## 实分析0-10

Terence Tao 的实分析课程已经完成了一多半，我到今天正好将他已经贴出来的帖子全部看完了，习题在脑子里都做了一遍，之还有一两个没有想通。到目前为止，收获不小，想到此做一个小结。这里是课程主页的链接

Explore posts in the same categories: Math.AP, Maths

1. [...] s Weblog placed an interesting blog post on å® [...]

2. PDEbeginner Says:

Dear Xiaochuan,

Have you done the second part of exercise 17? Prof. Tao has given me the answer, but I still didn’t understand it.

If it is right, we can obtain $(L^{\infty}([0,1]))^{*} \neq L^{1}([0,1])$ because [0,1] include (0,1/2), (1/2,3/4), (3/4,7/8) … positive measure set.

Could you help to try to find out my problem? Thanks a lot!

3. liuxiaochuan Says:

PDEbeginner：
Exercise 17 of which note?

4. PDEbeginner Says:

oops, sorry, when I was issuing this question, nearly went to bed and not had a fresh brain. Note 6. Thanks a lot

5. liuxiaochuan Says:

Dear PDEbeginner,
You are more than welcome to present any questions here. I am glad to discuss with you, especially questions from Professor Tao’s blog.

As to this question, since we alredy have a apace containing a countably infinite sequence of disjoint sets of positive measure. Suppose these sets are $E_1,E_2,\cdots$. Then for any element $\{a_n\}_1^\infty\in l^\infty(\Bbb N)$, we can define the funtions in $L^\infty$ as $\frac{1}{m(E_n)}I_{E_n}$ accordingly, taking values $\frac{1}{m(E_n)}$ in $E_n$ and vanishes otherwise. By this bijection, we actually get a closed subspace within $L^\infty$, which, by the first part of this exercise, lead to a contradiction.

What I am interested is, how do you do the first two questions, I am not quite sure that I fully get the proof of them, at least not satisfied.

6. PDEbeginner Says:

Dear Xiaochuan,

Thank you very much for your answer! I think there is some problem in your construction: if $lim_{n \rightarrow \infty} m(E_n)=0$, then the closure of $span\{latex 1/m(E_n) 1_{E_n}\}$ is not a supspace of $L^{\infty}$.

If your argument is true, then $(L^{\infty}([0,1]))^{*} \neq L^{1}([0,1])$ by taking (0,1/2), (1/2,3/4), (3/4,7/8) …. But it is well known that $(L^{\infty}([0,1]))^{*} = L^{1}([0,1])$.

As for the first two questions: The second one is easy, by taking the dual of reflexive identity. The first one is a little tricky: Suppose $Y \subset X$, clearly $Y^{*} \subset X^{*}$. Take any functional $l$ on $Y^{*}$, by H-B thm, there exists at least one extension of $l$, denoted by $\bar{l}$. Since X is reflexive, there exists some $x \in X$ such that $l(x)=\bar{l}$. Decompose $x=x_1+x_2$ with $x_1 \in Y$ and $||x_2||=\inf_{y_1 \in Y}||x-y_1||$ (actually such decomposition is unique), then apply some easy argument to identify $l(x_1)=l$.

7. liuxiaochuan Says:

Dear PDEbeginner,

My answer to the first question is quite similar to yours, though not so clear as yours.

As for the last one, there was indeed some problems with my previous answer. But I don’t think that $(L^{\infty}([0,1]))^{*} = L^{1}([0,1])$ is correct. Let $C([0,1])$ be the subspace of all the continuous functions in $L^\infty$, and dirac functional $\delta_x$ so that $\delta_x(f)=f(x)$. So $\delta_x$ is a continuous linear functional but don’t belong to $L^1$.

I modify the proof as following. Suppose we alredy have a space containing a countably infinite sequence of disjoint sets of positive measure. Suppose these sets are $E_1,E_2,\cdots$. Then for any element $\{a_n\}_1^\infty \in l^1(\Bbb N)$, we can define the funtion in $L^1$ as $\sum\frac{a_n}{m(E_n)}I_{E_n}$ accordingly. By this bijection, we actually get a closed subspace within $L^1$, and hence a copy in $L^\infty$.

8. PDEbeginner Says:

Dear Xiaochuan,

I am now reading the lecture note 8 of real analysis on Prof. Tao’s blog. I have some problem on the claim in Example 12, i.e. f is a right continuous function iff f is a continuous map from $({\bf R}, \mathcal{F}_r)$ to $({\bf R}, \mathcal{F})$. I thought it over and over again, but still not found any hint to prove it. I was wondering if you could help to give an answer!

Thanks a lot in advance anyway!

9. liuxiaochuan Says:

Dear PDEbeginner：
For one direction, our purpose is to prove that for any $x_0\in Bbb {R}$, the map $f(x)$ is right-continuous, provided f is continuous from $({\bf R}, \mathcal{F}_r)$ to $({\bf R}, \mathcal{F})$. We rewrite this as ‘$\forall \varepsilon >0$, there exists some $\delta>0$ so that $|f(x)-f(x_0)|<\varepsilon$ for all $x\in [x_0,x_0+\delta)$‘,which is true since the inverse of an intervel $(f(x_0)-\varepsilon,f(x_0+\varepsilon))$ always contain some $[x_0,l)$ for some l.

As for the other direction, suppose for contradiction f is not continuous from $({\bf R}, \mathcal{F}_r)$ to $({\bf R}, \mathcal{F})$. Then one can find an intervel (a,b) whose inverse cannot contain any intervel (also nonempty).Then we can find a $x_0$ with its image in (a,b) which is not right-continuous, thus a contradiction.

10. PDEbeginner Says:

Dear Xiaochuan,

To be honest, I don’t quite understand your argument. But finally I figured out the proof.

I think the key point for this exercise is to understand the open sets in $({\bf R}, \mathcal{F}_r)$: given any open set A in $({\bf R}, \mathcal{F}_r)$, if $x \in A$, then there exists some $\delta>0$ such that [$x,x+\delta$) $\subset A$. With this observation, I think the proof is quite easy.

11. PDEbeginner Says:

There seems some formula not being shown in the above post, which is ‘$[x, x+\delta)$ is the subset of $A$‘.

12. waterloo2005 Says:

请问这本书有电子版的吗?

13. liuxiaochuan Says:

书还没有出版，可以链接到陶的博客上，直接打印。

14. waterloo2005 Says:

就是博文吧，不是pdf之类的吧

15. waterloo2005 Says:

听说Terry年仅7岁的时候就养成了读数学书，题目全做的习惯。想想自己曾经有过的错误想法，汗颜。从现在开始倒还不晚。

现在有些数学书，课后的习题都是东拼西凑来的，和内容的呼应太差，用那章的知识就是高斯也作不出来。现在国内这样的书太多了。

16. abc Says:

在verycd上可以搜到的

17. Qiang Says:

Hi, Liu

I’m a graduate in ECE and recently become quite interested in Real Analysis which I think is useful to my own research. However, when I tried to approach those exercises in Tao’s lecture notes, it’s so hard for me (apparently :)). Do you have solutions to those exercises ?

Thank you, I quite enjoy your blog !

• liuxiaochuan Says:

Dear Qiang:

Thank you for your comment. Sorry for the delay. I indeed did almost all the exercises. But I didn’t write them down. If you are interested in any specific problems, I would be more than happy to discuss them with you.

• Qiang Says:

Hi, Xiaochuang

So glad to hear from you ! I have several questions concerning about Exercise 17 in lecture note 1 (Caratheodory). For i->ii, since for element set, $m(A)=m^{*}(A)$, so $m(A)=m^*(A)=m^*((A\bigcap E)\bigcup(A\bigcap E^c))$, because apparently elementary set A is Lebesgue measurable, so $A\bigcap E$ and $A\bigcap E^c$ are measurable, therefore by finite additivity of Lebesgue outer measure, we get $m^*((A\bigcap E)\bigcup(A\bigcap E^c))=m^*(A\bigcap E)+m^*(A\setminus E)$ as desired. For ii->iii, any box $B$ is also an elementary set with $m(B)=|B|$, so we get $|B|=m^*(B\bigcap E)+m^*(B\setminus E)$. Can you give me hints on iii->i ?

And also I don’t know how to prove part 2 and 3 of Exercise 13

Thanks