Ngô Quốc Anh

May 28, 2008

Lebesgue’s density theorem

Filed under: Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 3:25

In mathematics, Lebesgue’s density theorem states that for any Lebesgue measurable set A, the “density” of A is 1 at almost every point in A. Intuitively, this means that the “edge” of A, the set of points in A whose “neighborhood” is partially in A and partially outside of A, is negligible.

Let μ be the Lebesgue measure on the Euclidean space Rn and A be a Lebesgue measurable subset of Rn. Define the approximate density of A in a ε-neighborhood of a point x in Rn as

$d_\varepsilon(x)=\frac{\mu(A\cap B_\varepsilon(x))}{\mu(B_\varepsilon(x))}$

where Bε denotes the closed ball of radius ε centered at x.

Lebesgue’s density theorem asserts that for almost every point of A the density

$d(x)=\lim_{\varepsilon\to 0} d_{\varepsilon}(x)$

exists and is equal to 1.

In other words, for every measurable set A the density of A is 0 or 1 almost everywhere in Rn. However, it is a curious fact that if μ(A) > 0 and μ(Rn\A) > 0, then there are always points of Rn where the density is neither 0 nor 1.

For example, given a square in the plane, the density at every point inside the square is 1, on the edges is 1/2, and at the corners is 1/4. The set of points in the plane at which the density is neither 0 nor 1 is non-empty (the square boundary), but it is negligible.

May 10, 2008

An integral inequality

Filed under: Các Bài Tập Nhỏ, Giải Tích 2 — Ngô Quốc Anh @ 3:48

Let $[0,1]\rightarrow \mathbb{R}^{ + }$ be a continuous function and let  be a natural number. Prove that

$\int\limits_{0}^{1}f^{n}(x^{n})dx\geq \frac {1}{n}\left(\frac {n +1}{n }\right)^{n - 1}\left(\int\limits_{0}^{1}f(x)dx\right)^{n}$.

Solution. Recall Hôlder inequality

, , $\frac {1}{a} + \frac {1}{b} = 1$,

$|\int_{K}h.g|\leq(\int_{K}|h|^{a})^{1/a}(\int_{K}|g|^{b})^{1/b}$.

Then

$\int_{0}^{1}f(x)dx = \int_{0}^{1}y^{\frac {n - 1}{n^2}}(f(y)y^{\frac {1 - n}{n^2}})dy\leq (\int_{0}^{1}y^{\frac {1}{n}}dy)^{\frac {n - 1}{n}}(\int_{0}^{1}f^{n}(y)y^{\frac {1}{n} - 1}dy)^{\frac {1}{n}}$

taking the n-power and in the second integral at RHS , do change variable  will give the inequality.

May 1, 2008

Two times derivable real function

Filed under: Giải Tích 1 — Ngô Quốc Anh @ 16:41

Two times derivable real function
RMO 2008, 11th Grade, Problem 3

Let  be a function, two times derivable on  for which there exist  such that$\frac { f(b)-f(a) }{b-a} \neq f'(c) ,$ for all . Prove that .

Nondecreasing function

Filed under: Các Bài Tập Nhỏ, Giải Tích 1 — Ngô Quốc Anh @ 16:39

Nice nondecreasing function
RMO 2008, 11th Grade, Problem 1

Let  be a continous function such that the sequences  are nondecreasing for any real number . Prove that  is nondecreasing.

Proof. It’s trivial too see that if , then $f\left( \frac mn x \right) \geq f(x)$, for all , so in particular  for all rational numbers . Now let , and let , . If , let . Because  is continuous, there exists a rational number ,  such that $f(x_0)= f(y) + \frac {\delta}{2}$. But , so  contradiction. This means that if , and , such that  then .

Now let  be any two real numbers. Then there exists  such that , and using the above claim it follows that  which proves that  is nondecreasing.

Primitives…

Filed under: Giải Tích 2 — Ngô Quốc Anh @ 16:37

Darboux property
RMO 2008, Grade 12, Problem 1

Let  and  be a continuous function on  and having Darboux property on . Prove that if  and for all nonnegative  we have  then  admits primitives on .

Proof. The title condition is almost entirely redundant- it simply says that .

Obviously,  is an antiderivative away from zero. The only interesting question is the one-sided derivative . Now, $\frac{d}{dx}\frac{F(x)}{x}=\frac{xf(x)-F(x)}{x^2}\ge 0$ by hypothesis. We also have $\liminf{x\to 0}\frac{F(x)}{x}\le \liminf_{x\to 0}f(x)=0$ by the same integral inequality. Since  is increasing (and bounded below), it has a limit at zero. Since the  is zero, this limit is zero, and  has the proper one-sided derivative at zero.

An integral inequality involving differentiable functions with continous derivatives

Filed under: Giải Tích 2 — Ngô Quốc Anh @ 16:31

In this entry, we address the following question

Let $f : [0,1] \to \mathbb R$ be a derivable function, with a continuous derivative $f'$ on $[0,1]$. Prove that if $f\left(\frac{1}{2}\right)=0$, then

$\displaystyle\int^1_0 \left( f'(x) \right)^2 dx \geq 12 \left( \int^1_0 f(x) dx \right)^2$.

Solution. The best possible constant comes from the $2$-norm of $\frac{1}{2}-x$. Since

$\displaystyle \int_0^1 \left(\frac{1}{2}-x\right)^2 = \frac{1}{12},$

this constant is indeed $12$. We get equality if and only if $f'(x)$ is a real multiple of $\frac{1}{2}-x$.

First, write

$\displaystyle f(t)=\int_\frac{1}{2}^t f'(x)dx.$

Define $g(x)=f'(x)$. The LHS is

$\displaystyle \int_0^1 (g(x))^2dx,$

and the RHS becomes

$\displaystyle\left(\int_0^1 \int_{\frac12}^t g(x)dxdt\right)^2= \left(\int_0^1 \int_x^{\frac12}g(x)dtdx\right)^2 =\left(\int_0^1 \left(\frac12-x\right)g(x)dx\right)^2.$

In the language of inner products, we want to show that

$\displaystyle \|g\|^2 \geqslant 12 (g \cdot h)^2,$

where

$\displaystyle h(x)=\frac{1}{2}-x$.

The standard C-S inequality gives

$\displaystyle (g\cdot h)^2\leqslant \|g\|^2\|h\|^2=\frac{1}{12}\|g\|^2,$

and we are done.

Source: RMO 2008, Grade 12, Problem 2.