Stein 3장 연습문제

여기서도 4k+2만 푼다. (+ Prob 8)

숙제 문제는 6,7,9,17,19,24,32 + Problem 8.

 

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2. (Kernel의 조건이 바뀌면 어떻게 될까?)

Suppose $\{K_\delta\}$ is a family of kernels that satisfies:

1. $|K_\delta(x)|\leq A\delta^{-d}$ for all $\delta>0$
2. $|K_\delta(x)|\leq A\delta/|x|^{d+1}$ for all $\delta>0$.
3. $\int K_\delta = {0}$ (Not 1!!)

Show that if $f$ is integrable on $\mathbb R^d$, then

$$(f\ast K_\delta)(x)\to 0 \textrm{  for a.e. $x$, as $\delta \to 0$.}$$

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6. (숙제문제)

In one dimension there is a version of the basic inequality

$$m(\{x\in \mathbb R: f^\ast>\alpha\})\leq\frac{A}{\alpha}\Vert f\Vert_1$$

in the form of an identity. We define the "one-sided" maximal function

 

$$f^\ast_+(x)=\sup_{h>0}\frac{1}{h}\int_x^{x+h}|f(y)|dy.$$

 

If $E^+_\alpha=\{x\in \mathbb R: f^\ast_+>\alpha\}$, then

 

$$m(E^+_\alpha)=\frac{1}{\alpha}\int_{E^+_\alpha}|f(y)|dy.$$

 

[Hint: Apply Lemma 3.5 to $F(x)=\int_0^x|f(y)|dy-\alpha x$. Then $E^+_\alpha$ is the union of disjoint intervals $(a_k,b_k)$ with $\int_{a_k}^{b_k}|f(y)|dy=\alpha(a_k-b_k)$.]

Proof

 

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10.

Construct an increasing function on $\mathbb R$ whose set of discontinuities is precisely $\mathbb Q$.

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14.

The following measurability issues arose in the discussion of differentiability of functions.

1. Suppose $F$ is continuous on $[a,b]$. Show that $$D^+(F)(x)=\limsup_{h\to 0+}\frac{F(x+h)-F(x)}{h}$$ is measurable.
2. Suppose $J(x)=\sum^\infty_{n=1}\alpha_n j_n(x)$ is a jump function as in Section 3.3. Show that $$\limsup_{h\to 0}\frac{J(x+h)-J(x)}{h}$$ is measurable.

[Hint: For (a), the continuity of $F$ allows one to restrict to countably many $h$ in takeing the limsup. For (b), given $k>m$, let $F^N_{k,m}=\sup_{1/k\leq|h|\leq1/m}\left|\frac{J_N(x+h)-J_N(x)}{h}\right|$, where $J_N(x)=\sum^N_{n=1} \alpha_n j_n(x)$. Note that each $F_{k,m}^N$ is measurable. Then, successively, let $N\to \infty, k\to \infty$, and finally $m\to \infty$.]

 

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18.
Verify the agreement between the two definitions given for the Cantor-Lebesgue function in Exercise 2. Chapter 1 and in Section 3.1 of this chapter.

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22.

Suppose that $F$ and $G$ are absolutely continuous in $[-\pi,\pi]$. Show that their product $FG$ is also absolutely continuous. This has the following consequences.

1. $$\int^\pi_{-\pi} F'(x)G(x)dx=-\int^\pi_{-\pi} F(x)G'(x)dx+[F(x)G(x)]^\pi_{-\pi}$$
2. Let $F(\pi)=F(-\pi)$. Show that if $$a_n=\frac{1}{2\pi} \int^\pi_{-\pi}F(x)e^{-inx}dx,$$ such that $F(x)\sim \sum a_n e^{inx}$, then $$F'(x)\sim \sum ina_ne^{inx}.$$
3. What happens if $F(-\pi) \neq F(\pi)$? [Hint: Consider $F(x)=x$.]

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26.

외측도를 Cube가 아니라 Ball로 정의하면 어떻게 될까? 그래도 똑같은데, 원래 정의를 $m_\ast$로, Ball로 만든 정의를 $m_\ast^\mathcal B$라고 하면 $m_\ast(E)\leq m_\ast^\mathcal B(E)$는 자명하고, Reverse Inequality를 보여야 한다. 다음을 보여라. $$\sum_j m(B_j)\leq m_\ast(E)+\epsilon$$

Note that for any preassigned $\delta$, we can choose the balls to have diameter $<\delta$. 

[Hint: $E$를 열린집합으로 싸매고 적당히 근사되게 finite ball 고르고, 나머지 부분은 cube로 덮고 그걸 ball로 바꾼다.]

 

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30.

$F$가 bounded이고 임의의 Bounded suinterval $[a,b]$에서 유계변동이고 $\sup_{a,b}T_F(a,b)<\infty$이면, $\mathbb R$에서 유계변동이라고 한다. 다음을 보여라.

• $\int_\mathbb{R}|F(x_h)-F(x)|dx\leq A|h|$, for some constant $A$ and all $h\in \mathbb R$.
• $|\int_\mathbb{R} F(x)\varphi'(x)dx|\leq A,$ where $\varphi$ ranges over all $C^1$ functions of bounded support with $\sup|\varphi(x)|\leq 1$.

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Prob 8.  (숙제문제)

Let $\mathcal R$ denote the set of all rectangles in $\mathbb R^2$ that contain the origin and with sides parallel ot the coordinate axis. Conseider the maximal operator associated to this family, namely

$$f^\ast_{\mathcal R}=\sup_{R\in \mathcal R}\frac{1}{m(R)}\int_R|f(x-y)|dy.$$

1. Then, $f\mapsto f^\ast_\mathcal R$ does not satisfy the weak type inequality $$m(\{x:f^\ast_\mathcal R (x)>\alpha\})\leq \frac{A}{\alpha}\Vert f\Vert_1$$
2. Using this, one can show that there exists $f\in L^1(\mathbb R)$ so that for $R\in \mathcal R$ $$\limsup_{\textrm{diam}(R)\to 0}\frac{1}{m(R)}\int_R f(x-y)dy=\infty \textrm{ for a.e. $x$.}$$

[Hint: For part (a), let $B$ be the unit ball, and consider the function $\phi(x)=\chi_B(x)/m(B)$. For $\delta>0$, let $\phi_\delta (x) = \delta^{-2}\phi(x/\delta)$. Then $$(\phi_\delta)^\ast_\mathcal R (x)\to \frac{1}{|x_1||x_2|} \textrm{ as $\delta \to 0$,}$$ for every $(x_1,x_2)$, with $x_1x_2 \neq 0.$ If the weak type inequality held, then we would have $$m(\{|x|\leq 1: |x_1x_2|^{-1}>\alpha\})\leq\frac{A}{\alpha}.$$ This is a contradiction since the left-hand side is of the order of $(\log\alpha)\alpha$ as $\alpha$ tends to infinity.]