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Tuesday, November 19, 2013

Linear Algebra

Problem 1

Prove that if TL(V) is normal, then
nullTk=nullT and rangeTk=rangeT

Solution:
easy to check that nullTnullTk
to prove an inclusion in the other direction, suppose that vnullTk
Tkv=T(Tk1v)=0Tk1vnullT and Tk1vrangeTk1rangeT
but V=nullT(nullT)=nullTrangeT=nullTrangeT
the third equality holds because T is normal
Tk1vnullTrangeTTk1v=0, inductively, we have Tv=0
This shows that nullTknullT, completing the proof that nullT=nullTk

Now we are going to show that rangeTk=rangeT
rangeT=rangeT=(nullT)=(nullTk)=range(Tk)=rangeTk
first equality holds because T is normal, third equality is come from the previous proof, the last equality holds because Tk is normal

Thursday, November 7, 2013

Lebesgue Measure

Problem 1

Let An be any sequence of subsets of R
(i) if AnA then λ(An)λ(A)
(ii)λ(lim infAn)lim infnλ(An), where lim infAn={x:xAnultimately}

solution:
(i)
Let  Fn=ni=1Ei, F=En,A=An, where En is measurable such that AnEn, and EnAn is negligible (it is easy to prove such En exists), clearly AF,
FnF, therefore λ(Fn)λ(F)
λ(An)λ(Fn)λ(An)+λ(FnAn)λ(An)+ni=1λ(EiAi)=λ(An)
λ(Fn)=λ(An)λ(A)λ(F), and since FnF, we have λ(An)λ(A)
(ii)
Let Bn=knAk, then BnB=lim infAn
λ(Bn)λ(An)lim infnλ(Bn)=λ(lim infAn)lim infnλ(An)
The equality follows from (i)

Problem 2

For A[a,b]R, λ(A) is also called the exterior measure of A in [a,b] and is denoted λe(A), the interior measure of A in [a,b], denoted λi(A), is defined by the formula
λi(A)=(ba)λe([a,b]A)in general, λi(A)λe(A) ; the set A is measurable if and only if λi(A)=λe(A)

solution:
is easy
We use λ instead of λe denote I=[a,b], let J be any bounded interval λ(I)=λ(IA)+λ(IA)=λ(IAJ)+λ(IAJ)+λ(IAJ)+λ(IAJ)=λ(IAJ)+λ(IAJ)+λ(IAJ)+λ(IAJ)λ(IJ)+λ(IJ)=λ(I)
Therefore we have λ(IAJ)+λ(IAJ)=λ(IJ), and it is easy to see that λ(AJ)+λ(AJ)=λ(J) Suppose S is a subset of R for every ε>0, there exists U,SU=Jn and λ(Jn)<λ(S)+ε
λ(AS)+λ(AS)λ(AU)+λ(AU)λ(AJn)+λ(AJn)=(λ(AJn)+λ(AJn))=λ(Jn)<λ(S)+ε
Since ε is arbitrary, we can conclude that A is measurable 

Problem 3

If A is a Lebesgue-measurable subset of R such that λ(A)>0, then the difference set D={ab:a,cA} is a neighborhood of 0, that is, (δ,δ)D for some δ>0.

Tuesday, October 15, 2013

Sequences of Functions

Problem 1
Let {gn} be a sequence of real valued functions such that gn+1(x)gn(x) for each x in T and for every n=1,2,.... If {gn} is uniformly bounded on T and if fn(x) converges uniformly on T, then fn(x)gn(x) also converges uniformly on T.

Problem 2
Let an be a decreasing sequence of positive terms. Prove that the series ansinnx converges uniformly on R if, and only if, nan0 as n

Only if direction is easy so we omit the proof here, we only prove the if direction
Suppose nan0 as n, let Bn(x)=nk=1sinkx nk=1aksinkx=Bnan+1nk=1Bk(ak+1ak) |mk=n+1aksinkx|=|Bnan+1Bmam+1+mk=n+1Bk(ak+1ak)|=|Bnan+1Bmam+1+B(am+1an+1)|, wheremin{Bn+1,Bn+2,...,Bm}Bmax{Bn+1,Bn+2,...,Bm}|(BnB)an+1|+|(BBm)am+1|<2Nan+1+2Mam+1<ε

problem 1 can be proved in a similar way

Problem 3
Given a power series n=0anzn whose coefficients are related by an equation of the form an+Aan1+Ban2=0 (n=2,3,...)
Show that for any x for which ther series converges, its sum is a0+(a1+Aa0)x1+Ax+Bx2

Problem 4
Show that the binomial series (1+x)^{\alpha}=\sum_{n=0}^{\infty} {\alpha \choose k} x^n exhibits the following behavior at the points x=\pm 1
a) If x=-1, the series converges for \alpha\geq 0 and diverges for \alpha<0
b) If x=1, the series diverges for \alpha \leq -1, converges conditionally for \alpha in the inteval -1<\alpha<0, and converges absolutely for \alpha\geq0