ช่วยทำข้อสอบ analysisของจุฬาให้หน่อยครับ

[mayalone]

15 ข้อ ข้อละ 8 คะแนน ครับ ช่วยคิดหน่อยงับ

[passer-by]

1. We'll prove that $ f(x)= 0 $ for all irrational numbers

Let $x$ be an irrational number

Since Q is dense in R , we can find sequence $ (x_n) \subset Q $ converges to irrational $x$

And by continuity of $f$ , then we have $ f(x_n) $ converges to $ f(x)$

But from definition of $f$ , we have $ f(x_n)= 0 $ for all $n$

Hence, $ f(x)=0 $ for all irrational $x$

NOTE: This problem leads to interesting corollary: If f,g are continuous on R and f(r)=g(r) for all rational numbers r ,then f(x)=g(x) for all real numbers x


2. Clearly, such $f$ is continuous on R

Since $f(1) < 0$ and $f(2) >0$ then ,by intermediate value theorem, there exists $ r \in (1,2) $ such that $ f(r)=0$

Use this theorem again on $(-8,0)$ since $f(-8) >0$ and $f(0)<0$


3. Define $ g(x)= f(x) - \beta $ Then g is continuous on R and $g(x_0) <0 $

By definition of continuous function, there exists a neighborhood of $x_0$ such that

$ \mid g(x)-g(x_0) \mid < \frac{-g(x_0)}{2} $

Equivalently, $ g(x) < \frac{g(x_0)}{2} < 0$

So $ f(x) < \beta $ in a neighborhood of $x_0$

4. (Like proof for the case f,g are continuous)

5. Not always.

For example, $ f(x)=\left\{\begin{array} &-1& x \leq 0 \\ x-1& x>0 \end{array}\right. $
$ g(x)=\left\{\begin{array} &x& x \leq 0 \\ 0& x>0 \end{array}\right. $

But $ (fg)(x)=\left\{\begin{array} &-x& x \leq 0 \\ 0& x>0 \end{array}\right. $ which is decreasing function.

[M@gpie]

ยากจังคับ แหะๆ สงสัยให้ผมไปสอบคงจะไม่รอด

[mayalone]

ขอบคุณครับ

[passer-by]

6. By definition of f , then we have

$ \mid f(x)-g_{\frac{\epsilon}{3}}(x) \mid < \frac{\epsilon}{3} $ ...(1)
And $ \mid f(y)-g_{\frac{\epsilon}{3}}(y) \mid < \frac{\epsilon}{3} $...(2)

Since $g$ is uniformly continuous on A ,then there exists $ \delta >0 $ such that

$ \mid g_{\frac{\epsilon}{3}}(x)-g_{\frac{\epsilon}{3}}(y) \mid < \frac{\epsilon}{3} $ (if $ \mid x-y \mid < \delta $) ...(3)

So if we combine (1) to (3) then we have

$\mid f(x)-f(y) \mid \leq \,\mid f(x)-g_{\frac{\epsilon}{3}}(x) \mid + \mid g_{\frac{\epsilon}{3}}(x)-g_{\frac{\epsilon}{3}}(y) \mid + \mid g_{\frac{\epsilon}{3}}(y)-f(y) \mid < \frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3} =\epsilon $(if $ \mid x-y \mid < \delta $ and all x,y in A)

Hence, f is uniformly continuous on A.

7. Neither open nor closed set.

Let $ x \in Q$

Since every neighborhood of x contains irrational numbers ,then Q can't be open set.

Moreover, we can construct sequence of points in Q converging to irrational number,say ,y , so y is a limit point of Q, but not in Q. It follows that Q isn't closed set.

NOTE: In mathematics, set seems different from door. Sometimes it's both closed and open simultaneously and sometimes it's neither closed nor open simultaneously.

8. No.

For example, $ K_n= [n,n+1] $ which is compact in R. So $$ \bigcup_{n=1}^{\infty}=[1,\infty)$$ Such countable union isn't compact (since it's unbounded in R).


9. (I 'll do this problem based on compact in R only)

Since compact implies bounded ,then inf K and sup K exist.

The remaining is to prove that both are in K. We'll prove for sup K only and inf K is done in similar way.

Using characterization of supremum, for all $\epsilon >0$ there exists $x \in K $ such that
$\text{sup K} - \epsilon < x $...(1)

Since $ x \leq \, \text{sup K} < \text{sup K} +\epsilon $...(2)

Combine (1),(2) we obtain$ \mid \text{sup K} - x \mid < \epsilon $

Equivalently, all neighborhoods of sup K contains element in K and thus sup K is a limit point of K.

But K is compact , so it's closed and hence K contains all limit points,including sup K.

10. Let $ x,y \in I $ and suppose that x<y

By Mean Value Theorem of f on (x,y),

so there exists $ c \in (x,y) $ such that $\frac{f(y)-f(x)}{y-x} = f'(c) < 0 $

Thus, f(y)-f(x) < 0 ,that is , f is strictly decreasing.

[mayalone]

ขอบคูณมากเลยครับ อีกนิดนะครับ

[passer-by]

11. See #18 of this link Solution

13. Since $ f_n$ converges uniformly to $f$ , then

$ \exists N $ such that $ \mid f_N(x)-f(x) \mid < 1 $ for all $ x \in A $

And since $f_n$ is bounded on A,say ,by M.

It follows that $ \mid f(x) \mid \leq \,\, \mid f(x)-f_N(x) \mid + \mid f_N(x) \mid < 1+M \quad \forall x \in A $

NOTE: เว้นวรรค คำว่า "ขอบเขต" กับคำว่า "บน A" ในโจทย์ด้วยก็ดีนะครับ

14. Use Weierstrass M-Test

15. Let $S_k$ be partial sum of $ \sum f_n $

For $c = 0$ , Obvious !

For $ c \neq 0 $

Since $ \sum f_n $ converges uniformly to $ f $ then

$ \forall \epsilon >0 \,\, \exists N $ such that $ \mid S_k(x)-f(x) \mid <\frac{\epsilon}{\mid c\mid} \quad (k \geq N , \forall x \in A)$

After this, it's not difficult to follow.

Edit : Add Link of Question 11.