Linh Tinh « Ngô Quốc Anh

Tháng Mười 30, 2009

A characteristic of essentially bounded functions

Chuyên mục: Các Bài Tập Nhỏ, Giải Tích 6 (MA5205), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 11:53

In this topic, we prove the following statement

Statement: Let $(X,\mathcal B, m)$ be a probability space. Let $h \in L^2(m)$ . Then $h$ is essentially bounded iff $h \cdot f \in L^2(m)$ for all $f \in L^2(m)$ .

Proof. If $h$ is bounded, then by using the Holder inequality one has

$\displaystyle\int_X {{{\left| {h \cdot f} \right|}^2}dm}\leq \underbrace {\sqrt {\int_X {{{\left| h \right|}^2}dm} } }_{ \leqslant c}\sqrt {\int_X {{{\left| f \right|}^2}dm} }<+\infty$

for all $f \in L^2(m)$ . Conversely, we suppose $h$ is such that $h \cdot f \in L^2(m)$ whenever $f \in L^2(m)$ . Let

$\displaystyle X_n = \{ x \in X : n-1 \leq |h(x)| < n\}, \quad \forall n>0$ .

Then $\{X_n\}_1^\infty$ partitions $X$ . Let

$\displaystyle f\left( x \right) =\sum\limits_{n = 1}^\infty{\frac{1}{{n\sqrt {m\left( {{X_n}} \right)} }}{\chi_{{X_n}}}\left( x \right)} ,$

where it is understood that the $n$ -term is omiited if $m(X_n)=0$ . Then

$\displaystyle\int_X {{{\left| f \right|}^2}dm}=\int_X {{{\left({\sum\limits_{n = 1}^\infty {\frac{1}{{n\sqrt {m\left( {{X_n}}\right)} }}{\chi _{{X_n}}}\left( x \right)} } \right)}^2}dm}\leq \sum\limits_{n = 1}^\infty{\frac{1}{{{n^2}}}}<\infty$

which implies $f \in L^2(m)$ . Since

$\displaystyle\int_X {{{\left| {hf} \right|}^2}dm}=\sum\limits_{n \in F} {\int_{{X_n}} {{{\left| {hf} \right|}^2}dm}}\geq\sum\limits_{n \in F}{{{\left( {\frac{{n - 1}}{n}}\right)}^2}}$

where $F = \left\{ {n:m\left( {{X_n}} \right) \ne 0} \right\}$ . Sincc $h \cdot f \in L^2(m)$ we have that $F$ is finite and therefore $h$ is essentially bounded.

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Tháng Mười 28, 2009

The weak and weak* topologies: A few words

Chuyên mục: Giải tích 7 (MA4247), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 2:48

The weak and weak* topologies are the weakest in which certain linear functionals are continuous.

We start with a normed linear space $X$ . The dual space of $X$ , denoted by $X'$ , is the collection of all continuous linear functionals, i.e., the set of all mapping $\ell : X \to \mathbb R$ satisfying

$\ell(ax)=a \ell (x)$ , $\ell(x+y)=\ell(x)+\ell(y)$

and

$\displaystyle\lim_{n \to \infty} \ell(x_n) = \ell(x)$ when $\displaystyle\lim_{n \to \infty} \|x_n - x\|=0$ .

Definition 1. In $X$ , the strong topology is the norm topology, i.e., we can talk about an open set of $X$ , for example $U$ in the following sense: $U \subset X$ is said to be open if and only if for each $x_0 \in U$ , there exists $\varepsilon>0$ such that $\{ x \in X: \|x-x_0\|<\varepsilon\} \subset U$ .

Claim 1. Bounded linear functionals are continuous in the strong topology.

Proof. We first recall that a linear functional $\ell$ is said to be bounded if there is a positive number $c$ such that $|\ell (x)| \leq c\|x\|$ for all $x \in X$ .

Now we assume $\ell$ is continuous but not bounded; then for any choice of $c=n$ , one has $\ell(x_n) > n \|x_n\|$ . Clearly, $x_n$ can be replaced by any multiple of $x_n$ ; if we normalize $x_n$ so that

$\displaystyle \|x_n\|=\frac{1}{\sqrt{n}}$

then $x_n \to 0$ but $\ell (x_n) \to \infty$ . This shows the lack of boundedness implies the lack of continuity.

Now we assume $\ell$ is bounded. For arbitray $x_n$ and $x$ , one gets

$|\ell(x_n)-\ell (x)| = |\ell (x_n-x)| \leq c\|x_n-x\|$ ;

this shows that boundedness implies continuity.

Definition 2. In $X$ , the weak topology is the weakest topology in which all bounded linear functionals are continuous.

The open sets in the weak topology are unions of finite intersections of sets of the form

$\{ x : a< \ell(x) < b\}$ .

Clearly, in an infinite-dimensional space the intersection of a finite number of sets of the above form is unbounded. This shows that every set that is open in the weak topology is unbounded. In particular, the balls

$\{ x : \|x\|<R\}$

opens in the strong topology, are not open is the weak topology.

Definition 3. In $X'$ the dual space of $X$ , the weak* topology is the crudest topology in which all linear functionals

$x: X' \to \mathbb R, x(\ell) := \ell(x)$

are continuous.

If $X'$ is nonreflexive, the weak* topology is genuinely coarser than the weak topology, as will be clear from the following theorem due to Alaoglu

Theorem (Alaoglu). The closed unit ball in $X'$ is compact in the weak* topology.

We end this topic by the following theorem

Theorem. The closed unit ball in $X$ is compact in the weak topology if and only if $X$ is reflexive.

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Tháng Chín 18, 2009

Moving plane method via a small example inspired by Gidas, Ni, and Nirenberg

Chuyên mục: Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 1:41

In this section we will use the moving plane method to discuss the symmetry of solutions. The following result was first proved by Gidas, Ni and Nirenberg.

Theorem. Suppose $u \in C(\bar B_1) \cap C^2(B_1)$ is a positive solution of

$-\Delta u = f(u)$ in $B_1$ and $u=0$ on $\partial B_1$

where $f$ is locally Lipschitz in $\mathbb R$ . Then $u$ is radialy symmetric in $B_1$ and $\frac{\partial u}{\partial r}(x)<0$ for $x \ne 0$ .

The original proof requires that solutions be $C^2$ up to the boundary. Here we give a method which does not depend on the smoothness of domains nor the smoothness of solutions up to the boundary.

Statement. Suppose that is a bounded domain which is convex in $x_1$ direction and symmetric with respect to the plane $\{x_1 = 0\}$ . Suppose $u \in C^2(\Omega)\cap C(\bar\Omega)$ is a positive solution of

$-\Delta u = f(u)$ in $\Omega$ and $u=0$ on $\partial \Omega$

where $f$ is locally Lipschitz in $\mathbb R$ . Then $u$ is radialy symmetric in $x_1$ and $D_{x_1}u(x)<0$ for $x \ne 0$ .

Idea of proof. Write $x=(x_1, y)\in\Omega$ for $y \in \mathbb R^{n-1}$ . We will prove that

$u(x_1,y)<u(x_1^\star,y)$ for any $x_1>0$ and $x_1^\star<x_1$ with $x_1^\star+x_1>0$ .

Then by letting $x_1^\star \to -x_1$ , we get $u(x_1,y)\leq u(-x_1,y)$ for any $x_1$ . Then by changing the direction $x_1 \to -x_1$ we get the symmetry.

Proof. Let $a=\sup x_1$ for $(x_1, y)\in\Omega$ . For $0<\lambda<a$ , define

$\Sigma_\lambda=\{x \in \Omega: x_1>\lambda\}$

$T_\lambda = \{x_1=\lambda\}$

$\Sigma'_\lambda=$ the reflection of $\Sigma_\lambda$ with the respect to $T_\lambda$

$x_\lambda=(2\lambda-x_1,x_2,...,x_n)$ for $x=(x_1,x_2,...,x_n)$ .

In $\Sigma_\lambda$ we define

$w_\lambda(x)=u(x)-u(x_\lambda)$ for $x\in \Sigma_\lambda$ .

Then we have by Mean Value Theorem

$\Delta w_\lambda+c(x,\lambda)w_\lambda=0$ in $\Sigma_\lambda$

$w_\lambda \leq 0$ and $w_\lambda \not \equiv 0$ on $\partial \Sigma_\lambda$ .

where $c(x,\lambda)$ is a bounded function in $\Sigma_\lambda$ .

We need to show $w_\lambda<0$ in $\Sigma_\lambda$ for any $\lambda \in (0,a)$ . This implies in particular that $w_\lambda$ assumes along $\partial \Sigma_\lambda \cap \Omega$ its maximum in $\Sigma_\lambda$ . By Hopf Lemma we have for any such $\lambda \in (0,a)$

$D_{x_1}w_\lambda\bigg|_{x_1=\lambda} = 2D_{x_1}u\bigg|_{x_1=\lambda}<0$ .

For any $\lambda$ close to $a$ , we have $w_\lambda<0$ by the maximum principle for narrow domain. Let $(\lambda_0, a)$ be the largest interval of values of $\lambda$ such that $w_\lambda<0$ in $\Sigma_\lambda$ . We want to show that $\lambda_0=0$ . If $\lambda_0>0$ , by continuity, $w_\lambda \leq 0$ in $\Sigma_{\lambda_0}$ and $w_{\lambda_0} \not \equiv 0$ on $\partial \Sigma_{\lambda_0}$ . Then the Strong Maximum Principle implies $w_{\lambda_0}<0$ in $\Sigma_{\lambda_0}$ . We will show that for any small $\varepsilon>0$

$w_{\lambda_0-\varepsilon} <0$ in $\Sigma_{\lambda_0-\varepsilon}$ .

Fix $\delta>0$ (to be determined). Let $K$ be a closed subset in $\Sigma_{\lambda_0}$ such that $|\Sigma_{\lambda_0} - K| <\frac{\delta}{2}$ . The fact that $w_{\lambda_0}<0$ in $\Sigma_{\lambda_0}$ implies

$w_{\lambda_0}(x)\leq -\eta <0$ for any $x \in K$ .

By continuity we have

$w_{\lambda_0-\varepsilon}<0$ in $K$ .

For $\varepsilon>0$ small, $|\Sigma_{\lambda_0-\varepsilon}-K|<\delta$ . We choose $\delta$ in such a way that we may apply the Maximum Principle for domain with small volume to $w_{\lambda_0-\varepsilon}$ in $\Sigma_{\lambda_0-\varepsilon}-K$ . Hence we get

$w_{\lambda_0-\varepsilon} \leq 0$ in $\Sigma_{\lambda_0-\varepsilon}-K$

and then

$w_{\lambda_0-\varepsilon}(x) <0$ in $\Sigma_{\lambda_0-\varepsilon} -K$ .

Therefore we obtain for any small $\varepsilon>0$

$w_{\lambda_0-\varepsilon}(x) <0$ in $\Sigma_{\lambda_0-\varepsilon}$ .

This contradicts the choice of $\lambda_0$ .

latex

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Tháng Chín 13, 2009

An algebraic identity for 9th level

Chuyên mục: Các Bài Tập Nhỏ, Linh Tinh — Ngô Quốc Anh @ 21:16

In this topic I will show you how to prove

$\sqrt[3]{2} + \sqrt[3]{{20}} - \sqrt[3]{{25}} = 3\sqrt {\sqrt[3]{5} - \sqrt[3]{4}}$

In order to prove that fact, we just do as following: by using

$(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)$

we obtain

${\Big( {\sqrt[3]{2} + \underbrace {\sqrt[3]{{20}}}_{\sqrt[3]{{{2^2}}}\sqrt[3]{5}} - \underbrace {\sqrt[3]{{25}}}_{\sqrt[3]{{{5^2}}}}} \Big)^2}=\sqrt[3]{{{2^2}}} + \sqrt[3]{{{2^4}}}\sqrt[3]{{{5^2}}} + \sqrt[3]{{{5^4}}} + 2\left( {\underbrace {\sqrt[3]{2}\sqrt[3]{{{2^2}}}}_2\sqrt[3]{5} - \sqrt[3]{2}\sqrt[3]{{{5^2}}} - \sqrt[3]{{{2^2}}}\underbrace {\sqrt[3]{5}\sqrt[3]{{{5^2}}}}_5} \right).$

Therefore

${\left( {\sqrt[3]{2} + \sqrt[3]{{20}} - \sqrt[3]{{25}}} \right)^2} = \sqrt[3]{{{2^2}}} + 2\sqrt[3]{2}\sqrt[3]{{{5^2}}} + 5\sqrt[3]{5} + 2\left( {2\sqrt[3]{5} - \sqrt[3]{2}\sqrt[3]{{{5^2}}} - 5\sqrt[3]{{{2^2}}}} \right).$

Clearly

$\sqrt[3]{{{2^2}}} + 2\sqrt[3]{2}\sqrt[3]{{{5^2}}} + 5\sqrt[3]{5} + 2\left( {2\sqrt[3]{5} - \sqrt[3]{2}\sqrt[3]{{{5^2}}} - 5\sqrt[3]{{{2^2}}}} \right) = 9\left( {\sqrt[3]{5} - \sqrt[3]{{{2^2}}}} \right).$

Thus

${\left( {\sqrt[3]{2} + \sqrt[3]{{20}} - \sqrt[3]{{25}}} \right)^2} = 9\left( {\sqrt[3]{5} - \sqrt[3]{4}} \right)$

which yields

$\left| {\sqrt[3]{2} + \sqrt[3]{{20}} - \sqrt[3]{{25}}} \right| = 3\sqrt {\sqrt[3]{5} - \sqrt[3]{4}} .$

Finally, by using the fact that $(a+b)^3=a^3+3a^2b+3ab^2+b^3$ we get

${\left( {\sqrt[3]{2} + \sqrt[3]{{20}}} \right)^3} - 25 = \underbrace {\left( {22 + 3\sqrt[3]{{{2^2}}}\sqrt[3]{{20}} + 3\sqrt[3]{2}\sqrt[3]{{{{20}^2}}}} \right) - 25}_{3\left( {\sqrt[3]{{{2^2}}}\sqrt[3]{{20}} + \sqrt[3]{2}\sqrt[3]{{{{20}^2}}} - 1} \right)} > 0$

which implies

${\sqrt[3]{2} + \sqrt[3]{{20}} - \sqrt[3]{{25}}}>0.$

In other words,

$\sqrt[3]{2} + \sqrt[3]{{20}} - \sqrt[3]{{25}} = 3\sqrt {\sqrt[3]{5} - \sqrt[3]{4}}.$

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QE in Department of Mathematics, National University of Singapore, August 2009

Chuyên mục: Giải Tích Cổ Điển, Giải Tích Hiện Đại, Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 14:53

I have just passed QE held in August 2009 for my first attendance, I hereby show you the analysis paper

Question 1 [10 marks]. Suppose $f$ and $g$ are both measurable functions on the interval $(0,1)$ such that for all $t \in \mathbb R^1$

$| \{ x \in(0,1):f(x)\geq t\}| = |\{ x\in (0,1):g(x)\geq t\}|$

Assume that $f$ and $g$ both are monotone decreasing and continuous from left. Can you conclude that $f(x)=g(x)$ for all $x \in (0,1)$ ? Give the reason to support your answer.

Question 2 [10 marks]. Compute the volume of the region bounded by

${\left( {{a_{11}}x + {a_{12}}y + {a_{13}}z} \right)^2} + {\left( {{a_{21}}x + {a_{22}}y + {a_{23}}z} \right)^2} + {\left( {{a_{31}}x + {a_{32}}y + {a_{33}}z} \right)^2} = 1$

where the determinant of the $3 \times 3$ matrix $(a_{ij})$ is NOT equal to zero.

Question 3 [10 marks]. Let $D$ be a measureable set in $\mathbb R^n$ with finite measure. Suppose $\phi(x,t)$ is a real valued continuous function on $D \times \mathbb R^1$ such that for almost every $x \in D$ , $\phi(x,t)$ is a continuous function of $t$ and for every real number $t$ , $\phi(x,t)$ is measurable function of $x$ . If $\{f_n\}$ is a sequence of measurable functions on $D$ that converges to $f$ in measure, show that $\{\phi(x,f_n(x))\}$ converges to $\phi(x,f(x))$ in measure.

Question 4 [10 marks]. Find the function

$I\left( y \right) =\displaystyle\int\limits_0^\infty {{e^{ - a{x^2}}}\cos \left( {yx} \right)dx}$

if $a>0$ is a constant. Justify your answer.

Question 5 [10 marks]. Compute the intergal

$\displaystyle\int\limits_0^\pi {\frac{{x\sin x}}{{1 + {a^2} - 2a\cos x}}dx}$

where $a>0$ is a constant.

Question 6 [10 marks]. Supposet $f(z)$ is a holomorphic function on the complex plane $\mathbb C$ . If $f$ locally keeps the area invariant, what will the function $f$ be?

Question 7 [10 marks]. Is there an analytic function $f$ on $\Delta$ (unit disk in the complex plane with center $0$ ) such that $|f(z)|<1$ for $|z|<1$ with $f(0)=\frac{1}{2}$ and $f'(0)=\frac{3}{4}$ ? If so, find such an $f$ . Is it unique?

Question 8 [10 marks]. Let $m<n$ be two positive integers and $\Omega$ and $G$ be open subsets in $\mathbb R^n$ and $\mathbb R^m$ respectively. Does there exist a map $f :\Omega \to G$ such that $f$ and the inverse of $f$ are both $C^1$ ? Justify your answer.

Question 9 [10 marks]. Is there a square integrable function $f$ on $[0,\pi]$ such that both inequalities

$\displaystyle\int\limits_0^\pi{{{\left( {f\left( x \right)-\sin x} \right)}^2}dx}\leq\frac{4}{9}$

and

$\displaystyle\int\limits_0^\pi{{{\left( {f\left( x \right)-\cos x} \right)}^2}dx}\leq\frac{1}{9}$

hold? Justify your answer.

Question 10 [10 marks]. Let $\alpha_k$ for $k=1,2,...,n$ be $n$ real numbers such that $0<\alpha_k<\pi$ for any $k$ . Define

$\alpha=\displaystyle\frac{1}{n}\sum\limits_{k=1}^{n}{\alpha_k}.$

Show that

$\displaystyle{\left( {\prod\limits_{k = 1}^n {\frac{{\sin {\alpha _k}}}{{{\alpha _k}}}} } \right)^{\frac{1}{n}}} \leq \frac{{\sin \alpha }}{\alpha }.$

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Tháng tám 29, 2009

On a polynomials of degree n, having all its zeros in the unit dis

Chuyên mục: Các Bài Tập Nhỏ, Giải tích 7 (MA4247), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 21:07

I found a very useful inequality involving a polynomials of degree $n$ , having all its zeros in the unit disk from the following paper, doi:10.1016/j.jmaa.2009.07.049, published in J. Math. Anal. Appl.

Statement. If $P(z)$ is a polynomial of degree $n$ , having all its zeros in the disk $|z| \leq 1$ , then

$\left| {zP'(z)} \right| \geq \displaystyle\frac{n} {2}\left| {P(z)} \right|$

for $|z|=1$ .

Proof. Since all the zeros of $P(z)$ lie in $latex|z| \leq 1$. Hence if $z_1, z_2,...,z_n$ are the zeros of $P(z)$ , then $|z_j| \leq 1$ for all $j =1,2,...,n$ . Clearly,

$\displaystyle\Re \frac{{{e^{i\theta }}P'\left( {{e^{i\theta }}} \right)}}{{P\left( {{e^{i\theta }}} \right)}} = \sum\limits_{j = 1}^n {\Re \frac{{{e^{i\theta }}}}{{{e^{i\theta }} - {z_j}}}} ,$

for every point $e^{i\theta}$ , $0 \leq \theta < 2 \pi$ which is not a zero of $P(z)$ . Note that

$\displaystyle\sum\limits_{j = 1}^n {\Re \frac{e^{i\theta }}{e^{i\theta } - z_j}}\geq\sum\limits_{j = 1}^n{\frac{1}{2}}=\frac{n}{2}.$

This implies

$\displaystyle\left| {\frac{{{e^{i\theta }}P'\left( {{e^{i\theta }}} \right)}}{{P\left( {{e^{i\theta }}} \right)}}} \right| \geq \Re \frac{{{e^{i\theta }}P'\left( {{e^{i\theta }}} \right)}}{{P\left( {{e^{i\theta }}} \right)}} \geq \frac{n}{2},$

for every point $e^{i \pi}$ , $0 \leq \theta < 2\pi$ . Hence

$\left| {zP'(z)} \right| \geq \displaystyle \frac{n}{2}\left| {P(z)} \right|$

for $|z|=1$ and this completes the proof.

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Tháng tám 18, 2009

An other limit supremum of sin function

Chuyên mục: Các Bài Tập Nhỏ, Giải Tích 1, Giải Tích 6 (MA5205), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 12:35

In the topic we showed that for any irrational $\alpha$ the limit $\mathop {\lim }\limits_{n \to \infty } \sin \left( {n\alpha \pi } \right)$ does not exist. In this topic, we consider the following limit

$\mathop {\overline {\lim } }\limits_{n \to \infty } \sin \left( {nx} \right) .$

To be precise, we prove that

$\mathop {\overline {\lim } }\limits_{n \to \infty } \sin \left( {nx} \right) = 1$

for almost every $x \in [0,2\pi]$ .

Solution. Let

$A = \left\{ {x \in \left( {0,2\pi } \right): \frac{x} {\pi } \notin \mathbb{Q}} \right\}.$

Then is a measurable set of measure $2\pi$ . Moreover, for any $x \in A$ ,

$\mathop {\overline {\lim } }\limits_{n \to \infty } \sin \left( {nx} \right) = 1.$

Indeed for any $x \in A$ , since

$\left\{ {k\frac{x} {\pi } - 2l: l \in \mathbb{Z}} \right\}$

is dense subgroup of $\mathbb R$ there are sequences $\{k_n\}$ and $\{l_n\}$ of $\mathbb Z$ such that

$\mathop {\lim }\limits_{n \to \infty } \left( {{k_n}\frac{x} {\pi } - {l_n}} \right) = \frac{1} {2}.$

Since

$\frac{1} {2} \notin \left\{ {k\frac{x} {\pi } - 2l: k,l \in \mathbb{Z}} \right\}$

$\{k_n\}$ admits a subsequence $\{k'_n\}$ either increasing to $+\infty$ or decreasing to $-\infty$ . If $\mathop {\lim }\limits_{n \to \infty } {{k'}_n} = + \infty$ then

$\mathop {\lim }\limits_{n \to \infty } \sin \left( {{{k'}_n}x} \right) = \mathop {\lim }\limits_{n \to \infty } \sin \left( {...$

Otherwise $\mathop {\lim }\limits_{n \to \infty } \left( { - 3{{k'}_n}} \right) = + \infty$ and

$\mathop {\lim }\limits_{n \to \infty } \sin \left( { - 3{{k'}_n}x} \right) = \mathop {\lim }\limits_{n \to \infty } \sin \lef...$

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Tháng tám 17, 2009

The use of equivalence in measure

Chuyên mục: Các Bài Tập Nhỏ, Giải Tích 6 (MA5205), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 14:19

In mathematics, and specifically in measure theory, equivalence is a notion of two measures being “the same”. Two measures are equivalent if they have the same null sets.

Definition. Let $(X, \Sigma)$ be a measurable space, and let $\mu, \nu : \Sigma \to [0, +\infty]$ be two measures. Then $\mu$ is said to be equivalent to $\nu$ if

$\mu (A) = 0 \iff \nu (A) = 0$

for measurable sets in $\Sigma$ , i.e. the two measures have precisely the same null sets. Equivalence is often denoted $\displaystyle{\mu \sim \nu}$ or $\mu \approx \nu$ .

In terms of absolute continuity of measures, two measures are equivalent if and only if each is absolutely continuous with respect to the other:

$\mu \sim \nu \iff \mu \ll \nu \ll \mu.$

Equivalence of measures is an equivalence relation on the set of all measures $\Sigma \to [0, +\infty]$ .

Examples.

Gaussian measure and Lebesgue measure on the real line are equivalent to one another.
Lebesgue measure and Dirac measure on the real line are inequivalent.

Application. Let $\mu$ be a finite measure on $\mathbb R$ , and define

$f\left( x \right) = \int\limits_{ - \infty }^{ + \infty } {\frac{{\ln \left| {x - t} \right|}} {{\sqrt {\left| {x - t} \right...$

Show that f(x) is finite a.e. with respect to the Lebesgue measure on $\mathbb R$ .

Proof. Let

$g\left( x \right) = \left\{ \begin{gathered} \frac{{\ln \left| x \right|}} {{\sqrt {\left| x \right|} }},x \ne 0, \hfill \\...$

then $g \in L^1(\mathbb R, d\nu)$ where

$d\nu \left( x \right) = \frac{{dx}} {{1 + {x^2}}}$

and

$f\left( x \right) = \int_{ - \infty }^{ + \infty } {g\left( {x - t} \right)d\mu \left( t \right)} ,x \in \mathbb{R}.$

Clearly

$\int_{ - \infty }^{ + \infty } {\left| {f\left( x \right)} \right|d\mu \left( x \right)} \leqslant \int_{ - \infty }^{ + \in...$

and by Fubini’s Theorem

$\int_{ - \infty }^{ + \infty } {\left( {\int_{ - \infty }^{ + \infty } {\left| {g\left( {x - t} \right)} \right|d\mu \left( t...$

then

$\int_{ - \infty }^{ + \infty } {\left| {f\left( x \right)} \right|d\mu \left( x \right)} \leqslant \int_{ - \infty }^{ + \in...$

Since

$\begin{gathered} \int\limits_{ - \infty }^{ + \infty } {\left( {\int\limits_{ - \infty }^{ + \infty } {\left| {g\left( {x -...$

Thus the following function

$x \mapsto \int\limits_{ - \infty }^{ + \infty } {\left| {g\left( {x - t} \right)} \right|d\mu \left( t \right)}$

finite a.e. with respect to the measure $\nu$ . The conclusion follows from that the measure $\nu$ and the Lebesgue measure are equivalent.

Source: http://en.wikipedia.org/wiki/Equivalence_(measure_theory)

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Tháng tám 16, 2009

On the stability of the Runge-Kutta 4 (RK4)

Chuyên mục: Các Bài Tập Nhỏ, Giải tích 9 (MA5265), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 14:11

In the literature, the so-called RK4 is given as following: we first define the following coefficients

$\begin{gathered} {K_1} = f\left( {{t_i},{y_i}} \right), \hfill \\ {K_2} = f\left( {{t_i} + \frac{1} {2}\Delta t,{y_i} + \...$

then

${y_{i + 1}} = {y_i} + \frac{{\Delta t}} {6}\left( {{K_1} + 2{K_2} + 2{K_3} + {K_4}} \right).$

This is the most important iterative method for the approximation of solutions of ordinary differential equations

$y' = f(t, y), \quad y(t_0) = y_0.$

This technique was developed around 1900 by the German mathematicians C. Runge and M.W. Kutta.

In order to study its stability, we use the model problem

$y' = \lambda y, \quad \Re \lambda < 0.$

In other words, we replace f(t,y) by $\lambda y$ . Then the stability condition for time step $\Delta$ comes from the following condition

$\left| \frac{y_{i+1}}{y_i} \right| \leq 1.$

Applying the above discussion to RK4 method, we see that

$\begin{gathered} {K_4} = \lambda \left( {{y_i} + \Delta t{K_3}} \right), \hfill \\ {K_3} = \lambda \left( {{y_i} + \frac{...$

which implies

$\begin{gathered} {K_4} = \lambda \left( {{y_i} + \Delta t\lambda \left( {{y_i} + \frac{1} {2}\Delta t\lambda \left( {{y_i} ...$

which yields

${y_{i + 1}} = {y_i} + \frac{{\Delta t}} {6}\left( {\lambda {y_i} + 2\lambda {y_i}\left( {1 + \frac{1} {2}\Delta t\lambda } \r...$

Thus

$\frac{{{y_{i + 1}}}} {{{y_i}}} = 1 + \frac{{\Delta t}} {6}\left[ {\lambda + 2\lambda \left( {1 + \frac{1} {2}\Delta t\lambda...$

which is nothing but

$\frac{{{y_{i + 1}}}} {{{y_i}}} = 1 + \Delta t\lambda + \frac{{{{\left( {\Delta t\lambda } \right)}^2}}} {2} + \frac{{{{\left...$

Therefore, the stability condition is given as follows

$\left| {1 + z + \frac{{{z^2}}} {2} + \frac{{{z^3}}} {6} + \frac{{{z^4}}} {{24}}} \right| \leq 1, \quad \Re z < 0.$

Để lại phản hồi

The eigenvalues of a common tridiagonal matrix

Chuyên mục: Các Bài Tập Nhỏ, Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 3:36

The eigenvalues of then $N \times N$ tridiagonal matrix

$\left( {\begin{array}{*{20}{c}} a & b & {} & {} & {} & {} & {} \\ c & a & b & {} &...$

are

${\lambda _s} = a + 2\sqrt {bc} \cos \frac{{s\pi }} {{N + 1}}$

and its conresponding eigenvector is

${v_s}^T = \left( {{{\left( {\frac{c} {b}} \right)}^{\frac{1} {2}}}\sin \frac{{s\pi }} {{N + 1}},{{\left( {\frac{c} {b}} \righ...$

Proof. Let $\lambda$ present an eigenvalue of the given matrix (denoted by ) and the corresponding eigenvector with components v_1,...,v_N . Then

$\left( {\begin{array}{*{20}{c}} a & b & {} & {} & {} & {} & {} \\ c & a & b & {} &...$

If we defined $v_0 = v_{N+1}=0$ , then we have

$c{v_{i - 1}} + \left( {a - \lambda } \right){v_i} + b{v_{i + 1}} = 0,i = \overline {1,N} .$

The solution of the above equation is of the form

${v_i} = Bm_1^i + Cm_2^i$

where B, C are constants and m_1, m_2 are the roots of the equation

$c+(a-\lambda)m + bm^2=0.$

Since $v_0 = v_{N+1}=0$ , then 0=B+C and $0 = Bm_1^{N + 1} + Cm_2^{N + 1}$ . Hence,

${\left( {\frac{{{m_1}}} {{{m_2}}}} \right)^{N + 1}} = 1 = {e^{\sqrt { - 1} i2\pi }}$

$\frac{{{m_1}}} {{{m_2}}} = {e^{\frac{{\sqrt { - 1} i2\pi }} {{N + 1}}}}.$

We also have

${m_1}{m_2} = \frac{c} {b}, \quad {m_1} + {m_2} = \frac{{\lambda - a}} {b}.$

Finally,

$\lambda = a + b\sqrt {\frac{c} {b}} \left( {{e^{\frac{{\sqrt { - 1} i2\pi }} {{N + 1}}}} + {e^{ - \frac{{\sqrt { - 1} i2\pi ...$

The -component of the eigenvector is

${v_j} = Bm_1^j + Cm_2^j = B\sqrt {{{\left( {\frac{c} {b}} \right)}^j}} \left( {{e^{\frac{{\sqrt { - 1} i2\pi }} {{N + 1}}}} -...$

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The eigenvalues of a common tridiagonal matrix

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