Giải Tích 6 (MA5205) « Ngô Quốc Anh

Tháng Mười Một 30, 2009

A property of the essentially bounded function 2

Chuyên mục: Các Bài Tập Nhỏ, Giải Tích 6 (MA5205) — Ngô Quốc Anh @ 22:52

This topic is a companion to the following topic. In this topic, we consider the case when $E$ is the whole space, i.e. $E = \mathbb R^n$ . We also add an extra function $g$ to $a_n$ . To be precise, we have

Question. Suppose $g>0$ on $\mathbb R^n$ is in $L^1(\mathbb R^n)$ in Lebesgue sense. Let $f \in L^\infty(\mathbb R^n)$ such that $\| f\|_\infty > 0$ . Define

$\displaystyle {a_n} = \int_{\mathbb R^n} {{{\left| f \right|}^ng}}$

for $n=1,2,3,...$ Show that

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \frac{{{a_{n + 1}}}} {{{a_n}}} = {\left\| f \right\|_\infty }$ .

Solution. For any $\alpha$ with $0<\alpha < \|f\|_\infty$ , let

$\displaystyle{E_\alpha } = \left\{ {x \in E: f\left( x \right) \geqslant \alpha } \right\}$

and

$\displaystyle {F_\alpha } = E\backslash {E_\alpha }$

then $\infty> |E_\alpha|>0$ . Clearly when $\alpha$ is sufficiently closed to $\|f\|_\infty$ , $\int_{E_\alpha}g>0$ . For any $k \in \mathbb N$ ( $k$ can be zero), note that

$\displaystyle\int_{{E_\alpha }} {{{\left| f \right|}^n}g} \geqslant {\alpha ^n}\int_{{E_\alpha }} g$

and

$\displaystyle\int_{{F_\alpha }} {{{\left| f \right|}^{n + k}}g} \leqslant \left\| f \right\|_\infty ^k\int_{{F_\alpha }} {{{\left| f \right|}^n}g}$ .

Then

$\displaystyle\frac{{\int_{{F_\alpha }} {{{\left| f \right|}^{n + k}}g} }}{{\int_{{E_\alpha }} {{{\left| f \right|}^n}g} }} \leqslant \frac{{\left\| f \right\|_\infty ^k\int_{{F_\alpha }} {{{\left| f \right|}^n}g} }}{{{\alpha ^n}\int_{{E_\alpha }} g }} = \frac{{\left\| f \right\|_\infty ^k}}{{\int_{{E_\alpha }} g }}\int_{{F_\alpha }} {{{\left| {\frac{f}{\alpha }} \right|}^n}g}$ .

By the Dominated Convergence Theorem, one gets

$\displaystyle 0 \leqslant \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{\int_{{F_\alpha }} {{{\left| f \right|}^{n + k}}g} }}{{\int_{{E_\alpha }} {{{\left| f \right|}^n}g} }}} \right) \leqslant \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{\left\| f \right\|_\infty ^k}}{{\int_{{E_\alpha }} g }}\int_{{F_\alpha }} {{{\left| {\frac{f}{\alpha }} \right|}^n}g} } \right) = 0$ .

Hence

$\displaystyle\begin{gathered}\mathop {\lim \inf }\limits_{n \to \infty } \left( {\frac{{\int\limits_E {{{\left| f \right|}^{n + 1}}g} }}{{\int\limits_E {{{\left| f \right|}^n}g} }}} \right) \geqslant \mathop {\lim \inf }\limits_{n \to \infty } \left( {\frac{{\int\limits_{{F_\alpha }} {{{\left| f \right|}^{n + 1}}g} + \int\limits_{{E_\alpha }} {{{\left| f \right|}^{n + 1}}g} }}{{\int\limits_{{E_\alpha }} {{{\left| f \right|}^n}g} + \int\limits_{{F_\alpha }} {{{\left| f \right|}^n}g} }}} \right) \hfill \\\qquad \geqslant \mathop {\lim \inf }\limits_{n \to \infty } \left( {\frac{{\int\limits_{{F_\alpha }} {{{\left| f \right|}^{n + 1}}g} + \alpha \int\limits_{{E_\alpha }} {{{\left| f \right|}^n}g} }}{{\int\limits_{{E_\alpha }} {{{\left| f \right|}^n}g} + \int\limits_{{F_\alpha }} {{{\left| f \right|}^n}g} }}} \right) \hfill \\\qquad = \mathop {\lim \inf }\limits_{n \to \infty } \left( {\frac{{\int\limits_{{F_\alpha }} {{{\left| f \right|}^{n + 1}}g} }}{{\int\limits_{{E_\alpha }} {{{\left| f \right|}^n}g} }} + \alpha } \right)/\left( {1 + \frac{{\int\limits_{{F_\alpha }} {{{\left| f \right|}^n}g} }}{{\int\limits_{{E_\alpha }} {{{\left| f \right|}^n}g} }}} \right) \hfill \\ \qquad = \alpha . \hfill \\ \end{gathered}$

Letting $\alpha \nearrow {\left\| f \right\|_\infty }$ , we get that

$\displaystyle\mathop{\lim }\limits_{n\to\infty }\left({\frac{{\int_{E}{{{\left| f\right|}^{n+1}g}}}}{{\int_{E}{{{\left| f\right|}^{n}g}}}}}\right) ={\left\| f\right\|_\infty }$ .

As an application, if we put $a_0 = 1$ , then from

$\displaystyle {a_{n + 1}} = \frac{{{a_1}}} {{{a_0}}}.\frac{{{a_2}}} {{{a_1}}} \cdots\frac{{{a_{n + 1}}}} {{{a_n}}}$

we deduce that

$\displaystyle\mathop{\lim }\limits_{n\to\infty }\sqrt[n]{{{a_{n}}}}=\mathop{\lim }\limits_{n\to\infty }\frac{{{a_{n+1}}}}{{{a_{n}}}}={\left\| f\right\|_\infty }$ .

In other words,

$\displaystyle\mathop{\lim }\limits_{n\to\infty }{\left({\int_{E}{{{\left| f\right|}^{n}g}}}\right)^{\frac{1}{n}}}={\left\| f\right\|_\infty }$ .

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Tháng Mười Một 10, 2009

A trivial identity of probability measures

Chuyên mục: Các Bài Tập Nhỏ, Giải Tích 6 (MA5205) — Ngô Quốc Anh @ 14:36

Let us consider a probability space $(X,\mathcal B,\mu)$ , i.e., $(X,\mathcal B,\mu)$ is a measurable space together with $\mu(X)=1$ . We assume further that $A, B \in \mathcal B$ are such that $\mu(A)=\mu(B)=1$ . Then we conclude that $\mu(A \cap B)=1$ .

Indeed, since $A \subset A \cup B \subset X$ then $1=\mu(A\cup B)$ . We write $A \cup B$ in the following way

$A\cup B = A\backslash B \quad \bigcup \quad A \cap B \quad\bigcup \quad B\backslash A$ .

We then see that $\mu(A\backslash B)=0$ since $A\backslash B \subset X\backslash B$ . Similarly, $\mu(B\backslash A)=0$ . Hence, $\mu(A \cap B)=1$ .

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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 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 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 15, 2009

Several questions involving the Vitali set

Chuyên mục: Các Bài Tập Nhỏ, Giải Tích 6 (MA5205), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 14:36

We denote by the Vitali set which is defined as follows:

We say that $x, y \in [0, 1)$ are equivalent, and write $x \sim y$ , if and only if is a rational number. This equivalence relation partitions into an uncountable family of disjoint equivalence classes. By the axiom of choice there is a set which contains exactly one element from each equivalence class.

Now let $\{ \tau_n\}$ be a sequence of all rationals in [0, 1) with $\tau_0 =0$ and define $V_n = V + \tau_n$ (mod 1).

Now we show that the V_n are pairwise disjoint and

$\bigcup\limits_n {V_n } = \left[ {0,1} \right)$ .

Indeed, if $x \in V_i \cap V_j$ , then $x = v_i+\tau_i$ (mod 1) and $x = v_j+\tau_j$ (mod 1), with v_i and v_j belonging to V = V_0 . Consequently, $v_i - v_j \in \mathbb Q$ , which means that $v_i \sim v_j$ and therefore i = j . This shows that $V_i \cap V_j = 0$ if $i \ne j$ . Since each $x \in [0, 1)$ is in some equivalence class, differs modulo 1 from an element in by a rational number, say , in [0, 1) . Thus $x \in V_k$ , which proves that

$[0, 1) \subset \bigcup\limits_n {V_n }$ .

The opposite inclusion is obvious.

Question 1. Show that there exist sets E_1, E_2, ...,E_k,... such that $E_k \searrow E$ , and $|E_k|_e < \infty$ and

$\lim_{k \to \infty} |E_k|_e > |E|_e$

with strict inequality.

Solution. We put $E_n = \bigcup\limits_{k \geq n} {V_k }$ . Clearly, $\{E_n\}$ is a decreasing sequence. Since the V_k are pairwise disjoint, we see that $E: = \bigcap\limits_n {E_n } = \emptyset$ and $E_n \searrow E$ . Moreover,

$\left| {E_n } \right|_e \geq \left| {V_n } \right|_e = \left| V \right|_e > 0$

(the last inequality comes from the fact that is not measurable). It is now enough to show that

$\mathop {\lim }\limits_{n \to \infty } \left| {E_n } \right|_e \geq \left| V \right|_e > 0 = \left| E \right|_e$

and the proof is complete.

Question 2. Show that there exist disjoint E_1, E_2, ...,E_k,... such that

$\left| { \bigcup_k E_k } \right|_e < \sum_k {\left| {E_k } \right|_e }$

with strict inequality.

Solution. We put E_n = V_n then E_n are pairwise disjoint and obviously

$\left| {\bigcup_n {E_n } } \right|_e = \left| {\left[ {0,1} \right)} \right|_e = 1$ .

Moreover, all the E_n are of the same outer measure. Thus $\sum_n {\left| {E_n } \right|_e } = + \infty$ which completes the proof.

Question 3. Show that each of the sets

${E_n} = \bigcup\limits_{k = 0}^n {{V_k}}$

is non-measurable.

Question 4. Show that if is a measurable subset of the Vitali set , then |E|=0 .

Question 5. Show that there exist sets and such that

${\left| {A \cup B} \right|_i} = {\left| A \right|_i} + {\left| B \right|_i}$

but

${\left| {A \cup B} \right|_e} < {\left| A \right|_e} + {\left| B \right|_e}.$

Question 6. Show that any set of positive outer measure contains a non-measurable subset.

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Tháng Bảy 31, 2009

A property of the essentially bounded function

Chuyên mục: Các Bài Tập Nhỏ, Giải Tích 6 (MA5205), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 5:41

Let $E$ be a subset of $\mathbb R^n$ with $|E| < \infty$ in Lebesgue sense. Suppose $f \in L^\infty(E)$ and $\| f\|_\infty > 0$ . Set

$\displaystyle {a_n} = \int_E {{{\left| f \right|}^n}}$

for $n=1,2,3,...$ Show that

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \frac{{{a_{n + 1}}}} {{{a_n}}} = {\left\| f \right\|_\infty }$ .

Solution. For any $\alpha$ with $0<\alpha < \|f\|_\infty$ , let

$\displaystyle{E_\alpha } = \left\{ {x \in E: f\left( x \right) \geqslant \alpha } \right\}$

and

$\displaystyle {F_\alpha } = E\backslash {E_\alpha }$

then $|E_\alpha|>0$ . For any $k \in \mathbb N$ , by the Dominated Convergence Theorem,

$\displaystyle\mathop{\lim }\limits_{n\to\infty }\left({\dfrac{{\int_{{F_\alpha }}{{{\left| f\right|}^{n+k}}}}}{{\int_{{E_\alpha }}{{{\left| f\right|}^{n}}}}}}\right)\leqslant\underbrace{\mathop{\lim }\limits_{n\to\infty }\frac{1}{{\left|{{E_\alpha }}\right|}}\int_{{F_\alpha }}{{{\left|{\frac{f}{\alpha }}\right|}^{n}}\left\| f\right\|_\infty^{k}}}_{0}$ .

Hence

$\displaystyle\mathop{\lim\inf }\limits_{n\to\infty }\left({\frac{{\int_{E}{{{\left| f\right|}^{n+1}}}}}{{\int_{E}{{{\left| f\right|}^{n}}}}}}\right)\geqslant\mathop{\lim\inf }\limits_{n\to\infty }\left({\frac{{\alpha\int_{{E_\alpha }}{{{\left| f\right|}^{n}}}+\int_{{F_\alpha }}{{{\left| f\right|}^{n+1}}}}}{{\int_{{E_\alpha }}{{{\left| f\right|}^{n}}}+\int_{{F_\alpha }}{{{\left| f\right|}^{n}}}}}}\right) =\alpha$ .

Letting $\alpha \nearrow {\left\| f \right\|_\infty }$ , we get that

$\displaystyle\mathop{\lim }\limits_{n\to\infty }\left({\frac{{\int_{E}{{{\left| f\right|}^{n+1}}}}}{{\int_{E}{{{\left| f\right|}^{n}}}}}}\right) ={\left\| f\right\|_\infty }$ .

As an application, if we put $a_0 = 1$ , then from

$\displaystyle {a_{n + 1}} = \frac{{{a_1}}} {{{a_0}}}.\frac{{{a_2}}} {{{a_1}}} \cdots\frac{{{a_{n + 1}}}} {{{a_n}}}$

we deduce that

$\displaystyle\mathop{\lim }\limits_{n\to\infty }\sqrt[n]{{{a_{n}}}}=\mathop{\lim }\limits_{n\to\infty }\frac{{{a_{n+1}}}}{{{a_{n}}}}={\left\| f\right\|_\infty }$ .

In other words,

$\displaystyle\mathop{\lim }\limits_{n\to\infty }{\left({\int_{E}{{{\left| f\right|}^{n}}}}\right)^{\frac{1}{n}}}={\left\| f\right\|_\infty }$ .

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A property of the essentially bounded function

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