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.

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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.

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