## A little encouragement

If you know who is David Aldous, you will laugh really hard after reading this extract from the report of a referee who rejected his paper. Even big name’s papers can be rejected, so don’t be discouraged young mathematicians

http://www.stat.berkeley.edu/~aldous/least_interesting_paper_ever.txt

Categories: Uncategorized

## A useful identity in convergence of random series

Let $X_1, X_2,\ldots,$ be independent and let $S_{m,n} = X_{m+1} + \cdots + X_n.$ Then for any $a>0$

$P(\max_{m 2a)\min_{ma).$

The identity can be used to prove the following results:

1) P. Levy: Let $X_1, X_2,\ldots$ be independent and let $S_n = X_1+\cdots+X_n$. If $\lim_{n\rightarrow \infty} S_n$ exists in probability, then it also exists almost surely.

2) Let  $X_1, X_2,\ldots$  be i.i.d. and $S_n = X_1+\cdots+X_n.$ If $\lim_{n\rightarrow \infty} S_n/n \rightarrow 0$ in probability, then $(\max_{1\leq m\leq n} S_m)/n \rightarrow 0$ in probability.

Categories: Probability Theory

An application of Fourier analysis (in Vietnamese),

Author: Prof. Ngo Bao Chau

Categories: Uncategorized

## Lecture 3: The outer measure generated by a measure

I’ll post lecture 1 and 2 soon.

In lecture 1 and 2, I defined a measure $\mu$ on a semiring $S$ and an outer measure on the power set of $X$. Our goal is that we want to define a measurable set $E\subseteq X.$ As we recall, a measurable set $E\subseteq X$ satifies the following condition: $\mu(A) =\mu(A\cap E) + \mu(A\cap E^c)$ for any $A\subseteq X.$ Notice that $\mu$ here is an outer measure.

In this lecture, a measure $\mu$ is extended to an outer measure $\mu^*.$ As proved before, the collection $\Lambda$ of measurable sets with respect to this outer measure is a $\sigma-algebra.$ We will show that $S\subseteq\Lambda.$ Thus, semirings are a basic collections of sets which a measure theory can be built on.

Let a measure space $(X,S,\mu)$ be fixed. The outer measure $\mu^*$ is defined as follows:

$\mu^*(A) = \inf \{\sum_{i=1}^{\infty} \mu(A_i): A_i \in S \text{ and } A\subseteq \bigcup_{i=1}^\infty A_i \}.$

Theorem: $\mu^*$ is an outer measure.

Proof:

i) Let $A_i=\emptyset$ for $i\geq 1.$ Then $\mu^*(\bigcup_{i=1}^\infty A_i)=\mu^*(\emptyset) \leq \sum_{i=1}^\infty \mu(\emptyset)=0.$ Thus $\mu*(\emptyset)=0.$

ii) Let $A,B\in X$ and $A\subseteq B.$ We consider a sequence $\{A_i\}$ of sets in $S$ such that $B\subseteq \bigcup_{i=1}^\infty A_i.$ Then clearly, $A\subseteq\bigcup_{i=1}^\infty A_i.$ Thus, $\mu^*(A)\leq\sum_{i=1}^\infty \mu(A_i)$ Further, we have $\mu^*(B)=\inf \{\sum_{i=1}^{\infty} \mu(A_i): A_i \in S \text{ and } B\subseteq \bigcup_{i=1}^\infty A_i \}.$ This follows that $\mu^*(A)\leq \mu^*(B).$ Hence, $\mu^*$ is monotone.

iii) Let $\{A_i\}$ be a sequence of pairwise disjoint sets in $X.$ If $\sum_{i=1}^\infty \mu^*(A_i)=\infty$. Then clearly, $\mu^*(\bigcup_{i=1}^\infty A_i) \leq \sum_{i=1}^\infty \mu^*(A_i).$

Thus we can assume $\sum_{i=1}^\infty \mu^*(A_i)<\infty.$ Since each $\mu^*(A_i) <\infty,$ there exists a sequence $\{A_i\}^{n}$ of sets in $S$ such that $A_i\subseteq \bigcup_{n=1}^\infty A_i^n$ and $\sum_{n=1}^\infty \mu(A_i^n)\le \mu^* (A_i)+2^{-i}\epsilon.$ Hence, $\bigcup_{i=1}^\infty A_i \subseteq \bigcup_{i=1}^\infty \bigcup_{n=1}^\infty A_i^n.$ Furthermore,

$\mu^* (\bigcup_{i=1}^\infty A_i)\leq \sum_{i=1}^\infty\sum_{n=1}^\infty\mu(A_i^n) 0,$ we have $\mu^* (\bigcup_{i=1}^\infty A_i) \leq \sum_{i=1}^\infty \mu^*(A_i) .$ This proves the $\sigma$-subadditivity of $\mu^*. \square$

The next theorem will show that $\mu^*$ is indeed an extension of $\mu$ from $S$ to $\mathcal{P}(X).$

Theorem: If $E\in S,$ then $\mu^*(E)=\mu(E).$

Proof:

Let $A_1=E$ and $A_i=\emptyset$ for $i>1.$ Then $E\subseteq \bigcup_{i=1}^\infty A_i.$ Thus, $\mu^*(E)\leq \mu(\bigcup_{i=1}^\infty A_i)=\mu(E).$

Consider a sequence $\{A_i\}$ of sets in $S$ satisfying $A\subseteq \bigcup_{i=1}^\infty A_i.$ Since $\mu$ is a measure on semiring $S.$ It is $\sigma$-subadditive on $S.$ Therefore, we have $\mu(A)\leq\sum_{i=1}^\infty \mu(A_i).$ Thus, $\mu(A)\leq\mu^*(A). \square$

The outer measure generated by a measure have some nice characteristics which can be shown by the following theorem.

Theorem: Let $E\subseteq X.$ The followings are equivalent:

(1) $E$ is $\mu^*$-measurable.

(2) $\mu(A)=\mu^*(A\cap E)+\mu^*(A\cap E^c)$ with any $A\in S$ and $\mu(A)<\infty.$

(3) $\mu(A)\geq \mu^*(A\cap E)+\mu^*(A\cap E^c)$ with any $A\in S$ and $\mu(A)<\infty.$

(4) $\mu^*(A)\geq \mu^*(A\cap E)+\mu^*(A\cap E^c)$ with any $A\subseteq X.$

Proof:

Categories: Uncategorized

## Polynomial in a Field of Characteristic p

June 30, 2011 1 comment

I’ll go ahead and it post it here as well:
Hey Ken, I’m trying to check if this proof is correct(it may not be the only way, but its one I thought of)

To Prove: Let $F$ be a field with $charF=p$. Let $q(x)=x^p-x-c$ where $c\in F$. Show if $q$ is reducible, then $q$ has a root in $F$ (actually all).

This is actually an iff in the book, but this is the nontrivial direction.

My idea is the following. Let $u$ be a root of $q$. Let $f,g$ be nonunit irreducible factors of $q$. Now it is easy to see the set of all roots is the set $\{u+i:i=1,\ldots,p\}$. Thus, $Z(f)=u+I, Z(g)=u+J$ for some finite sets $I,J\subset \mathbb{Z}.$

Notice this means $\prod_{i\in I}u+i, \prod_{i\in J} u+i$ are both in $F$. Let $f_1=\prod_{i\in I}u+i$ and $f_2=\prod_{i\in J} u+i$. Then $u$ is a root for both $f_1-\prod_{i\in I}{x+i}$ and $f_2-\prod_{i\in J}x+i$ which are in $F[x]$. But since the degrees of these are $|I|,|J|$; respectively, the only way this is possible is if $|I|=|J|$. But we could do this for any irreducible factor so each irreducible factor must have the same degree.

Hence, if $q=q_1\cdots q_k$ with each $q_i$ irreducible, then $p=deg(q)=k\cdot n$ for some $n\geq 1$. Since $p$ is prime either $k$ or $n$ must be $1$. If $n=1$ then each root is in $F$. If $k=1$ then $q$ is irreducible. Since $p$ is reducible the result follows.

After writing this out carefully it feels completely correct (of course I left out some easy details).

Categories: Algebra questions

## Complex #34: sneaky use of Rouche’s theorem

Problem #34 is: Prove that there does not exist a polynomial of the form $p(z)= z^n+a_{n-1}z^{n-1}+\ldots+a_0$ such that $|p (z)| < 1$ for all $z$ such that $|z|=1$.

Proof: Suppose that there does exist such a polynomial. Note that $|z^n|>|-p(z)|$ for $|z|=1$. By Rouche’s theorem, $z^n-p(z)=-a_{n-1}z^{n-1}-\ldots-a_0$ has the same number of zeros (with multiplicities) as $z^n$ inside the unit disk. But that is impossible, since $z^n$ has $n$ zeros and $z^n-p(z)$ has at most $n-1$ zeros. QED.

Categories: Complex questions

## Complex #27: integrating over a long rectangle

Problem 27: Show that $\int_{\gamma} e^{iz}e^{-z^2}dz$ has the same value on every straight line path $\gamma$ (oriented in the $+x$ direction) parallel to the real axis.
Proof: Consider the rectangle $R=[-M,M]\times [0,Y]$, where $Y$ is some given number and $M$ is a very large number. Since the integrand is entire, the integral over the boundary of $R$ is zero by the Cauchy Integral Theorem, and as $M\to\infty$, the horizontal pieces correspond to $\int_{\gamma} e^{iz}e^{-z^2}dz$ with $\gamma$ being the line $Im (z)=0$ or the line $Im(z)=Y$, with one being negative and the other positive. Thus, the result is proved if we can show that the contributions from the vertical pieces go to zero as $M$ goes to $\infty$.
Observe that the absolute value of the integrand at $z=x+iy$ is $e^{-x^2+y^2-y}$, which on each vertical boundary piece is $e^{-M^2+y^2-y}$ for $0\le y\le Y$. Integrating this over $y$ will give a value bounded by a constant times $e^{-M^2}$, which clearly goes to zero as $M\to\infty$.