Các Bài Tập Nhỏ « Ngô Quốc Anh

Tháng Một 5, 2010

On a property of log-function appeared in conformally invariant equations

Chuyên mục: Các Bài Tập Nhỏ, Giải Tích 6 (MA5205), Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 23:45

Suppose $f \in L^1(\mathbb R^n) \cap L_{loc}^\infty (\mathbb R^n)$ with $f \geq 0$ . Define

$\displaystyle Sf\left( x \right) = \int_{\mathbb{R}^n } {\log \frac{{\left| y \right|}}{{\left| {x - y} \right|}}f\left( y \right)dy}$ .

Show that $Sf(x)$ is finite for all $x \in \mathbb R^n$ and $Sf \in L_{loc}^1(\mathbb R^n)$ .

Proof. We consider the equivalent form of $Sf(x)$ as follows (this is nothing but $-Sf(x)$ )

$\displaystyle S'f\left( x \right) = \int_{\mathbb{R}^n } {\left( {\log \frac{{\left| {x - y} \right|}}{{\left| y \right|}}} \right)f\left( y \right)dy}$ .

We firstly consider the case when $|x| \geq 4$ . From the identity

$\displaystyle \mathbb{R}^4= \underbrace {\left\{ {y:\left| {x - y} \right| \geq\frac{{\left| x \right|}}{2}} \right\}}_A \cup \underbrace {\left\{ {y:\left| {x - y} \right| \leq\frac{{\left| x \right|}}{2}} \right\}}_B$

if $y \in B$ , then $\left| y \right| \geq \left| x \right| - \left| {x - y} \right| \geq\frac{{\left| x \right|}}{2} \geq\left| {x - y} \right|$ which implies $\log \frac{{\left| {x - y} \right|}}{{\left| y \right|}} \leq0$ . Therefore

$\displaystyle S'f\left( x \right) \leq\int_A {\left( {\log \frac{{\left| {x - y} \right|}}{{\left| y \right|}}} \right)f\left( y \right)dy}$ .

Since $|x - y|\leq |x|+|y| \leq |x| |y|$ provided $|x|, |y| \geq 2$ and $\log |x-y| \leq \log |x| + C$ for $|x| \geq 4$ and $|y| \leq 2$ , then we have

$\displaystyle \begin{gathered}\int\limits_A {\left( {\log \frac{{\left| {x - y} \right|}}{{\left| y \right|}}} \right)f\left( y \right)dy}= \int\limits_{A \cap \left\{ {y:\left| y \right| \geqslant 2} \right\}} {\left( {\log \frac{{\left| {x - y} \right|}}{{\left| y \right|}}} \right)f\left( y \right)dy}+ \int\limits_{A \cap \left\{ {y:\left| y \right| < 2} \right\}} {\left( {\log \frac{{\left| {x - y} \right|}}{{\left| y \right|}}} \right)f\left( y \right)dy}\hfill \\ \qquad\qquad\qquad\qquad\leq\log \left| x \right|\left( {\int\limits_{A \cap \left\{ {y:\left| y \right| \geqslant 2} \right\}} {f\left( y \right)dy} } \right) + {\rm const}. \hfill \\ \end{gathered}$

Thus, $S'f(x)$ is bounded from above for all $|x| \geq 4$ . For $0< |x| \leq 4$ , consider $S'f\left(\frac{4^2}{x} \right)$ . More precisely,

$\displaystyle\begin{gathered} + \infty> S'f\left( {\frac{4}{{\left| x \right|}}x} \right) = \int_{{\mathbb{R}^n}} {\left( {\log \frac{{\left| {\frac{4}{{\left| x \right|}}x - y} \right|}}{{\left| y \right|}}} \right)f\left( y \right)dy}\hfill \\ \qquad= \frac{4}{{\left| x \right|}}\int_{{\mathbb{R}^n}} {\left( {\log \frac{{\left| {x - \frac{{\left| x \right|}}{4}y} \right|}}{{\left| {\frac{{\left| x \right|}}{4}y} \right|}}} \right)\widetilde f\left( {\frac{{\left| x \right|}}{4}y} \right)d\left( {\frac{{\left| x \right|}}{4}y} \right)}\hfill \\ \qquad= \frac{4}{{\left| x \right|}}S'f\left( x \right) \hfill \\ \end{gathered}$

where $\widetilde f\left( y \right): = f\left( {\frac{4}{{\left| x \right|}}y} \right)$ . Note that

$\displaystyle 0 \leq\int_{\mathbb{R}^n } {\widetilde f\left( y \right)dy}= \frac{{\left| x \right|}}{4}\int_{\mathbb{R}^n } {f\left( {\frac{4}{{\left| x \right|}}y} \right)d\left( {\frac{4}{{\left| x \right|}}y} \right)}<+ \infty$ .

If $|x| = 0$ , then $Sf'(0) = 0$ . Thus, $S'f(x)$ is bounded from above in $\mathbb R^n$ which implies that $Sf(x)$ is bounded from below in $\mathbb R^n$ . Similar, we can prove that $Sf(x)$ is bounded from above in $\mathbb R^n$ .

Finally, from the above estimates, clearly $S'f$ is of class $S'f \in L_{loc}^1(\mathbb R^n)$ , and thus, so is $Sf$ .

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Tháng Mười Hai 17, 2009

Schwarz’s Lemma, Schwarz-Pick theorem, and some applications involving inequalities

Chuyên mục: Các Bài Tập Nhỏ, Giải tích 7 (MA4247), Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 10:52

In mathematics, the Schwarz lemma, named after Hermann Amandus Schwarz, is a result in complex analysis about holomorphic functions defined on the open unit disk.

Schwarz’s Lemma: Let $D=\{z : |z|<1\}$ be the open unit disk in the complex plane $\mathbb C$ . Let $f : D \to \overline D$ be a holomorphic function with $f(0)=0$ . The Schwarz lemma states that under these circumstances $|f(z)| \leq |z|$ for all $z \in D$ , and $|f'(0)| \leq 1$ . Moreover, if the equality $|f(z)|=|z|$ holds for any $z \ne 0$ , or $|f'(0)|=1$ then $f$ is a rotation, that is, $f(z)=az$ with $|a=1$ .

This lemma is less celebrated than stronger theorems, such as the Riemann mapping theorem, which it helps to prove; however, it is one of the simplest results capturing the “rigidity” of holomorphic functions. No similar result exists for real functions, of course. To prove the lemma, one applies the maximum modulus principle to the function $\frac{f(z)}{z}$ .

Proof: Let $g(z)=\frac{f(z)}{z}$ . The function $g(z)$ is holomorphic in $D$ (excluding $0$ ) since $f(0)=0$ and $f$ is holomorphic. Let $D_r$ be a closed disc within $D$ with radius $r$ . By the maximum modulus principle,

$\displaystyle |g(z)| = \frac{|f(z)|}{|z|} \leq \frac{|f(z_r)|}{|z_r|} \le \frac{1}{r}$

for all $z$ in $D_r$ and all $z_r$ on the boundary of $D_r$ . As $r$ approaches $1$ we get $|g(z)| \leq 1$ . Moreover, if there exists a $z_0$ in $D$ such that $g(z_0)=1$ . Then, applying the maximum modulus principle to $g$ , we obtain that $g$ is constant, hence $f(z)=kz$ , where $k$ is constant and $|k|=1$ . This is also the case if $|f'(0)|=1$ .

A variant of the Schwarz lemma can be stated that is invariant under analytic automorphisms on the unit disk, i.e. bijective holomorphic mappings of the unit disc to itself. This variant is known as the Schwarz-Pick theorem (after Georg Pick):

Schwarz-Pick theorem: Let $f : D \to D$ be holomorphic. Then, for all $z_1, z_2 \in D$ ,

$\displaystyle\left|\frac{f(z_1)-f(z_2)}{1-\overline{f(z_1)}f(z_2)}\right| \le \frac{\left|z_1-z_2\right|}{\left|1-\overline{z_1}z_2\right|}$

and, for all $z \in D$

$\displaystyle\frac{\left|f'(z)\right|}{1-\left|f(z)\right|^2} \le \frac{1}{1-\left|z\right|^2}$ .

The expression

$\displaystyle d(z_1,z_2)=\tanh^{-1}\left(\frac{\left|z_1-z_2\right|}{\left|1-\overline{z_1}z_2\right|}\right)$

is the distance of the points $z_1, z_2$ in the Poincaré metric, i.e. the metric in the Poincaré disc model for hyperbolic geometry in dimension two. The Schwarz-Pick theorem then essentially states that a holomorphic map of the unit disk into itself decreases the distance of points in the Poincaré metric. If equality holds throughout in one of the two inequalities above (which is equivalent to saying that the holomorphic map preserves the distance in the Poincaré metric) , then $f$ must be an analytic automorphism of the unit disc, given by a Möbius transformation mapping the unit disc to itself.

An analogous statement on the upper half-plane $\mathbb H$ can be made as follows:

Let $f: \mathbb H \to \mathbb H$ be holomorphic. Then, for all $z_1, z_2 \in \mathbb H$ ,

$\displaystyle\left|\frac{f(z_1)-f(z_2)}{\overline{f(z_1)}-f(z_2)}\right| \le \frac{\left|z_1-z_2\right|}{\left|\overline{z_1}-z_2\right|}$ .

This is an easy consequence of the Schwarz-Pick theorem mentioned above: One just needs to remember that the Cayley transform

$\displaystyle W(z) = \frac{z-i}{z + i}$

maps the upper half-plane $\mathbb H$ conformally onto the unit disc $D$ . Then, the map $W \circ f \circ W^{-1}$ is a holomorphic map from $D$ onto $D$ . Using the Schwarz-Pick theorem on this map, and finally simplifying the results by using the formula for $W$ , we get the desired result. Also, for all $z \in \mathbb H$ ,

$\displaystyle\frac{\left|f'(z)\right|}{\mbox{Im }f(z)} \le \frac{1}{\mbox{Im }(z)}$ .

If equality holds for either the one or the other expressions, then $f$ must be a Möbius transformation with real coefficients. That is, if equality holds, then

$\displaystyle f(z)=\frac{az+b}{cz+d}$

with $a$ , $b$ , $c$ , $d$ being real numbers, and $ad-bc>0$ .

Proof: The proof of the Schwarz-Pick theorem follows from Schwarz’s lemma and the fact that a Möbius transformation of the form

$\displaystyle\frac{z-z_0}{\overline{z_0}z-1}$ where $|z_0|<1$

maps the unit circle to itself. Fix $z_1$ and define the Möbius transformations

$\displaystyle M(z)=\frac{z_1-z}{1-\overline{z_1}z}$ and $\displaystyle\phi(z)=\frac{f(z_1)-z}{1-\overline{f(z_1)}z}$ .

Since $M(z_1)=0$ and the Möbius transformation is invertible, the composition $\varphi(f(M^{-1}(z)))$ maps $0$ to $0$ and the unit disk is mapped into itself. Thus we can apply Schwarz’s lemma, which is to say

$\displaystyle |\phi(f(M^{-1}(z)))|=\left|\frac{f(z_1)-f(M^{-1}(z))}{1-\overline{f(z_1)}f(M^{-1}(z))}\right| \leq |z|$ .

Now calling $z_2=M^{-1}(z)$ (which will still be in the unit disk) yields the desired conclusion

$\displaystyle\left|\frac{f(z_1)-f(z_2)}{1-\overline{f(z_1)}f(z_2)}\right| \le \left|\frac{z_1-z_2}{1-\overline{z_1}z_2}\right|$ .

To prove the second part of the theorem, we just let $z_2$ tend to $z_1$ .

Application 1 (QE Berkeley Spring 1991). Let the function $f$ be analytic in the unit disc, with $|f(z)| \leq 1$ and $f(0)=0$ . Assume that there is a number $r \in (0,1)$ such that $f(r)=f(-r)=0$ . Prove that

$\displaystyle\left| {f\left( z \right)} \right| \leqslant \left| z \right|\left| {\frac{{{z^2} - {r^2}}} {{1 - {r^2}{z^2}}}} \right|$ .

Solution. Schwartz’s lemma implies that the function $f_1(z)=\frac{f(z)}{z}$ satisfies $|f_1(z)| \leq 1$ . The linear fractional map $z \mapsto \frac{z-r}{1-rz}$ sends the unit disc onto itself. Applying Schwartz’s lemma to the function

$\displaystyle f_2(z)=f_1\left( \frac{z-r}{1-rz} \right)$

we conclude that the function

$\displaystyle f_3(z)=\frac{f_1(z)}{\left(\frac{z-r}{1-rz}\right)}$

satisfies $|f_3(z)| \leq 1$ . Similarly, the map $z \mapsto \frac{z+r}{1+rz}$ sends the unit disc onto itself, and Schwartz’s lemma applied to the function

$\displaystyle f_4(z)=\frac{f_3(z)}{\left( \frac{z+r}{1+rz} \right)}$

implies that the function $f_5(z) =f_3\left( \frac{z+r}{1+rz} \right)$ satisfies $|f_5(z)| \leq 1$ . All together, then

$\displaystyle\left| {f\left( z \right)} \right| \leqslant \left| z \right|\left| {\frac{{z - r}} {{1 - rz}}} \right|\left| {\frac{{z + r}} {{1 + rz}}} \right|\left| {{f_5}\left( z \right)} \right| \leqslant \left| z \right|\left| {\frac{{z - r}} {{1 - rz}}} \right|\left| {\frac{{z + r}} {{1 + rz}}} \right|$ .

which is the desired inequality.

Application 2 (QE NUS Spring 2009). Suppose $f$ is analytic in $D := \{ z \in \mathbb C : |z|<1\}$ with $|f(z)|<1$ . Show that

$\displaystyle\frac{{\left| {f\left( 0 \right)} \right| - \left| z \right|}} {{1 - \left| {f\left( 0 \right)} \right|\left| z \right|}} \leqslant \left| {f\left( z \right)} \right| \leqslant \frac{{\left| {f\left( 0 \right)} \right| + \left| z \right|}} {{1 + \left| {f\left( 0 \right)} \right|\left| z \right|}}$

for all $z \in D$ .

Application 3 (QE NUS Fall 2009). 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?

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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

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

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

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

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 26, 2009

How to calculate limit by using definition of definite integral?

Chuyên mục: Các Bài Tập Nhỏ, Giải Tích 2 — Ngô Quốc Anh @ 14:23

Let $f : [a, b] \to \mathbb R$ be a continuous function, not necessarily nonnegative. Partition $[a, b]$ into $n$ consecutive sub-intervals $[x_{i-1}, x_i]$ ( $i = 1, 2, ..., n$ ) each of length $\Delta x = \frac{b-a}{n}$ , where we set $a=x_0$ , $b=x_n$ and $x_1, x_2,...,x_{n-1}$ to be successive points between $a$ and $b$ with $x_k-x_{k-1}=\Delta x$ . Let $c_k$ be any intermediate point in the sub-interval $[x_{k-1},x_k]$ . Then the sum

$\displaystyle\sum\limits_{k = 1}^n {f\left( {{c_k}} \right)\Delta x}$

is called a Riemann sum for $f$ on $[a, b]$ .

Suppose we let the number of partition in $P$ tends to infinity.

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {f\left( {{c_k}} \right)\Delta x} = I.$

We call $I$ the Riemann integral (or definite integral) of $f$ over $[a, b]$ and we write

$\displaystyle I = \int_a^b {f(x)dx} .$

In other words,

$\displaystyle\int_a^b {f(x)dx} = \frac{{b - a}}{n}\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {f\left( {{c_k}} \right)}$

if the limit on the right side exists.

If we put $c_k=x_{k-1}$ we the obtain

$\displaystyle\int_a^b {f(x)dx} = \frac{{b - a}}{n}\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {f\left( {a + (k - 1)\frac{{b - a}}{n}} \right)} .$

Example 1. Find

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{n + 1}} + \cdots + \frac{1}{{2n - 2}} + \frac{1}{{2n - 1}}} \right).$

Solution. Clearly

$\displaystyle\frac{1}{n} + \frac{1}{{n + 1}} + \cdots + \frac{1}{{2n - 2}} + \frac{1}{{2n - 1}} = \sum\limits_{k = 1}^n {\frac{1}{{n + (k - 1)}}} = \frac{1}{n}\sum\limits_{k = 1}^n {\frac{1}{{1 + \frac{{k - 1}}{n}}}} .$

Then if we choose $a=0$ , $b=1$ we then get

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{n + 1}} + \cdots + \frac{1}{{2n - 2}} + \frac{1}{{2n - 1}}} \right) = \int_0^1 {\frac{{dx}}{{1 + x}}} .$

With this it is easy to see that

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{n + 1}} + \cdots + \frac{1}{{2n - 2}} + \frac{1}{{2n - 1}}} \right) = \ln 2$

since

$\displaystyle\int_0^1 {\frac{{dx}}{{1 + x}}} = \ln 2.$

If we put $c_k=x_k$ we the obtain

$\displaystyle\int_a^b {f(x)dx} = \frac{{b - a}}{n}\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {f\left( {a + k\frac{{b - a}}{n}} \right)} .$

Example 2. Find

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \left( {\frac{n}{{{n^2} + {1^2}}} + \frac{n}{{{n^2} + {2^2}}} + \cdots + \frac{n}{{{n^2} + {{(n - 1)}^2}}} + \frac{n}{{{n^2} + {n^2}}}} \right).$

Solution. Clearly,

$\displaystyle\frac{n}{{{n^2} + {1^2}}} + \frac{n}{{{n^2} + {2^2}}} + \cdots + \frac{n}{{{n^2} + {n^2}}} = \sum\limits_{k = 1}^n {\frac{n}{{{n^2} + {k^2}}}} = \frac{1}{n}\sum\limits_{k = 1}^n {\frac{1}{{1 + {{\left( {\frac{k}{n}} \right)}^2}}}}$

which yields

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \left( {\frac{n}{{{n^2} + {1^2}}} + \frac{n}{{{n^2} + {2^2}}} + \cdots + \frac{n}{{{n^2} + {n^2}}}} \right) = \int_0^1 {\frac{{dx}}{{1 + {x^2}}}} = \frac{\pi }{4}.$

Remark. It is worth mentioning that in general it is not true that

$\displaystyle\lim \left( {summation} \right) = summation\left( {\lim } \right).$

For example, we all know that for each fixed $k$

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \frac{1}{{n + k}} = 0$

but

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \left( {\sum\limits_{k = 1}^n {\frac{1}{{n + k}}} } \right) = \ln 2 \ne 0 = \sum\limits_{k = 1}^n {\left( {\mathop {\lim }\limits_{n \to \infty } \frac{1}{{n + k}}} \right)} .$

The point is

$\displaystyle\lim \left( {summation} \right) = summation\left( {\lim } \right)$

holds true only for finite summation.

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

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How to calculate limit by using definition of definite integral?

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An algebraic identity for 9th level

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