Singular integrals and differentiability properties of functions-Elias M. Stein - Princeton University Press (1970)
1-1
The most important theorem: Hardy Littlewood Maximal function Inequality.
Calderón-Zygmund Decomposition
2-2
Thm 1:
Condition (b): \[ |\nabla K(x)| \leq B/|x|^{n+1} \qquad \forall x \neq 0 \] Proof Steps of weak 1-1: 1. Calderon-Zygmund decomposition. The good function can be controled easily by T is weak 2-2. 2. The obstacle is the bad function, which is 0 in the "small set" \(F\), and integration is 0 in any cube \(Q_k\). Decompose \(b\) to each of these cube \(Q_i\) as \(b_i\) . Considering any \(x\in F\), we can modify the convolution \(T(f)\) by a constant times \(\int b_j(x)\) . 3. Use Mean value thm, and by porportional fact, we get 4. Marcinkiewicz integral 5. We get weak 1-1 !
2-3
Comparing to Thm with Condition (b'), which is \[ \int_{|x|\geq2|y|} |K(x-y)-K(x)|dx\leq B' \qquad |y|> 0 \] Consider slightly bigger cubes \(Q_j^*\) which share the same center \(y^j\) with cube \(Q_j\), but exapnd its diameter \(2n^{\frac{1}{2}}\) times.
Just because we have the \(|x|\geq2|y|\) in condition (b')
By Fubini we can evaluate \(\int_{Q_j}\leq\int_{Q_j^*}\) with condition (b'), and get to (15) in the original proof.
Think: what is the relation between (b) and (b')?
Now also change the condition (a) \[ |\hat{K}(x)|\leq B \] to condition (a') \[ |K(x)| \leq B|x|^{-n} \] ???
We try to get Thm 2: Suppose the kernel \(K(x)\) satisfies the conditions
\[ \begin{align} &|K(x)| \leq B|x|^{-n} \qquad &|x|>0 \\ & \int_{|x|\geq2|y|} |K(x-y)-K(x)|dx\leq B \qquad &|y|>0 \end{align} \] > then second line is called a "regularity property", for it discribe some kind of smoothness
and the cancellation property \[ \int_{R_1<|x|<R_2} K(x)dx=0 \qquad 0<R_1<R_2<\infty \] For \(f\in L^p(\mathbb{R}^n),\; 1<p<\infty\) , let \[ T_\varepsilon(f)(x)= \int_{|y|\geq \varepsilon}f(x-y)K(y)dy \] then we have \[ \|T_\varepsilon(f)\|_p \leq A_p \|f\|_p \] with \(A_p\) independent of \(f\) and \(\varepsilon\) . And \(\forall f\in L^p(\mathbb{R}^n),\; 1<p<\infty\), we have \[ \lim_{\varepsilon \to 0}T_\varepsilon(f) =: T(f) \] exists in \(L^p\) norm. The operator \(T\) so defined also satisfies the inequality above.
Steps of proof: 1. We take Thm1 as granted, and the modification of condition(b') as granted; 2. We show that if we truncated the operator \(K\) to some \(K_\varepsilon\) , then we have a Lemma(in 2-3.3) that \(K_\varepsilon \in L^2\) , and \[\sup_{y} |\hat{K}_\varepsilon(y)| \leq CB \qquad \forall \varepsilon >0\] where \(C\) depends only on the dimension \(n\). - the proof of lemma is some kind of tricky: firstly we do only for \(K_1\) - Then we - - 3. the \(K_\varepsilon \in L^2\) and fixed \(C\) give us the condition (a) in Thm1, so we can just use it to obtain a uniform \(L^p\) bound for \(T_\varepsilon\) ; 4. For the existance of limit, we first consider smooth function with compactly support: seperate \(T_\varepsilon(f)\) to \(\int_{|y|>1}\) and \(\int_{1 \geq |y|\geq \varepsilon}\) , the first part is fixed and is a \(L^p\) function, modify the second part with \(f_1(x)\) times the cancellation property integral, then by Mean Value Thm we get it has a limit; 5. For arbitrary \(f\in L^p(\mathbb{R}^n)\) , we use the approximation by smooth function with compactly support. The approximation is in norm, so by the inequality \(\|T_\varepsilon(f)\|_p \leq A_p \|f\|_p\) , the leftover can be dropped; 6. So finally the limiting poerator T is well defined, and by taking limit on both sides of the inequality \(\|T_\varepsilon(f)\|_p \leq A_p \|f\|_p\) we get inequality \(\|T(f)\|_p \leq A_p \|f\|_p\) . Finished!
Operator topology
https://en.wikipedia.org/wiki/Operator_topologies
2-4
2023/05/16- 2023/05/22 meeting 2023/05/23
singular integral operators which commute with dilations, and also with translations.
commute with dilations: homogeneous of degree -n
commute with transformations: T can be realized in terms of a multiplier \(m\), s.t. \(T(f)\hat{} = \hat{f} \cdot m\)
So we know that any singular integral operators which commute with dilations and translations will fall into the situation mentioned in following Theorem 3:
Theorem 3: Let \(\Omega\) be homogeneous of degree 0, and suppose that \(\Omega\) satisfies this cancellation property
\[ \int_{S^{n-1}} \Omega(x) d\sigma = 0 \]
and the smoothness property \[ \text{let } \omega(\delta) = \sup_{|x-x'|\leq \delta,\; |x|=|x'|=1} \int_0^1 \frac{\omega(\delta) d\delta}{\delta} < \infty \] then for \(1<p<\infty\), \(f\in L^p(\mathbb{R}^n)\), let \[ T_\varepsilon (f) (x) = \int_{|y|\geq \varepsilon} \frac{\Omega(y)}{|y|^n}f(x-y) dy \] (a) Then there exists a bound \(A_p\) independent of \(f\) or \(\varepsilon\) , so that \[ \|T_\varepsilon(f)\|_p \leq A_p \|f\|_p \] (b) The limit of \(T_\varepsilon(f)\) exists in \(L^p\) norm, and by defining \(T(f) = \lim_{\varepsilon \to 0} T_\varepsilon(f)\), we have \[ \|T(f)\|_p \leq A_p \|f\|_p \] (c) If \(f\in L^2(\mathbb{R}^n)\), then the Fourier transforms of \(f\) and \(T(f)\) are related by \((Tf)\hat{} (x) = m(x) \hat{f}(x)\), where \(m\) is a homogeneous function of degree 0. We can write out the explicit form of \(m\) (....).
Cauchy principal value:
Actually it is a tempered distribution. https://en.wikipedia.org/wiki/Cauchy_principal_value#Distribution_theory
If we change the way to choose end point, what will we have?