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.

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.

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.

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

  1. g_n \to 0 a.e.
  2. j(f) \in L^1.
  3. \displaystyle\int \varphi_\varepsilon(g_n(x))d\mu(x) \leq C < \infty for some constant C, independent of \varepsilon and n.
  4. \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 }.

    The fact that u_n \to u strongly in L^p implies that \lim_{n\to \infty} \int |u_n|^p d\mu = \int |u|^p d\mu. Therefore,

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

    As a consequence, one has the following result

    \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.

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.

Tháng Mười 11, 2009

An example of sectional curvature of sphere

Chuyên mục: Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 15:34

In \mathbb S^n we consider the metric

{g_{ij}} = {\left( {\displaystyle\frac{2}{{1 + {{\left| y \right|}^2}}}} \right)^2}{\delta _{ij}}.

We now find the sectional curvature of g.

Since

{g_{ij}} = \displaystyle{\left( {\frac{2}{{1 + {{\left| y \right|}^2}}}} \right)^2}{\delta _{ij}}

then

{g^{ij}} = \displaystyle{\left( {\frac{{1 + {{\left| y \right|}^2}}}{2}} \right)^2}{\delta _{ij}}.

Now we need to calculate

\Gamma _{ij}^k = \displaystyle\frac{1}{2}{g^{kl}}\left( {\frac{{\partial {g_{il}}}}{{\partial {y_j}}} + \frac{{\partial {g_{lj}}}}{{\partial {y_i}}} - \frac{{\partial {g_{ij}}}}{{\partial {y_l}}}} \right).

Clearly,

\displaystyle\frac{{\partial {g_{il}}}}{{\partial {y_j}}} = \frac{\partial }{{\partial {y_j}}}\left( {{{\left( {\frac{2}{{1 + {{\left| y \right|}^2}}}} \right)}^2}{\delta _{il}}} \right) = 2\frac{2}{{1 + {{\left| y \right|}^2}}}\frac{{ - 2}}{{{{\left( {1 + {{\left| y \right|}^2}} \right)}^2}}}\left( {2{y_j}} \right){\delta _{il}} = - 2{\left( {\frac{2}{{1 + {{\left| y \right|}^2}}}} \right)^3}{y_j}{\delta _{il}}.

Similarly, one has the following

\displaystyle\frac{{\partial {g_{lj}}}}{{\partial {y_i}}} = - 2{\left( {\frac{2}{{1 + {{\left| y \right|}^2}}}} \right)^3}{y_i}{\delta _{lj}}, \quad \frac{{\partial {g_{ij}}}}{{\partial {y_l}}} = - 2{\left( {\frac{2}{{1 + {{\left| y \right|}^2}}}} \right)^3}{y_l}{\delta _{ij}}.

Therefore,

\displaystyle \Gamma _{ij}^k= \frac{{ - 2}}{{1 + {{\left| y \right|}^2}}}\left( {{y_j}{\delta _{ik}} + {y_i}{\delta _{kj}} - {y_k}{\delta _{ij}}} \right).

Next we need to calculate coefficients R^m_{lij} of the curvature tensor. To this purpose, we use

\displaystyle R_{lij}^m = \frac{{\partial \Gamma _{lj}^m}}{{\partial {y_i}}} - \frac{{\partial \Gamma _{li}^m}}{{\partial {y_j}}} + \Gamma _{in}^m\Gamma _{jl}^n - \Gamma _{jn}^m\Gamma _{il}^n.

We have

\displaystyle\frac{{\partial \Gamma _{lj}^m}}{{\partial {y_i}}} = \frac{\partial }{{\partial {y_i}}}\left( {\frac{{ - 2}}{{1 + {{\left| y \right|}^2}}}\left( {{y_j}{\delta _{ml}} + {y_l}{\delta _{mj}} - {y_m}{\delta _{lj}}} \right)} \right)

which yields

\displaystyle\frac{{\partial \Gamma _{lj}^m}}{{\partial {y_i}}} ={\left( {\frac{2}{{1 + {{\left| y \right|}^2}}}} \right)^2}\left( {{y_i}{y_j}{\delta _{ml}} + {y_i}{y_l}{\delta _{mj}} - {y_i}{y_m}{\delta _{lj}}} \right) + \frac{{ - 2}}{{1 + {{\left| y \right|}^2}}}\left( {{\delta _{ij}}{\delta _{ml}} + {\delta _{il}}{\delta _{mj}} - {\delta _{im}}{\delta _{lj}}} \right).

Similarly, one has

\displaystyle\frac{{\partial \Gamma _{li}^m}}{{\partial {y_j}}} = {\left( {\frac{2}{{1 + {{\left| y \right|}^2}}}} \right)^2}\left( {{y_i}{y_j}{\delta _{ml}} + {y_j}{y_l}{\delta _{mi}} - {y_j}{y_m}{\delta _{li}}} \right) + \frac{{ - 2}}{{1 + {{\left| y \right|}^2}}}\left( {{\delta _{ji}}{\delta _{ml}} + {\delta _{jl}}{\delta _{mi}} - {\delta _{jm}}{\delta _{li}}} \right).

Therefore,

\displaystyle\frac{{\partial \Gamma _{lj}^m}}{{\partial {y_i}}} - \frac{{\partial \Gamma _{li}^m}}{{\partial {y_j}}} = - \frac{4}{{1 + {{\left| y \right|}^2}}}\left( {{\delta _{il}}{\delta _{mj}} - {\delta _{im}}{\delta _{lj}}} \right).

On the other hands,

\displaystyle\Gamma _{in}^m\Gamma _{jl}^n - \Gamma _{jn}^m\Gamma _{il}^n = {\left( {\frac{2}{{1 + {{\left| y \right|}^2}}}} \right)^2}\left\{ \begin{gathered} {y_i}{y_j}{\delta _{ml}} + {y_i}{y_l}{\delta _{mj}} - {y_i}{y_m}{\delta _{jl}} + \hfill \\ {y_l}{y_j}{\delta _{im}} + {y_j}{y_l}{\delta _{im}} - y_n^2{\delta _{jl}}{\delta _{im}} - \hfill \\ {y_m}{y_j}{\delta _{il}} - {y_m}{y_l}{\delta _{ij}} + {y_m}{y_i}{\delta _{jl}} \hfill \\ \hfill \\ - {y_i}{y_j}{\delta _{ml}} - {y_j}{y_l}{\delta _{mi}} + {y_j}{y_m}{\delta _{il}} - \hfill \\ {y_l}{y_i}{\delta _{jm}} - {y_i}{y_l}{\delta _{jm}} + y_n^2{\delta _{il}}{\delta _{jm}} + \hfill \\ {y_m}{y_i}{\delta _{jl}} + {y_m}{y_l}{\delta _{ij}} - {y_m}{y_j}{\delta _{il}} \end{gathered} \right\}

which yields

\displaystyle\Gamma _{in}^m\Gamma _{jl}^n - \Gamma _{jn}^m\Gamma _{il}^n = {\left( {\frac{2}{{1 + {{\left| y \right|}^2}}}} \right)^2}{\left| y \right|^2}\left( {{\delta _{il}}{\delta _{jm}} - {\delta _{jl}}{\delta _{im}}} \right).

Thus,

\displaystyle R_{lij}^m = - \frac{4}{{1 + {{\left| y \right|}^2}}}\left( {{\delta _{il}}{\delta _{mj}} - {\delta _{im}}{\delta _{lj}}} \right) + {\left( {\frac{2}{{1 + {{\left| y \right|}^2}}}} \right)^2}{\left| y \right|^2}\left( {{\delta _{il}}{\delta _{jm}} - {\delta _{jl}}{\delta _{im}}} \right) = {\left( {\frac{2}{{1 + {{\left| y \right|}^2}}}} \right)^2}\left( {{\delta _{im}}{\delta _{lj}} - {\delta _{il}}{\delta _{mj}}} \right).

Finally, one obtains

\displaystyle K\left( {{e_i},{e_j}} \right) = \frac{{\left\langle {R\left( {{e_i},{e_j}} \right){e_j},{e_i}} \right\rangle }}{{\left\langle {{e_i},{e_i}} \right\rangle \left\langle {{e_j},{e_j}} \right\rangle - {{\left\langle {{e_i},{e_j}} \right\rangle }^2}}} = \frac{{R_{jij}^m\left\langle {{e_m},{e_i}} \right\rangle }}{{{{\left( {\frac{2}{{1 + {{\left| y \right|}^2}}}} \right)}^4}}} = \frac{{R_{jij}^m{{\left( {\frac{2}{{1 + {{\left| y \right|}^2}}}} \right)}^2}{\delta _{mi}}}}{{{{\left( {\frac{2}{{1 + {{\left| y \right|}^2}}}} \right)}^4}}} = 1.

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.

Tháng Chín 25, 2009

An introduction to Yamabe problem

Chuyên mục: Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 15:51

Hidehiko Yamabe, in his famous paper entitled On a deformation of Riemannian structures on compact manifolds, Osaka Math. J. 12 (1960), pp. 21-37,  wanted to solve the Poincaré conjecture. For this he thought, as a first step, to exhibit a metric with constant scalar curvature. He considered conformal metrics (the simplest change of metric is a conformal one), and gave a proof of the following statement:

“On a compact Riemannian manifold (M, g), there exists a metric g' conformal to g, such that the corresponding scalar curvature R' is constant”.

The Yamabe problem was born, since there is a gap in Yamabe’s proof. Now there are many proofs of this statement.

Let us recall the question. Let (M_n, g) be a compact C^\infty-Riemannian manifold of dimension n \geq 3, is its scalar curvature. The problem is:

“Does there exists a metric g', conformal to g, such that the scalar curvature R' of the metric g is constant?”.

Let us consider the conformal metric g'=e^fg with f \in C^\infty. If \Gamma'^l_{ij} and \Gamma_{ij}^l denote the Chrisoffel symbols relating to g' and g, respectively, then

\displaystyle\Gamma '^l_{ij} - \Gamma _{ij}^l = \frac{1} {2}\left( {{g_{kj}}\frac{{\partial f}} {{\partial {x_i}}} + {g_{ki}}\frac{{\partial f}} {{\partial {x_j}}} - {g_{ij}}\frac{{\partial f}} {{\partial {x_k}}}} \right){g^{kl}} = \frac{1} {2}\left( {\delta _j^l{\partial _i}f + \delta _i^l{\partial _j}f - {g_{ij}}{\nabla ^l}f} \right).

Clearly,

\displaystyle R'_{ij}=R'^k_{ikj}=R_{ij}-\frac{n-2}{2}\nabla _{ij}f+\frac{n-2}{4}\nabla _if\nabla _jf-\frac{1} {2}\left(\nabla _\nu^\nu f+\frac{n-2}{2}\nabla^\nu f\nabla _\nu f\right)g_{ij}

so

\displaystyle R' = {e^{ - f}}\left( {R - \left( {n - 1} \right)\nabla _\nu ^\nu f - \frac{{\left( {n - 1} \right)\left( {n - 2} \right)}} {4}{\nabla ^\nu }f{\nabla _\nu }f} \right).

If we consider the conformal deformation in the form g'=\varphi^\frac{4}{n-2}g (with \varphi \in C^\infty, \varphi>0), the scalar curvature satisfies the equation

\displaystyle \frac{{4\left( {n - 1} \right)}} {{n - 2}}\Delta \varphi + R\varphi = R'{\varphi ^{\frac{{n + 2}} {{n - 2}}}}

where \Delta \varphi = - {\nabla ^\nu }{\nabla _\nu }\varphi. So, Yamabe problem is equivalent to solving the above equation with R'=const and the solution \varphi must be smooth and strictly positive.

Link to PDF file of the paper Osaka Math. J. 12 (1960), pp. 21-37 can be found here http://projecteuclid.org/euclid.ojm/1200689814

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

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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