diam(K)^{-k} \left\|v\right\|_{L_2(K)}$$

for all $v\in P_X$ and where the constant $C$ does not depend on $diam(K)$.

There is a proof in The mathematical theory of finite element method By Susanne C. Brenner, L. Ridgway Scott p111 but they do not check if the constant $C$ is independent of $diam(K)$.

Thanks in advance.

Interior cone condition preserved on a small perturbation of the domain.

2011-02-03T11:24:37Z

I'm looking for a proof in the literature or just a proof of:

Let $\Omega\subset\mathbb{R}^d$ be an open and bounded domain with satisfying the interior cone condition with parameters $r$ and $\theta$. Let $\Omega_\delta$ be the $\delta$-interior of $\Omega$ that is

$ \Omega_\delta = ${$x\in\Omega : dist(x,\partial\Omega)>\delta$}$ $

There is a $\delta_0$ sufficiently small such that $\Omega_\delta$ for $0<\delta<\delta_0$ satisfies the interior cone condition with parameters $r/2$ and $\theta/2$.

Note: I'm also only interested when $\Omega$ lies on one side of its boundary.

Extension Operators for Sobolev spaces

2011-01-28T16:43:09Z

Let $\Omega\subset\mathbb{R}^d$ be a bounded domain with Lipschitz smooth boundary and $\delta>0$ sufficiently small so that $ \Omega_\delta = ${ $x\in\Omega : dist(x,\partial\Omega)>\delta $ }$ $ is also a domain with Lipschitz smooth boundary.

For sufficiently small $\delta<0$ and $\tau>d/2$ is there a linear extension operator $ E: H^\tau(\Omega_\delta) \rightarrow H^\tau(\Omega) $ such that

(i) $Ef(x) = f(x)$ for $x\in\Omega_\delta$

(ii) $||E f||_{H^\tau(\Omega_\delta)}\leq C||f||_{H^\tau(\Omega)}$

where the constant $C$ is independent of $\delta$?

In response to Tapio Rajala: $\Omega$ is a domain meaning it is an open connected subset of $\mathbb{R}^d$. From the definition of $\Omega_\delta$ and $\delta$ being sufficiently small this makes $\Omega_\delta$ a connected subset. I also want $\Omega_\delta$ to have a Lipschitz smooth boundary, that is locally a graph of a Lipschitz smooth function.

Bounding near the boundary for a Sobolev function.

2010-12-02T17:10:21Z

Let $f: \Omega\rightarrow \mathbb{R}$ where $\Omega\subset\mathbb{R}^d$ is bounded with lipschitz smooth boundary. Further suppose that $f\in\mathcal{H}^{\tau}(\Omega)$, $\tau>\frac{d}{2}$ (i.e. $f$ is continuous) and $f$ is zero on the boundary.

Let $$ \Omega_{\delta} = \{ x\in\Omega : \inf_{y\in\partial\Omega} \left\|x-y\right\|_2 > \delta \} .$$ Where $\delta>0$ is small enough to preserve smoothness in the boundary of $\Omega_{\delta}$. See: http://mathoverflow.net/questions/46628/shrinking-a-lipschitz-smooth-domain.

For sufficiently small $\delta>0$ it is true that:

$\left\|f\right\|_{L_2(\Omega\setminus\Omega_{\delta})} \leq C\delta^\alpha\left\|f\right\|_{L_2(\Omega)}$ with $\alpha\geq 1$ and $C$ is a constant not depending on $\delta$ or $f$.

If this is not possible what is the largest $\alpha\in(0,1)$ so that the above inequality holds.

Thanks in advance.

Note this is almost identical to my previous post http://mathoverflow.net/questions/46650/bounding-a-smooth-function-near-the-boundary but presented in a clearer fashion. I would be grateful for any comments or help.

Bounding a smooth function near the boundary

2010-11-19T17:04:51Z

Let $f: \Omega\rightarrow \mathbb{R}$ where $\Omega\subset\mathbb{R}^d$ is bounded with lipschitz smooth boundary. Further suppose that $f\in\mathcal{H}^{\tau}(\Omega)$, $\tau>0$ and that $f$ decays rapidly to $0$ on the boundary.

Are there any known bounds on $\left\|f\right\|_{L_2(\Omega\setminus\Omega_{\delta})}$? i.e. bounding $f$ near the boundary of $\Omega$.

Note: $ \left\|f\right\|_{L_2(\Omega)}= \left(\int_{\Omega} |f|^2\right)^{\frac{1}{2}} $

I ideally would like some bound of the form: Given $f\in\mathcal{H}^\tau(\Omega)$, $\tau>d/2$ which is zero on the boundary and $\delta$ sufficiently small then:

$\left\|f\right\|_{L_2(\Omega\setminus\Omega_{\delta})} \leq C\delta^\alpha\left\|f\right\|_{L_2(\Omega)}$ with $\alpha>0$ as large as possible (hopefully $\alpha=1$) and $C$ is a constant not depending on $\delta$ or $f$.

General Sobolev Inequalities

2010-11-22T10:00:07Z

In Partial Differential Equation by Lawerence Evan p284 there is this theorem stated:

Let $U$ be a bounded open subset of $\mathbb{R}^n$ with $C^1$ boundary. Suppose $u\in W^{k,p}$ then if $k>n/p$ we have

$ u\in C^{\alpha, \gamma}(\overline{U}) $ where $\alpha = k-\left[n/p\right]-1$ and $\gamma = \left[n/p\right]+1-n/p$ if $n/p$ is not an integer and any $0<\gamma<1$ if $n/p\in\mathbb{N}$.

I have two questions:

Does this result extend to $U$ being an open subset with only Lipschitz boundary?

2.Does the result also holds $k\not\in\mathbb{N}$? The author doesn't mention anyway that $k$ should be an integer but I just wanted to check.

Thank you in advance.

Shrinking a Lipschitz smooth domain.

2010-11-19T13:38:55Z

Let $\Omega\subset \mathbb{R}^d$ be an open and bounded domain with Lipschitz smooth boundary. Let $\delta>0$ and $ \Omega_\delta = \{ x\in\Omega : \inf_{y\in\partial\Omega} \left\|x-y\right\|_{L_2(\Omega)}\geq \delta \} $

Is there a $\hat{\delta}>0$ such that $\forall 0<\delta<\hat{\delta}$ the space $\Omega_{\delta}$ has a Lipschitz smooth boundary?

This statement seems like it should be true. I would really appreciate some help.

Thanks in advance.

Decay of the Fourier transform

2010-10-30T16:26:05Z

Suppose $f(z)$ is a function analytic in the strip $|Re(z)|\leq a$. Is the fourier transform $\hat{f}(w)=o(e^{-a|w|})$?

It seems plausible but I can't seem to prove it either.

There is similar result called the Paley-Wiener Theorem that states $e^{a|w|}\hat{f}(w)\in L_2(\mathbb{R})$, but I don't think that helps.

Thanks in advance.

Lipschitz smooth boundary definition

2010-10-29T05:39:04Z

Is the wikipedia definition of Lipschitz Euclidean domain correct?

See: http://en.wikipedia.org/wiki/Lipschitz_domain

i was wondering what stops me just showing the condition holds for one point and then just scale and translate that function $h_p$ for any point on the boundary... This doesn't seem right? What am I missing?

Thanks in advance.

Where does the Chebyshev polynomial notation come from?

2010-10-10T06:55:27Z

The $k$th Chebyshev polynomial is denoted by $T_k$ where

$T_k(x) = \cos(k\cos^{-1}(x))$

I was wondering where this notation came from. It has been suggested that it comes from Tschebyscheff (the Russian name for Chebyshev) but does anyone know the first use of this notation or verify this is the reason?

Where are the boundary layers?

2010-09-26T09:25:01Z

I am learning perturbation theory and would like to be able to determine where boundary layers are going to occur just by looking at the differential equation.

Let $n\in\mathbb{N}$ and $p_i(x)$, $0\leq i< n$ some sufficiently well-behaved functions.

Am I able to determine the boundary layers of the following problem just by looking at the $p_i(x)$ or some other easy to see property?:

$\epsilon \frac{d^n y}{dy^n} + \sum_{i=0}^{n-1} p_i(x) \frac{d^i y}{dy^i} =0$

$y(0)=a$ and $y(1)=b$

(I realise that I require more constraints to get a unique solution but I don't think this effects the existence of boundary layers)

Thanks in advance.

Answer by alext87 for Differences of near diagonal Ramsey numbers.

2010-09-24T08:19:30Z

Maybe try to add $O(n^2)$ vertices to the graph $K_{R(n,n)}$ which is R/B coloured without a blue $K_n$ without creating a $K_{n+1}$ in red.

Comment by alext87

2013-04-29T06:09:30Z

Yes, I also think the word "recursive" is better than "implicit". Changed. Thanks. I was thinking that the maximum made it "implicit" since the maximum probably does not have a closed form.

Comment by alext87

2011-11-08T14:01:32Z

Thank you very much for your responses! I am looking through them now.

Comment by alext87

2011-11-08T09:42:32Z

I am not assuming the curves to be smooth. For instance, the algebraic curve associated to $y^2 = x^3$ has a singularity at $0$ but does not abruptly end. I try and come up with a mathematical description in the question.

Comment by alext87

2011-08-18T08:13:24Z

Thanks for more information. That's great. I'm confused, by my lack of knowledge on percolation, and by the uniform lower bound on the probability of a black cross. I'm not sure how your answer fits together with Brendan McKay's answer...

Comment by alext87

2011-08-18T08:02:10Z

This is an interesting connection to percolation. Thank you! Unfortunately, MathOverFlow does not allow me to accept a comment as an answer.

Comment by alext87

2011-07-25T13:25:27Z

@fedja: True. I have changed the question to actually what I require. I'm hoping all this is well-defined now.

Comment by alext87

2011-06-16T07:20:51Z

Yes, Stein's result only requires Lipschitz for arbitrary high $m$. At least Dobrowolski's result proves it is true for balls and annuli. Actually, I'm not completely convinced because I couldn't find where he explicitly stated what the constant depends on. My high school German can't cope with all the maths terminology so I may have missed it.

Comment by alext87

2011-06-15T19:17:54Z

@Pietro Majer: Thanks for fixing the latex! :D @Dirk: I am a little confused by Satz 6.40 is it really stating only dependence on the Lipschitz condition. It seems to be imposing that the boundary it $C^1$ smooth.

Comment by alext87

2011-06-15T07:34:56Z

This approach works. Thanks. The answer was just a little vague in comparison to Nilima Nigma's response.

Comment by alext87

2011-06-15T07:26:27Z

$\phi$ can be smoother.

Comment by alext87

2011-06-08T16:27:00Z

Sure you can do this for integer $\tau$ just by using Leibniz rule. In fact the details are quite simple. I was more interested in the fractional sobolev space case where the details are escaping me.

Comment by alext87

2011-06-08T13:24:44Z

This is the Sobolev space $H^\tau=W^{\tau,2}$. en.wikipedia.org/wiki/Sobolev_space

Comment by alext87

2011-06-02T06:48:09Z

However, the problem of finding the roots of $z^2-2z+1=0$ can be computed exactly on a computer because the numbers $1$ and $2$ have exact representations as floating point numbers. Are there severely ill-conditioned problems which cause no concern when computing in finite precision?

Comment by alext87

2011-03-30T13:46:35Z

Sure. That is right.

Comment by alext87

2011-03-30T13:35:02Z

Assuming this is about matrices: The inverse of A-B does not exist if A and B have a common eigenvalue.