Giải Tích Hiện Đại « Ngô Quốc Anh

Tháng Mười Một 5, 2009

An equivalent criterion for absolutely continuous functions

Chuyên mục: Các Bài Tập Nhỏ, Giải Tích 6 (MA5205) — Ngô Quốc Anh @ 22:29

In mathematics, absolute continuity is a smoothness property which is stricter than continuity and uniform continuity.

Definition. A finite function $f$ on a finite interval $[a,b]$ is said to be absolute continuous if and only if for given $\varepsilon > 0$ , there exists $\delta > 0$ such that

$\displaystyle\sum_k |f(b_k) - f(a_k)| < \varepsilon$

for any collection (finite or not) $\{[a_k, b_k]\}$ of non-overlapping subintervals of $[a, b]$ with $\sum (b_k - a_k) < \delta$ .

Statement. Show that $f$ is absolutely continuous on $[a, b]$ if and only if given $\varepsilon > 0$ , there exists $\delta > 0$ such that

$\displaystyle \Big|\sum_k (f(b_k) - f(a_k)) \Big| < \varepsilon$

for any finite collection $\{[a_k, b_k]\}$ of non-overlapping subintervals of $[a, b]$ with $\sum (b_k - a_k) < \delta$ .

Proof. If $f$ is absolutely continuous on $[a, b]$ , then the result is easily obtained by using the definition and the fact that $|x + y| \leq |x| + |y|$ for every $x,y \in \mathbb R$ .

Now we prove that

for given $\varepsilon > 0$ , there exists $\delta > 0$ such that $\sum |f(b_k) - f(a_k)| < \varepsilon$ for any finite collection $\{[a_k, b_k]\}$ of non-overlapping subintervals of $[a, b]$ with $\sum (b_k - a_k) < \delta$ .

Indeed, we split the collection $\{[a_k, b_k]\}$ into two types:

type $A$ are all $k$ such that $f(b_k) - f(a_k) \geq 0$ and
type $B$ are all $k$ such that $f(b_k) - f(a_k) < 0$ .

For given $\varepsilon > 0$ , there exists $\delta > 0$ such that $| \sum (f(b_k) - f(a_k))| < \frac{\varepsilon}{3}$ for any finite collection $\{[a_k, b_k]\}$ with $\sum (b_k - a_k) < \delta$ . Then for $k \in A$ we also have

$\displaystyle\sum_{k \in A} \left( f(b_k) - f(a_k) \right) = \Big| \sum_{k \in A} (f(b_k) - f(a_k))\Big| < \frac{\varepsilon}{3}$ .

Similarly,

$\displaystyle\sum_{k \in B} \left( f(a_k) - f(b_k) \right) = \Big| \sum_{k \in B} (f(b_k) - f(a_k))\Big| < \frac{\varepsilon}{3}$ .

From the following inequality $a + b \leq |a-b| + b + b$ with $a,b \geq 0$ we deduce that

$\displaystyle\sum\limits_{k = 1}^n {\left| {f\left( {{b_k}} \right) - f\left( {{a_k}} \right)} \right|} = \underbrace {\sum\limits_{k \in A} {f\left( {{b_k}} \right) - f\left( {{a_k}} \right)} }_a + \underbrace {\sum\limits_{k \in B} {f\left( {{a_k}} \right) - f\left( {{b_k}} \right)} }_b$

with the fact that

$\displaystyle a+b \leqslant \left| {\sum\limits_{k \in A} {f\left( {{b_k}} \right) - f\left( {{a_k}} \right)} - \sum\limits_{k \in B} {f\left( {{a_k}} \right) - f\left( {{b_k}} \right)} } \right| + \sum\limits_{k \in B} {f\left( {{a_k}} \right) - f\left( {{b_k}} \right)} + \sum\limits_{k \in B} {f\left( {{a_k}} \right) - f\left( {{b_k}} \right)}.$

Note that the right hand side of the above inequality is bounded from above by

$\displaystyle\frac{\varepsilon }{3} + \frac{\varepsilon }{3} + \frac{\varepsilon }{3} = \varepsilon.$

Thus, we have proved that for given $\varepsilon >0$ , there exists $\delta>0$ such that

$\displaystyle\sum\limits_{k = 1}^n {\left| {f\left( {b_k } \right) - f\left( {a_k } \right)} \right|} < \varepsilon$

for any finite collection $\{[a_k, b_k]\}$ of non-overlapping subintervals of $[a, b]$ with $\sum (b_k - a_k) < \delta$ . Letting $n \to \infty$ we can claim that $f$ is absolutely continuous.

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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 Mười 17, 2009

The Brezis-Lieb lemma and several applications

Chuyên mục: Các Bài Tập Nhỏ, Giải Tích 6 (MA5205), Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 14:34

What we did in this topic was just an $L^p$ version of the Brezis-Lieb lemma. In this topic, we will discuss the generalization of this lemma.

Roughly speaking, what we are going to prove is the following: If $j : \mathbb C \to \mathbb C$ is a continuous function such that $j(0) = 0$ , then, when $f_n \to f$ a.e. and

$\displaystyle\int |j(f_n(x))| d\mu(x) \leq C < \infty$

we claim that

$\displaystyle\lim\limits_{n \to \infty} \int \left[ j(f_n) - j(f_n - f)\right] = \int j(f)$

under suitable conditions on $j$ and/or $\{f_n\}$ .

To be exact, in addition let $j$ satisfy the following hypothesis:

For every sufficiently small $\varepsilon>0$ , there exists two continuous, nonnegative functions $\varphi_\varepsilon$ and $\psi_\varepsilon$ such that

$\displaystyle |j(a+b)-j(a)| \leq \varepsilon \varphi_\varepsilon(a) + \psi_\varepsilon(b)$

for all $a, b \in \mathbb C$ .

Theorem. Let $j$ satisfy the above hypothesis and let $f_n = f+g_n$ be a sequence of measurable functions from $\Omega$ to $\mathbb C$ such that

$g_n \to 0$ a.e.
$j(f) \in L^1$ .
$\displaystyle\int \varphi_\varepsilon(g_n(x))d\mu(x) \leq C < \infty$ for some constant $C$ , independent of $\varepsilon$ and $n$ .
$\displaystyle\int \psi_\varepsilon(f(x)) d\mu(x) < \infty$ for all $\varepsilon >0$ .

Then, as $n \to \infty$ ,

$\displaystyle\lim\limits_{n \to \infty} \int \left| j(f+g_n) - j(g_n) - j(f) \right| d\mu =0.$

Proof. Fix $\varepsilon >0$ and let

$\displaystyle W_{\varepsilon, n} (x) = \Big[ \big|j(f_n(x)) -j(g_n(x)) - j(f(x))\big| - \varepsilon \varphi_\varepsilon (g_n(x))\Big]_+,$

where $[a]_+ = \max\{a,0\}$ . As $n \to \infty$ , $W_{\varepsilon, n} (x) \to 0$ a.e. On the other hand,

$\displaystyle \big| j(f_n) - f(g_n) - j(f)\big| \leq |j(f_n) - j(g_n)| + |j(f)| \leq \varepsilon \varphi_\varepsilon(g_n) + \psi_\varepsilon(f) + |j(f)|.$

Therefore, $W_{\varepsilon, n} \leq \psi_\varepsilon(f) + |j(f)| \in L^1$ . By the Lebesgue Dominated Convergence theorem, $\displaystyle\int W_{\varepsilon, n} d\mu \to 0$ as $n \to \infty$ . However,

$\displaystyle |j(f_n) - j(g_n) - j(f)| \leq W_{\varepsilon, n} +\varepsilon \varphi_\varepsilon(g_n)$

and thus

$\displaystyle I_n \equiv \int \big| j(f_n) - j(g_n) - j(f) \big| d\mu\leq \int \big[ W_{\varepsilon, n} + \varepsilon \varphi_\varepsilon(g_n)\big] d\mu .$

Consequently, $\limsup_{n \to \infty} I_n \leq \varepsilon C$ . Now let $\varepsilon \to 0$ .

Applications.

The simplest example is when we choose $j(x)=|x|^p$ where $0< p<\infty$ . In this situation, one has

$\displaystyle \int \Big(|f_n|^p - |f_n - f|^p - |f|^p \Big) d\mu \to 0.$

We now assume $u_n \rightharpoonup u$ in $W^{1, 2}$ . As a consequence and up to a subsequence, $u_n \to u$ in $L^\alpha$ for every $1<\alpha<2^\star := \frac{2n}{n-2}$ and $u_n \to u$ a.e. Therefore, for a fixed $q \in (2, 2^\star)$ , the fact that $u_n \to u$ in $L^q$ implies, by the Brezis-Lieb lemma, that
$\displaystyle u_n^{q-1} \to u^{q-1}$ in $L^\frac{q}{q-1}$ .

This is because $\{u_n^{q-1}\}_n \subset L^\frac{q}{q-1}$ is bounded, $u_n^{q-1} \to u^{q-1}$ a.e. and

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \int {\Big( {\underbrace {{{\left| {u_n^{q - 1}} \right|}^{\frac{q}{{q - 1}}}}}_{{{\left| {{u_n}} \right|}^q}} - {{\left| {u_n^{q - 1} - {u^{q - 1}}} \right|}^{\frac{q}{{q - 1}}}}} \Big)d\mu }=\int {\underbrace {{{\left| {{u^{q - 1}}} \right|}^{\frac{q}{{q - 1}}}}}_{{{\left| u \right|}^q}}d\mu }$ .

$u_n \to u$

$L^p$

$\lim_{n\to \infty} \int |u_n|^p d\mu = \int |u|^p d\mu$

$\displaystyle \mathop {\lim }\limits_{x \to \infty }\int {{{\left| {u_n^{q - 1} - {u^{q - 1}}} \right|}^{\frac{q}{{q - 1}}}}d\mu } = 0$ .

$\displaystyle \mathop {\lim }\limits_{x \to \infty } \int {\left( {u_n^{q - 1} - {u^{q - 1}}} \right)\left( {{u_n} - u} \right)d\mu } = 0$ .

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Tháng Mười 13, 2009

Strong convergence in L^p implies convergence a.e.

Chuyên mục: Các Bài Tập Nhỏ, Giải Tích 6 (MA5205), Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 21:49

This topic is to show how to prove the following statement: if $\{u_n\}_n$ converges strongly to some $u$ in $L^p(\Omega)$ , then up to a subsequence, $\{u_n\}_n$ converges almost everywhere to $u$ in $\Omega$ .

The proof relies on the so-called Tchebyshev’s inequality. To this end, we first observe that $\{u_n\}_n$ converges strongly to $u$ in $L^p(\Omega)$ means

$\displaystyle\lim\limits_{n \to \infty } \int_\Omega {{{\left| {{u_n} - u} \right|}^p}dx} = 0.$

We now apply the Tchebyshev’s inequality, indeed, for each $\varepsilon>0$ one has

$\displaystyle meas\left\{ {x:\left| {{u_n}(x) - u(x)} \right| >\varepsilon } \right\} \leqslant \frac{1}{{{\varepsilon ^p}}}\int_{\left\{ {x:\left| {{u_n}(x) - u(x)} \right| > \varepsilon } \right\}} {{{\left| {{u_n} - u} \right|}^p}dx} .$

The right hand side of the above inequality can be dominated by

$\displaystyle\frac{1}{{{\varepsilon ^p}}}\int_\Omega {{{\left| {{u_n} - u} \right|}^p}dx}$

which implies that

$\displaystyle 0 \leqslant \mathop {\lim }\limits_{n \to \infty } meas\left\{ {x:\left| {{u_n}(x) - u(x)} \right| > \varepsilon } \right\} \leqslant \mathop {\lim }\limits_{n \to \infty } \left( {\frac{1} {{{\varepsilon ^p}}}\int_\Omega {{{\left| {{u_n} - u} \right|}^p}dx} } \right) = 0.$

Thus $u_n$ converges to $u$ in measure. It turns out that up to a subsequence, $u_n$ converges to $u$ almost everywhere.

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Tháng Chín 13, 2009

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

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

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