Pesudo differential opertor and singular integral

I already understand this material 3days ago but it is a little difficult for me to type the latex…

 

1. Introduction

There is two space to understand a function’s behaviour, the physics space and the frequency space (Why thing going like this? Why there is such a duality?). Namely, we have:

\displaystyle \hat f(\xi)=\int_{{\mathbb R}^d}e^{2\pi i\xi x}f(x)dx \ \ \ \ \ (1)

 

The key point is, waves is a parameter group of scaling of definition of a constant fraquence wave, so it connected the multiplication and addition. Basically due to it can be look as the correlation of a function and the scaling of wave with carry all the information about {f}. A generation of this obeservation is the wavelet theory.

So as we well know, the key ingredient of Fourier transform is to image function as a sum of series waves. A famous theorem of Mikhlion said that a translation-invariant operator {T} on {R^n} could be represented by a multiplication operator on the Fourier transform side. translation is the meaning, {h\circ T=T\circ h, \forall h} is a translation.

In a formal level, consider it as distribution (compact distribution or temperature distribution is both OK). We have:

\displaystyle T(e^{2\pi ix\xi})=a(\xi)e^{2\pi ix\xi}, \forall \xi \in {\mathbb R}^n \ \ \ \ \ (2)

 

the meaning is if we consider {T} is a operator on distribution space, {T:S'\rightarrow S'}, then {\forall f\in S},

\displaystyle \int T(e^{2\pi ix\xi})f=\int a(\xi)e^{2\pi ix\xi}f

due to the linear combination of {e^{2\pi ix\xi}} will consititue a dense set in {S}. So this could extend to the whole distribution space by dual and give the definition of {T}, i.e.

\displaystyle (Tf)(x)=\int_{{\mathbb R}^n}a(x,\xi)e^{2\pi ix\xi}\hat f(\xi)d\xi \ \ \ \ \ (3)

 

Remark 1 {T} is bounded on {L^2({\mathbb R}^n)} when {a} is a bounded function, thanks to Parevel theorem. When {a} is a bounded function, the composition of two such operator could be defined, and the symbol of composition operator corresponding to the composite of their symbol, i.e.

\displaystyle T_a\circ T_b(e^{2\pi ix\xi})=b(\xi)a(\xi)e^{2\pi ix\xi} \ \ \ \ \ (4)

 

Remark 2 For parenval theorem, i.e. {\|\hat f\|_2=\|f\|_2}, there is two approach, heat kernel approximation approach and discretization.

We wish to investigate the operator given by multiplier, i.e.

\displaystyle (Tf)(x)=\int_{{\mathbb R}^n}a(x,\xi)e^{2\pi ix\xi}\hat f(\xi)d\xi \ \ \ \ \ (5)

When it is satisfied {\|T\|_{p-p}<\infty}?

Intuition, the following calculate is only morally true, not rigorous.

\displaystyle \begin{array}{rcl} \|Tf\|^p_p & = & \int_{{\mathbb R}^d }|\int_{{\mathbb R}^d}a(x,\xi) e^{2\pi ix\xi} \hat f(\xi)d\xi|^pdx \\ & \overset{\exists f}\sim & \int_{{\mathbb R}^d }\int_{{\mathbb R}^d}|a(x,\xi) e^{2\pi ix\xi} \hat f(\xi)|^p d\xi dx \\ & \overset{Fubini}\sim & \int_{{\mathbb R}^d}\int_{{\mathbb R}^d}|\widehat{a(x,\xi)}\hat f(\xi) |^pdxd\xi \\ & \sim & \int_{{\mathbb R}^n}\int_{{\mathbb R}^n}|{a(x,\xi)}^{\vee}*f(\xi)|^pdxd\xi \end{array}

 

So we need some restriction on { {a(x,\xi)}^{\vee}}, namely {\widehat{a(x,\xi)}}, so we need some decay condition on {|\partial_x^{\alpha}\partial_{\xi}^{\beta}a(x,\xi)|}, why this, just consider integral by part for {a(x,\xi)\in S}. The rigorozaton of this intuition inspirit us to the definition of symbol calss.

Definition 1 we say {a(x,\xi)} is in symbol class {S_m} iff,

\displaystyle |\partial_x^{\beta}\partial_{\xi}^{\alpha}a(x,\xi)|\leq A_{\alpha,\beta}(1+|\xi|)^{m-|\alpha|} \ \ \ \ \ (6)

for all {\alpha,\beta} is multi-indece.

Remark 3

  1. we note that all partial differential operator, whose coefficient, together with all their derivatives are bounded belong to this class, In this particular circumstance, the symbol is a polynomial in {\xi}, essentially the “characteristic polynomial” of the operator.
  2. The general operator of this class have a parallel description in terms of their kernels. That is, in a suitable sense,

    \displaystyle (Tf)(x)=\int_{{\mathbb R}_x}K(x,y)f(y)dy \ \ \ \ \ (7)

    besides enjoying a cancellation property, {K} is here characterized by differential inequalities “dual” to those for {a(x,\xi)}. In the key case where the order {m=0}, this kernel representation makes {T} a singular integral operator.

  3. The crucial {L^2} estimate, when {m=0}, is atelatively simple consequences of Plancherel’s theorem for the Fourier transform. With this, the {L^p} theory introduce in previous note is therefore applicable.
  4. The product identity that holds in the translation-invariant case generalized to the situation treated here as a symbolic calculus for the composition of operators. That is, there is an asymptotic formula for the composition of two such operators, whose main term is the point-wise product of their symbols.
  5. The succeeding terms of the formula are of decreasing orders. These orders measure not only the size of the symbols, but determine also the increasing smoothing properties of the corresponding operators. The smoothing properties are most neatly expressed in terms of the Sobolev space {W_k^p} and the Lipschitz space {\Lambda_{\alpha}}.

 

2. Pseudo-differential operator

“Freezing principle”: from variable coefficient differential equation to constant coefficient differential equation by approximation. divide into 2 steps:

  1. divide space into small cubes.
  2. take average of the coefficient of differential equation in every cubes.

Suppose we are interested in study the solution of the classical elliptic second order equation.

\displaystyle (Lu)(x)=\sum a_{ij}(x)\frac{\partial^2 u(x)}{\partial x_i\partial x_j}=f(x) \ \ \ \ \ (8)

 

Where the coefficient matrix {\{a_{ij}(x)\}} is assume to be real, symmetric, positive definite and smooth in {X}. Understanding {P}, such that,

\displaystyle PL=I \ \ \ \ \ (9)

 

Looking for a {P}. Such that {PL=I+E}. {E} is a error term which have good control. To do this, fix an arbituary point {x_0}, freeze the operator {L} at {x_0}:

\displaystyle L_{x_0}=\sum a_{ij}(x_0)\frac{\partial^2}{\partial x_i\partial x_j} \ \ \ \ \ (10)

 

In Fourier sense ({L^2} sence).

\displaystyle \begin{array}{rcl} L_{x_0}f(x) & = & \int e^{2\pi ix\xi}(\widehat{ \sum a_{ij}(x_0)\frac{\partial^2 f(\xi)}{\partial x_i\partial x_j}}) d\xi\\ & = & \int e^{2\pi ix\xi}\int e^{-2\pi i \xi y}\sum a_{ij}(x_0)\frac{\partial^2}{\partial x_i\partial x_j}f(y)dyd\xi\\ & = & \int e^{2\pi ix\xi}(-4 \pi^2)\sum_{i,j}a_{ij}(x_0)\xi_i\xi_j \end{array}

Remark 4 The remark is, morally speaking, for application of fourier transform in PDE. morally we could only solve the problem with linear differential equation (although we could consider the hyperbolic type). The main obstacle for Fourier transform application into PDE:

  1. it only make sense with Schwarz class or its dual, this is not main obstacle, in principle could be solved by rescaling.
  2. the main obstacle is it only compatible with linear differential equation.

 

Cut-off function: {\eta} vanish near the origin,

\displaystyle (P_{x_0}f)=\int_{{\mathbb R}^n}(-4\pi^2\sum_{i,j}a_{ij}(x_0)\xi_i\xi_j)^{-1}\eta(\xi)\hat {f(\xi)}e^{2\pi ix\xi}d\xi. \ \ \ \ \ (11)

 

then:

\displaystyle L_{x_0}P_{x_0}=I+E_{x_0}.

{E_{x_0}} is actually a smoothing operator, because it is given by convolution with a fixed test function. It should be seasonable when {x} near {x_0}, {(Pf)(x)} is well approximated by {(P_{x_0}f)(x)}, it is actually the case, define {((Pf)(x):=(P_xf)(x)}, i.e.

\displaystyle (Pf)(x)=\int_{{\mathbb R}^n}(-4\pi^2\sum_{i,j}a_{ij}(x)\xi_i\xi_j)^{-1}\eta(\xi)\hat f(\xi)e^{2\pi ix\xi}d\xi \ \ \ \ \ (12)

 

The operator {P} so given is a propotype of a pesudo-differential operator. Moreover, one has {LP=I+E}, where the error operator {E} is “smoothing of order 1”. That this is indeed the case is the main part of the symbolic calculus described.

Definition 2 (symbol class) A function {a(x,\xi)} belong to {S^m} and is said to be of order {m} of {a(x,\xi)} is a {C^{\infty}} function of {(x,\xi)\in {\mathbb R}^n\times {\mathbb R}^n} and satisfies the differential inequality:

\displaystyle |\partial_x^{\beta}\partial_{\xi}^{\alpha}a(x,\xi)|\leq A_{\alpha,\beta}(1+|\xi|)^{m-|\alpha|} \ \ \ \ \ (13)

 

For all {\alpha,\beta} are multi-indece.

Now we trun to the exact meaning of pesudo-differential operator, i.e. how them action on functions. Under some suffice given regularity condition, for {a\in S^m}, {T_a:S\rightarrow S}.

\displaystyle (Tf)(x)=\int_{{\mathbb R}^n}a(x,\xi)\hat f(\xi)e^{2\pi ix\xi}d\xi \ \ \ \ \ (14)

 

Remark 5 {T_a:S\rightarrow S} is continuous and for {a_k\rightarrow a} pointwise, {a_k\in S,\forall k\in {\mathbb N}^*}, {T_{a_k}(f)\rightarrow T_a(f)} in {S}.

then expense it, we get:

\displaystyle (T_af)(x)=\int\int a(x,\xi)e^{2\pi i\xi(x-y)}f(y)dyd\xi \ \ \ \ \ (15)

 

This could be diverge, even when {f\in S}. The key point is we do not have control with the second integral, morally speaking, this phenomenon is the weakness of Lesbegue integral which would not happen in Riemann integral, so sometime we need the idea from Riemann integral, this phnomenon is settle by multi a cut off function {\eta_{\epsilon}} and take {\epsilon\rightarrow \infty}, the same deal also occur as the introduced of P.V. integral in Hilbert transform. The precise method to deal with the obstacle is following: {a_{\epsilon}(x,\xi)=a(x,\xi)\gamma(\epsilon x,\epsilon \xi)}, if {a\in S^m}, {a_{\epsilon}\in S^m}. {T_{a_{\epsilon}}\rightarrow T_a} in the sense:

{\forall f\in S}, {T_{a_{\epsilon}}(f)\rightarrow T_a(f)},

\displaystyle (T_af)(x)=\lim_{\epsilon\rightarrow 0}\int\int a_{\epsilon}(x,\xi)e^{2\pi i\xi(x-y)}f(y)dyd\xi \ \ \ \ \ (16)

We also have:

\displaystyle <T_af,g>=<f,T_a^*g>, \forall f,g\in S. \ \ \ \ \ (17)

 

Then we have:

\displaystyle (T^*_ag)(y)=\lim_{\epsilon\rightarrow 0}\int\int \bar a_{\epsilon}(x,\xi)e^{2\pi i\xi(y-x)}g(x)dxd\xi \ \ \ \ \ (18)

and {<f,g>} denotes {\int_{{\mathbb R}^n}f(x)\bar g(x)dx}. Thus the pesudo-differential operator {T_a} initially defined as a mapping from {S} to {S}, extend via the identity 17 to a mapping from the space of temperatured distribution {S'} to itself {S'}. Notice also that {T_a} is automatically continuous in this space. \newpage

3. {L^p} bounded theorem

We first introduce a powerful tools, called dyadic decomposition,

Lemma 3 (dyadic decomposition) In eculid space {{\mathbb R}^n} there exists a function {\phi\in C^{\infty}({\mathbb R}^n)} such that,

\displaystyle \sum_{i\in {\mathbb Z}}\phi(2^{-i}x)=1 \ \ \ \ \ (19)

 

and {\forall x\in {\mathbb R}^n}, there is only two of {i\in {\mathbb Z}} such that {\phi(2^{-i}x)\neq 0}, and we can choose {\phi} to be radical and {\phi(x) \geq 0,\forall x\in {\mathbb R}^n}.

Remark 6

So for a given mutiplier {a(x,\xi)}, we will have {a(x,\xi)=\sum_{i\in {\mathbb Z}}a_i(x,\xi)=\sum_{i\in Z}\phi(2^{-ix})a(x)}.

Proof: The proof is easy, after rescaling we just need observed there is a bump function satisfied whole condition. \Box

Theorem 4 Suppose {a} is a symbol of order 0, i.e. that {a\in S^0} Then the operator {T_a}, initially defined on {S}, extends to a bounded operator from {L^2({\mathbb R}^n)} to itself.

Remark 7 Suffice to show {\|T_a(f)\|_{L^2}\leq A\|f\|_{L^2}, \forall f\in S} and by dual.

In fact we can directly proof a more general theorem:

Theorem 5 Let {m:{\mathbb R}^d-\{0\}\rightarrow {\mathbb C}} satisfy, for any multi-index {\gamma} of length {|\gamma|\leq d+2},

\displaystyle |\partial^{\gamma}m(\xi)|\leq B|\xi|^{-|\gamma|}

For all {\xi\neq 0}. Then, for any {0<p<\infty}, there is a constant {C=C(d,p)} such that,

\displaystyle \| (m\hat f)^{\vee}\|_p\leq C(p,d)\|f\|_p \ \ \ \ \ (20)

 

for all {f\in S}.

Proof: {a\in S^0}, so we have:

\displaystyle |\partial_x^{\beta}\partial_{\xi}^{\alpha}a(x,\xi)\leq a_{\alpha,\beta}(1+|\beta|)^{-|\alpha|} \ \ \ \ \ (21)

{\forall \alpha,\beta} are multi indeces. Then we consider dyadic decomposition, the is a function {\phi} satisfied the condition in 19, define {a_i(x,\xi)=\phi(2^{-i}x)a(x)}. then {supp a_i(x,\xi)} cpt, {\|a_i\|<\infty}. So {a_i\in L^p({\mathbb R}^n)}, we have,

\displaystyle \begin{array}{rcl} \|T_{a_i}f\|_p^p & = & \int_{{\mathbb R}^d}|K_i*f(x)|^pdx \\ & = & \int_{{\mathbb R}^d}|\int_{{\mathbb R}^d}K_i(x-y)f(y)dy|^pdydx\\ & \leq &\int_{{\mathbb R}^d}\int_{{\mathbb R}^d}|K_i(x-y)f(y)|^pdydx\\ & = & \|K_i\|_p^p\|f\|_p^p \end{array}

{\|K_i\|_p} have good decay estimate, thanks to {u_i=\phi(2^{-i}x)a(x)\in S^0}, this estimate is deduce morally along the same ingredient of “station phase”, it is come from a argument combine “counting point” argument and a rescaling argument. So,

\displaystyle \begin{array}{rcl} \|Tf\|_p & = & \|\sum T_if\|_p\\ & \leq & (\sum \|K_i\|_p^p)\|f\|_p \end{array}

But we have {\sum\|K_i\|_p^p\leq \infty}, ending the proof. \Box

Remark 8 this method also make sense of restrict the condition to be:

\displaystyle |\partial_x^{\beta}\partial_{\xi}^{\alpha}a(x,\xi)|\leq A_{\alpha,\beta}(1+|\xi|)^{-|\alpha|}, \forall |\alpha|\leq d+2. \ \ \ \ \ (22)

Where {d} is the dimension of the space, and we could change {2} to {1+\epsilon}.

Remark 9

\displaystyle \widehat{\frac{\partial^2 u}{\partial x_i\partial x_j}}|\xi|=\frac{\xi_i\xi_j}{|\xi|^2}\widehat \Delta u(\xi), m(\xi)=\frac{\xi_i\xi_j}{|\xi|^2} \ \ \ \ \ (23)

is a counter example for {p=1,\infty}.

Remark 10 The key point is the estimate

\displaystyle \int|\int e^{2\pi i\xi x}\phi(2^{-j}\xi)m(\xi)d\xi |^pdx \ \ \ \ \ (24)

Correlation of taylor expension and wavelet expension. This is also crutial for the theory of station phase.

4. Calculus of symbols

This calculus of symbols would imply there is some structure on this set.

Theorem 6 Suppose {a,b} are symbols belonging to {S^{m_1}} and {S^{m_2}} respectively. Then there is a symbol {c} in {S^{m_1+m_2}} so that:

\displaystyle Tc=T_a\circ T_b

Moreover,

\displaystyle c\sim \sum_{\alpha}\frac{(2\pi i)^{-|\alpha|}}{\alpha}(\partial_{\xi}^{\alpha}a)(\partial_x^{\alpha}b). \ \ \ \ \ (25)

in the sense that,

\displaystyle c-\sum_{|\alpha|<N}\frac{(2\pi i)^{-|\alpha|}}{\alpha !}\partial_{\xi}^{\alpha}\partial_x^{\alpha}b\in S^{m_1+m_2-N} \ \ \ \ \ (26)

For all {N>0}.

The following “proof” is not rigorous, we just calculate it formally, we could believe it is true rigorously, by some approximation process. Proof: We assume {a,b} have compact support so that our manipulations are justified. We use the alternate formula 15 to write,

\displaystyle (T_af)(y)=\int b(y,\xi)e^{2\pi i\xi(y-z)}f(z)dzd\xi \ \ \ \ \ (27)

Then we apply {T_a}, again in the form 15, but here with the variable {\eta} replacing in the integration. The result is,

\displaystyle T_a(T_bf)(x)=\int a(x,\eta)b(y,\xi)e^{2\pi i\eta(x-y)}e^{2\pi i\xi(y-z)}f(z)dzd\xi dyd\eta. \ \ \ \ \ (28)

This calculate is easy to derive, but the following is more tricky. Now {e^{2\pi i\eta(x-y)}\cdot e^{2\pi i\xi(y-z)}=e^{2\pi i(x-y)(\eta-\xi)}\cdot e^{2\pi i(x-z)\xi}}, so

\displaystyle T_a(T_bf)(x)=\int c(x,\xi)e^{2\pi i(x-z)\xi}f(z)dz d\xi \ \ \ \ \ (29)

with

\displaystyle c(x,\xi)=\int a(x,\eta)b(y,\xi)e^{2\pi i(x-y)(\eta-\xi)}dyd\eta \ \ \ \ \ (30)

 

we can also carry out the integration in the y-variable. This leads to the corresponding Fourier transform of {b} in that variable, and allows us to rewrite 30 as,

\displaystyle c(x,\xi)=\int a(x,\xi+\eta)\hat b(\eta,\xi)e^{2\pi i x\eta}d\eta. \ \ \ \ \ (31)

With this form in hand, use taylor expense to the symbol {a(x,\xi+\eta)}, i.e.

\displaystyle a(x,\xi+\eta)=\sum_{|\alpha|<N}\partial_{\xi}^{\alpha}a(x,\xi)\eta^{\alpha}+R_N(x,\xi,\eta) \ \ \ \ \ (32)

with a suitable error term {R_N}, due to

\displaystyle \frac{1}{\alpha !}\int \partial_{\xi}^{\alpha}a(x,\xi)\hat\eta(\eta,\xi)e^{2\pi ix\eta}d\eta=\frac{(2\pi i)^{|\alpha|}}{\alpha !}(\partial_{\xi}^{\alpha}a(x,\xi))(\partial_x^{\alpha}b(x,\xi)). \ \ \ \ \ (33)

we only need to proof {R_N\in S^{m_1+m_2-N}} and it is definitely the case, we get the theorem. \Box

Remark 11 We need replace {a,b} with {a_{\epsilon},b_{\epsilon}}, where

\displaystyle a_{\epsilon}(x,\xi)=a(x,\xi)\cdot \gamma(\epsilon ,\epsilon \xi), b_{\epsilon}(x,\xi)=b(x,\xi)\cdot \gamma(\epsilon ,\epsilon \xi). \ \ \ \ \ (34)

we note that {a_{\epsilon},b_{\epsilon}} satisfy the same differential inequalities that {a} and {b} do, uniformly in {\epsilon, 0<\epsilon\leq 1} .passage to the limit as {\epsilon\rightarrow 0} will then give us our desired result.

5. Estimate in {L^p} , Sobolev, and Lipchitz space

We now take up the regularity properties of our pesudo-differential operator as expressed in terms of the standard function spaces, we begin with the {L^p} boundedness of an operator of order {0}.

5.1. {L^p} estimate

Suppose {a} belongs to the symbol class {S^0}. Then, we can express {T=T_a} as

\displaystyle (Tf)(x)=\int K(x,y)f(y)dy=\int K(x,x-y)f(y)dy \ \ \ \ \ (35)

due to {a\in S^0}, we know, with some approximation argument and first do it with a cutoff symbol of {a}, i.e. {a_{\epsilon}}, that,

\displaystyle |K(x,y)|\leq A|x-y|^{-n} \ \ \ \ \ (36)

So that the integral coverage whenever {f\in S} and {x} is away from the support of {f}. Since we know that {T} is bounded on {L^2({\mathbb R})}, this representation extends to all {f\in L^2({\mathbb R})} for almost every {x\notin supp f}. More generally, we have,

\displaystyle |\partial_{x}^{\alpha}\partial_{y}^{\beta}K(x,y)|\leq A_{\alpha,\beta}|x-y|^{-n-|\alpha|-|\beta|} \ \ \ \ \ (37)

hence {K} satisfies,

\displaystyle \int_{|x-y|\geq 2\delta}|K(x,y)-K(x,\bar y)|dx\leq A, \ if\ |y-\bar y|\leq \delta, all \ \ \delta>0. \ \ \ \ \ (38)

Use the general singular integral theory we get the following {L^p} estimate.

Theorem 7 Suppose {T_a} is the pseudo-differential operator corresponding to a symbol {a} in {S^0}, then {T_a} extends to a bounded operator on {L^p({\mathbb R}^n)} to itself, for {1<p<\infty}.

5.2. Sobolev spaces

We first recall the definition of the Sobolev spaces {W_k^p}, where {k} is a positive integer. A function {f} belongs to {W_k^p({\mathbb R}^n)} if {f\in L^p({\mathbb R}^n)} and the partial derivatives {\partial_x^{\alpha}f}, taken in the sense of distribution, belong to {L^p({\mathbb R}^n)}, whenever {0\leq |\alpha|\leq k}. The norm in {W_k^p} is given by,

\displaystyle \|f\|_{W_k^p}=\sum_{|\alpha|\leq k}\|\partial_x^{\alpha}f\|_{L^p} \ \ \ \ \ (39)

the following result is the directly corollary of 7.

Theorem 8 Suppose {T_a} is a pseudo-differential operator whose symbol {a} belongs to {S^m}. If {m} is an integer and {k\geq m}, then {T_a} is a bounded mapping from {W_k^p} to {W_{k-m}^p}, whenever {1<p<\infty}.

Remark 12 This theorem remain valid for arbitrary real {k,m}.

5.3. Lipschitz spaces

Theorem 9 Suppose {a} is a symbol in {S^m}. Then the operator {T_a} is a bounded mapping from {\Lambda_{\gamma}} to {\Lambda_{\gamma-m}}, whenever {\gamma>m}.

Lemma 10 Suppose the symbol {a} belongs to {S^m}, and define {T_{a_j}=T_a\Delta_j}. Then, as operator from {\L^{\infty}({\mathbb R}^n)} to itself, the {T_{a_j}} have norms that satisfy

\displaystyle \|T_{a_j}\|\leq A2^{jm} \ \ \ \ \ (40)

We shall now point out a very simple but useful alternative characterization of {\Lambda_{\gamma}}. This is in terms of approximation by smooth functions; it is also closely connected with the definition of {\Lambda_{\alpha}} space as intermediate spaces, using the “real” method of interpolation.

Corollary 11 A function {f} belongs to {\Lambda_{\gamma}} if and only if there is a decomposition,

\displaystyle f=\sum_{j=0}^{\infty}f_j \ \ \ \ \ (41)

with {\|\partial_x^{\alpha}f_j\|_{L^infty}\leq A2^{-j\gamma}\cdot 2^{j|\alpha|}}, for all {0\leq |\alpha|\leq l}, where {l} is the smallest integer {>\gamma}.

When {f\in \Lambda_{/\gamma}}, the argument prove 10, with {T_a=I}, {f_j=F_j=\Delta_j(f)}, gives the required estimate for the {f_j}.

A second consequence of 9 is the following:

Corollary 12 The operator {(I-\Delta)^{\frac{m}{2}}} gives an isomorphism from {\Lambda_{\gamma}} to {\Lambda_{\gamma-m}}, whenever {\gamma>m}.

This is clear because {(I-\Delta)^{\frac{m}{2}}} is continuous from {\Lambda_{\gamma}} to {\Lambda_{\gamma-m}}, and its inverse, {(I-\Delta)^{\frac{-m}{2}}}, is continuous from {\Lambda_{\gamma-m}} to {\Lambda_{\gamma}}.

广告

Symplectic geometry

1. Introduction

This is the first note of a series of notes concert on semiclassical analysis. Given the basic material on symplectic geometry. Including the following material,

  1. The case at a point, or we can look it as the case in {{\mathbb R}^{2n}}.
  2. The standard material in symplectic geometry, i.e. Hamiltonian mechanics, two approach, global one concentrating on lie derivative, and a locally one concentrating on the power of Darboux theorem, i.e. the existence of a canonical coordinate.
  3. The basic facts on Poission bracket.
  4. The basic facts on Lagrange sub-manifold, and the involve of Liouville measure.

2. Case of a point, or {{\mathbb R}^{2n}}

Let {V: {\mathbb R}^n \rightarrow {\mathbb R}^n} be a vector field, at once we have a vector field, we could consider the associated flow of it,

\displaystyle \left\{ \begin{aligned} \overset{\cdot}\omega & = & V(\omega)\\ \omega(0) & =&z \end{aligned} \right. \ \ \ \ \ (1)

 

express the trajectory start from {z} along the vector field.

Remark 1 There {\overset{\cdot}\omega:=\frac{\partial \omega}{\partial t}}. One the other hand, due to the locally existence theorem of ODE, if the regularity of {V} is enough, then the solution exist and is uniqueness.

Definition 1 {\psi_t z=\omega(t,z)} or more convenient {\psi_t:= exp(tV)}. We call {\{\psi_t\}_{t\in {\mathbb R}}} the flow map or the exponential map generated by {V}.

Lemma 2 For flow map, we have following:

  1. {\psi_0z=z} for all {z\in {\mathbb R}^N}.
  2. {\psi_{t+s}=\phi_t\psi_s} for all {s,t\in {\mathbb R}}.
  3. for each time {t\in {\mathbb R}}, the mapping {\psi_t: {\mathbb R}^n\rightarrow {\mathbb R}^n} is a diffeomorphism with {(\psi_{t})^{-1}=\psi_{-t}}

 

So it is a group action on {{\mathbb R}^n}, with units as diffeomorphism of {{\mathbb R}^n}. Proof:

This lemma is the direct corollary of the theory of ODE. \Box

Now let us special to the case {{\mathbb R}^{2N}={\mathbb R}^n\times {\mathbb R}^n}. In local coordinate we have {z=(x,\xi)}, {x\in {\mathbb R}^n} express position of particle, {\xi \in {\mathbb R}^n} express momentum of particle.

Definition 3 {z=(x,\xi),w=(y,\eta )} in {{\mathbb R}^{2n}} define their symplectic product,

\displaystyle \sigma(z,w):=<\xi,y>-<x,\eta> \ \ \ \ \ (2)

In a matrix form, {\sigma} coincide with a {2n\times 2n} matrix

\displaystyle J=\begin{pmatrix} 0 & I \\ -I & 0\end{pmatrix} \ \ \ \ \ (3)

Following lemma given the basic property of {\sigma, J}.

Lemma 4 The following basic property are true.

  1. {\sigma(z,\omega)=<Jz,w>}, {\forall z,w \in {\mathbb R}^{2n}.}
  2. the bilinear form {\sigma} is antisymmetric, {\sigma(z,w)=-\sigma(w,z)} and degenerate, i.e. if {\sigma(z,w)=0} for all {w\in {\mathbb R}^{2n}}, then {z=0}.
  3. {J^2=-I}, {J^{T}=-J=J^{-1}}.

 

Proof:

  1. trivial calculate get this.
  2. trivial.
  3. {JJ^{-1}=-I}, by basic linear algebra everything follows.

\Box

3. Hamiltonian mechanics

Definition 5 Symplectic form: non-degenerate closed 2 form in a standard coordinate(Darboux coordinate, coordinate like {{\mathbb R}^{2n}}) looks like,

\displaystyle \begin{pmatrix} 0 & -I \\ I & 0 \end{pmatrix} \ \ \ \ \ (4)

{\forall x\in X}, map

\displaystyle T_xX\rightarrow T_x^*X

\displaystyle v\rightarrow w(\cdot, v)

is an isomorphism. {w} is called the symplectic form.

There is locally coordinate for {TM,T^*M}, i.e.,

\displaystyle TM:\frac{\partial}{\partial x_1},...,\frac{\partial}{\partial x_n}

\displaystyle T^*M:dx_1,...,dx_n

So {w=f^{ij}dx_i\wedge dx_j}, roughly we have {w(\frac{\partial}{\partial x_k},\frac{\partial}{\partial x_i})=f^{ik}-f^{ki}}, this is of course not true, but morally true. Now let us give the definition of symplectic manifold and the relationship of Hamiltonian mechanics.

Definition 6 We have the following definition,

  1. A symplectic manifold is a pair {(X,\omega)} where {X} is a smooth manifold and {w} is a closed two-form on {X} such that {\forall x\in X} the map,

    \displaystyle T_xX\rightarrow T_x^*X

    \displaystyle v\rightarrow w(\cdot,v)

    is an isomorphism, {w} is called the symplectic form.

  2. If {(X,w)} is symplectic, and {f:X\rightarrow {\mathbb R}} is differentiable, the hamiltonian vector field of {f} is the field {\Xi_f} on {X} whose image under the previous map is {df}. In other word, {\Xi_f} is characticed by the property,

    \displaystyle w(\cdot,\Xi_f)=df(\cdot)

  3. The flow of {\Xi_f} will be referenced to as the hamiltonian flow of {f}.

 

Lemma 7 If {X=R^{2n}} coordinate {(x_1,...,x_n,p_1,...,p_n)=(x,p)} and the symplectic form

\displaystyle w=\sum_{j=1}^ndp_j\wedge dx_j

then,

  1. If {f:{\mathbb R}^{2n}\rightarrow {\mathbb R}} is differentiable, then the integral curves of {\Xi_f} are the solutions to the system of ODEs,

    \displaystyle \left\{ \begin{aligned} \overset{\cdot}x_j & = & \frac{\partial f}{\partial p_j}\ \ \ j=1,2,...,n \\ \overset{\cdot}p_j & =& -\frac{\partial f}{\partial x_j} \ \ \ j=1,2,...,n \\ \end{aligned} \right. \ \ \ \ \ (5)

  2. Moreover, if

    \displaystyle f(x,p)=\frac{1}{2m}\|p\|^2+V(x). \ \ \ \ \ (6)

    Where {V} is a smooth solution (“potential”), and {(x(t),p(t))} is a trajectory of the Hamiltonian flow of {f}, then

    \displaystyle m\overset{\cdot\cdot}x=-\nabla \ \ \ \ \ (7)

    This is the Newton’s second law for the force {-\nabla V}.

Proof: {w=\sum_{j=1}^n dp_j\wedge dx_j}, due to we have {w(\cdot, \Xi_f=)=df}. So {\forall j} we have:

\displaystyle w(\frac{\partial}{\partial p_j},\Xi_f) = df(\frac{\partial}{\partial p_j})=\frac{\partial f}{\partial p_j} \ \ \ \ \ (8)

Assume {\Xi_f=\lambda^i \frac{\partial}{\partial x_i}+\gamma^i\frac{\partial}{\partial p_i}}. Then we have,

\displaystyle \begin{array}{rcl} \frac{\partial f}{\partial p_j} & = & w(\frac{\partial}{\partial p_j},\lambda^i\frac{\partial}{\partial x_i}+ \gamma^i\frac{\partial}{\partial p^i})\\ & = & \lambda^j\cdot (-1)^{\sigma(...)} \end{array}

and also,

\displaystyle \begin{array}{rcl} \frac{\partial f}{\partial x_j} & = & w(\frac{\partial}{\partial p_j},\lambda^i\frac{\partial}{\partial x_i}+ \gamma^i\frac{\partial}{\partial p^i})\\ & = & \gamma^j\cdot (-1)^{\sigma(...)} \end{array}

Combine with the definition of integral curve we derive the integral curves of {\Xi_f} i.e. {\gamma(t)} such that {\overset{\cdot}\gamma(t)=\Xi_f}, {\gamma(t)=(x_1(t),...,p_n(t))} is given by 5.

Now we begin to proof the Newton second law for the force {-\nabla V}. We consider the 2-dimensional case at first. We have,

\displaystyle \begin{array}{rcl} m\overset{\cdot\cdot}x & = & m(\overset{\cdot} {\frac{\partial f}{\partial p}})\\ & = & m\frac{\partial}{\partial t}(\frac{\partial f}{\partial p})\\ & = & m\frac{\partial}{\partial p}[\frac{\partial}{\partial t}(\frac{\|p\|^2}{m}+V(x))]\\ & = & m\frac{\partial }{\partial p}(\frac{2p\overset{\cdot}p}{m}+\frac{\partial V(x)}{\partial x}\cdot \frac{\partial f}{\partial p}) , due\ to \ -\frac{\partial f}{\partial x}=-\frac{\partial V(x)}{\partial x}.\\ & = & m\frac{\partial }{\partial p}(\frac{2p(-\frac{f}{\partial x})}{m}+\frac{\partial V(x)}{\partial x}\cdot \frac{\partial f}{\partial p})\\ & = &-\frac{\partial V(x)}{\partial x} \end{array}

The high dimension case is similar, thanks to the linearity of {V} and {\overset{\cdot\cdot}x}. \Box

Remark 2 Two make Newton’s second law to be true, the form 6 play a crucial role. Is there some generalization of this type of result to more general case, roughly speaking, it is reasonable to expect this could still be true if the hamiltonian function could be divide into potential energy part and kinetic energy part. And the describe of potential energy part is that it is given by a quadratic form.

Lemma 8 In general, for any Hamilton field {\Xi_f} one has:

  1. {\pounds_{\Xi_f}f=0}, conservation of energy. In orther word, {\Xi_f} is everywhere tangent to the level sets of {f}.
  2. {\pounds_{\Xi_f}\omega=0}, so the Hamiltonian flow of {f} consists of automorphism of {M,\omega}.

 

General speaking, to proof a theorem on manifold, there always have two choice, coordinate free proof and proof in a careful choose coordinate. If we choose to believe the Darboux theorem 9 is true, the meaning of it is that locally the symplectic manifold are the same.

Proof: If we believe the Darboux theorem 9 is true. then consider in a standard coordinate {(x_1,...,x_n,p_1,...,p_n)}, we have,

\displaystyle \Xi_f=\sum_{i}\frac{\partial f}{\partial x_i}\frac{\partial}{\partial p_i}-\sum_{i}\frac{\partial f}{\partial p_i}\frac{\partial}{\partial x_i} \ \ \ \ \ (9)

 

So of course {\pounds_{\Xi_f}f=\Xi f=0}. In general case, i.e. coordinate free proof, {\omega(\cdot,\Xi_f)=df}. Use identity if lie derivative. The second thing is also easy to proof by look in a local canonical coordinate, involve the indentity of lie derivative. \Box

Remark 3 I need more understanding on the lie derivative, see wiki.

Theorem 9 (Darboux theorem) Near any point there exist coordinate:{(x_1,...,x_n,p_1,...,p_n)} usually called Darboux coordinates, such that the sympletic form {w} has the form,

\displaystyle w=\sum_{j=1}^ndp_j \wedge dx_j. \ \ \ \ \ (10)

Remark 4 This theorem means there do not exist local invariant in symplectic manifold.

Proof: \Box

\newpage

Theorem 10 If {M} is any smooth manifold, then its cotangent bundle {X=T^*M} has a natural symplectic structure.

Proof: we have local coordinate on {T^M} derive from {(x_1,...,x_n,dx_1,...,dx_n)}, it is {(x_1,...,x_n,p_1,...,p_n)}. Remember we have Riemann metric: {g^{ij}dx_idx_j} on {T^*M\otimes T^*M}, the existence of Riemann metric involve a unit decomposition argument and bump function, I just recall it there. Now we move on, consider the relationship between {M,TM,T^*M}.

\displaystyle M{\longleftrightarrow} T^*M\overset{pairing, (X,f)=X(f)} {\longleftrightarrow} TM \ \ \ \ \ (11)

\Box

Remark 5 It need not be the case that {\alpha} non-degenerate {\Longrightarrow} {d\alpha} non-degenerate. This case in the lemma is a example to show that could be the case: {\alpha} degenerate {\Longrightarrow} {d\alpha} non-degenerate. We glue something together on the space pf differential operator to understand the topology of it but not deifferential structure or more refinement structure. Quntalization could be look as a way to glue, this could be down if there is a differential equation with some special condition (come from a flow take charge of it suffice).

Lemma 11 (The proof of {d\alpha} is non-degenerate) Let {(x_1,...,x_n)} be local coordinates on {U\subset M}. Define a coordinate system {(x_1,...,x_n,p_1,...,p_n)} on {T^*U} by the condition:

\displaystyle \forall \xi\in T^*_{x}U, p_j(\xi)=\xi(\frac{\partial}{\partial x_j}) \ \ \ \ \ (12)

Prove that in Darboux coordinate, {\alpha=\sum_{j=1}^n p_jdx_j} and therefore {\omega=\sum_{j=1}^n dp_j\wedge dx_j}.

Proof: \Box

Theorem 12 Let {(M,g)} be a smooth Riemann manifold and let {f:T^*M\rightarrow {\mathbb R}} be one half of the square of the Riemann norm, so that in local coordinate,

\displaystyle f(x,p)=\frac{1}{2}\sum_{i,j}g^{ij}p_ip_j \ \ \ \ \ (13)

 

Then the trajectorics of the hamiltonian flow of {f}, projected down to {M}, are geodesic aries in this fashion.

Newton’s second law+ energy vanish.

Proof:

\displaystyle m\overset{\cdot\cdot}x=\nabla V=0

So second variation formula describe of geodesic give us the fact that the trajective is geodesic. \Box

Remark 6 We could directly calculate in local coordinate.

4. Poisson brackets

{f} is the Hamiltonian generating the dynamic {g} is any smooth function on phase space (the symplectic manifold), then the rate of change of {g} along the trajectraries of d is the function

\displaystyle \overset{\cdot}g=\pounds_{\Xi_f}g=d_g(\Xi_f)=\omega(\Xi_f,\Xi_g) \ \ \ \ \ (14)

Definition 13 If {(X,\omega)} is symplectic and {f,g\in C^{\infty}(X)}, the poisson bracket of {f} and {g} is defined to be the function on {X}.

\displaystyle \{f,g\}:=\omega(\Xi_f,\Xi_g) \ \ \ \ \ (15)

Lemma 14 In canonical (Darboux) coordinate where {\omega=\sum_j dp_j\wedge dx_j}, one has,

\displaystyle \{f,g\}=\sum_{j=1}^n\frac{\partial f}{\partial p_j}\frac{\partial g}{\partial x_j}-\frac{\partial f}{\partial x_j}\frac{\partial g}{\partial p_j} \ \ \ \ \ (16)

In particular, {\{p_j,x_j\}=\delta_{ij}}.

Proof:

\displaystyle \begin{array}{rcl} \{f,g\} & = & \omega(\Xi_f,\Xi_g)\\ & = & \sum_{j=1}^n dp_j\wedge dx_j(-\frac{\partial f}{\partial p_j}\frac{\partial}{\partial x_j}+\frac{\partial f}{\partial x_j}\frac{\partial}{\partial p_j}, -\frac{\partial g}{\partial p_j}\frac{\partial}{\partial x_j}+\frac{\partial g}{\partial x_j}\frac{\partial}{\partial p_j} )\\ & = & \sum_{j=1}^n\frac{\partial f}{\partial p_j}\frac{\partial g}{\partial x_j}-\frac{\partial f}{\partial x_j}\frac{\partial g}{\partial p_j}. \end{array}

\Box

Theorem 15 If {(X,\omega)} is a symplectic manifold then {\{C^{\infty}(X),\{\cdot\}\}} is a Lie algebra.

Proof: Bilinearty, skew-symmetric come form,

\displaystyle \{f,g\}=\omega(\Xi_f,\Xi_g) \ \ \ \ \ (17)

Jacobi identity:

\displaystyle \{\{f,g\},h\}+\{\{g,h\},f\}+\{\{h,f\},g\}=0 \ \ \ \ \ (18)

 

could be proved by calculate under a local coordinate. \Box

 

Some interesting problems

There are some interesting problem, I post them at there in case I forget them. Excuse me if they are trivial, I have not took enough time to consider them about I think they are valuable to be consider.

Problem 1:

This problem is stated by graph coloring. there are two prat of it, in fact the first part I heard from someone else and I try to generate it to high dimension.

  1. there are finite lines \{l_i\}_{i\in I}, l_i\subset \mathbb R^2, crossing each other and the is a set J of crossing point. for technique reason, assume the position of lines are generic, i.e. no three of them intersect at one point. Then we could use 3 different colors to color  J make Neighbor points have different color. And to proof 3 is smallest.
  2. generate it to high dimension, to prove \mathbb R^n case, n+1 is the number.

This seems to be a graph problem, but the underlying structure is linear structure and some topological obstacle. I am not very sure. But it seems we can use an energy decrement argument with the obesevation:

The existence of a reasonable definition of “energy of correlation”.

the simplex arrive with the maximum of “correlation energy” in a very symmetric way, and this situation is easy to handle (coloring).

If make sense, this argument could also generate to high dimension.

Problem 2:

Let us consider some example of map between two metric space, a toy model is a line and two parallel lines, I called two parallel lines by X_1\cup X_2, the single line by X_3. The problem is try to find a tuple (d,f), where d is a metric define on X_1\cup X_2 and f: X_1\cup X_2\to X_3. such that the distortion of f^* d and the standard metric on X_3 arrive at a infimum, this of course could not be the case, such like the situation of Yamabe problem on manifold with conners. So, let us ask a more general problem, could we describe the behavior of f in some sense? what could we say with this kind of f?

A glimpse to the general theory

1. Introduction

We have talked about a very basic result in singular integral, i.e. if we have an additional condition, i.e. {q-q} bounded condition, then by interpolation theorem we only need to establish the weak {1-1} bound then we establish the {p-p} bound of {T}, {\forall 1< p< q }.

The category of of singular integral is very general, in fact the singular integral we interested in always equipped more special structure. We discuss following 3 types result which world be the central role in this further series note.

  1. Approximation of the identity.
  2. Singular integral with {L^2} bounded translation invariant operator.
  3. Maximal function, singular integral, and square functions.

The underlying object we consider in both the three case is some special singular integral, in the first case, it looks like a {T=sup_{t>0} \Phi_{t}*f}, this, among the other thing, has a close relationship with the maximal operator {Mf}. This is discussed in 2. For the singular integral with {L^2} bound, the Fourier transform or its discretization version, Fourier series is natural involved. And there is a “representation theorem” similar to the sprite of Reisz representation theorem, said, roughly speaking, if we consider the {L^2} bound operator adding the condition of transform invariant, then it is really coinside with the case of our image, the operator must behaviour as a Fourier multiple. This is the contant of famous Mikhlin multiplier theorem, and we discuss some technique difficulty in the process of establishing such a theorem, this is the contact of 3. At last we discuss some deep relationship between three basic underlying intution and objects in harmonica analysis, the Maximal function, singular integral, and square functions. They could all be understanding as tools to understanding the variant complicated emerging in singular integral. But there is definitely some common points. This is the theme of 4. Of course there are some further topic which are also interesting, but I do not want to discuss them here, maybe somewhere else.

2. Approximation of the identity

First topic, we discuss the approximation of the identity, this play a central role in understanding solution of PDE, why, I think a key point is this tools carry a lots of information about the scaling of the space, as it well known, analysis could roughly divide into two parts, “hard analysis” and “soft analysis”, approximation of the identity supply a way to transform a result form “hard analysis” side to “soft analysis” side and reverse. And when it shows its whole power always along with the involving of following Dominate convergence theorem:

Theorem 1 (DCT)

Let {\{f_n\}_{n=1}^{\infty}} be a series of function on measure space {(X,\Sigma,\mu)}, and {f_n \rightarrow f, a.e. x\in X}, and {\{f_n\}_{n=1}^{\infty}} satisfied a controlling condition, i.e. we can find a integrable function {g\in L^{1}(X)}, such that {|f_n(x)|\leq |g(x)|, a.e. x\in X, \forall n\in {\mathbb N}^*}, then we know,

\displaystyle \lim_{n\rightarrow \infty} \int_{X}f_n(x)d\mu\rightarrow \int f(x)d\mu \ \ \ \ \ (1)

 

In fact we have even stronger,

\displaystyle \lim_{n\rightarrow \infty} \int_{X}|f_n(x)-f(x)|d\mu=0 \ \ \ \ \ (2)

 

This is a standard theorem in real analysis, we give the proof.

Proof: {f} is the point-wise limit of {f_n} so we know f is measurable and also dominate by {g}, so by triangle inequality we have:

\displaystyle |f-f_n|\leq 2|g|

Then the 1 is trivially true, due to a diagonal taking subsequences trick. For more subtle result 2, we need use reverse Fatou theorem to show it is true, roughly speaking we have,

\displaystyle \limsup_{n\rightarrow \infty}\int_{X}|f_n-f|\leq \int_{X}\limsup_{n\rightarrow \infty}|f_n-f|=0

The key point is the first inequality above used the reverse Fatou theorem. \Box

Now we discuss of the main result of the approximation identity. So first we need to define what is a approximation identity. a key ingredient is scaling. i.e. we given a function {\Phi} and consider {\Phi_t=t^{-n}\Phi(\frac{x}{t})}, and we wish,

\displaystyle \lim_{t\rightarrow 0}(f*\Phi_t(x))=f(x), for a.e. x\ \in {\mathbb R}^n \ \ \ \ \ (3)

 

Whenever {f\in L^p, 1\leq p\leq \infty}, but there need some technique assume to make this intuition to be tight, this lead the following definition.

Definition 2 (Approximation of the identity) Suppose {\Phi} is a fixed function on {{\mathbb R}^n} that is appropriated small at infinity (have good enough decay rate), for example, take,

\displaystyle |\Phi(x)|\leq A(1+|x|)^{-n-\epsilon} \ \ \ \ \ (4)

 

Then we define {\{\Phi_t:\Phi_t(x)=t^{-n}\Phi(\frac{x}{t})\}} to be an approximation of the identity.

The key theorem is the following, related the approximation of the indentity with the maximal operator.

Theorem 3

\displaystyle \sup_{t>0}|(\Phi_t*f)(x)|\leq c_{\Phi}Mf(x) \ \ \ \ \ (5)

 

For heat kernel, the thing is more subtle.

Theorem 4 [Heat kernel estimate]

\displaystyle \|f-e^{t\Delta}f\|_2\leq \|\nabla f\|_2\sqrt{t} \ \ \ \ \ (6)

 

Remark 1 I know this theorem from Lieb’s book. The power of 4 combine with Plancherel theorem could use to establish the Sobolev inequality, at least for the index {p=2}.

There are 3 ingredients which cold be useful.

  1. the power of Rearrangement inequality involve in the Approximation of indentity operator. we could consider the relationship between {f*\Phi_t} and {f*\overline \Phi_t}, where {\overline \Phi} is constructed by take the average of {\Phi} on the level set but the foliation of scaling. Intuition seems some monotonic property natural emerge.
  2. There is a discretization model, i.e. the toy model on gragh, or we think it as correlation between particles, the key point is the rescaling deformation could be instead by semi group or renormalization property.
  3. We consider the more general case, now there is not only one {\Phi} but a group of them, i.e. {\Phi_k, k\in A}, this will involve some amenable theory I think.

We give two of the original and most important examples, First, if

\displaystyle \Phi(x)=c_n(1+|x|^2)^{\frac{-(n+1)}{2}}

where

\displaystyle c_n=\frac{\Gamma(\frac{n+1}{2})}{\pi^{\frac{n+1}{2}}}

then {\Phi_t(x)} is the possion kernel, and,

\displaystyle u(x,t)=(f*\Phi_t)(x)

Gives the solution of the Dirichlet problem for the upper half space,

\displaystyle {\mathbb R}^{n+1}_{+}=\{(x,t):x\in {\mathbb R}^n,t>0\}

Namely

\displaystyle \Delta u=(\frac{\partial^2}{\partial t^2}+ \sum_{j=1}^n\frac{\partial^2}{\partial x_j^2})u(x,t)=0,\ u(x,0)\equiv f(x) \ \ \ \ \ (7)

 

The second example is the Gaussian kernel,

\displaystyle \Phi(x)=(4\pi)^{-\frac{n}{2}}e^{-\frac{|x|^2}{4}}.

This time, if {u(x,t)=(f*\Phi_{t^{\frac{1}{2}}})(x)}, then {u} is a solution of the heat equation,

\displaystyle (\frac{\partial}{\partial t}- \sum_{j=1}^n\frac{\partial^2}{\partial x_j^2})u(x,t)=0,\ u(x,0)\equiv f(x) \ \ \ \ \ (8)

 

 

3. Singular integral with {L^2} bounded translation invariant operator

The main result proved in last note about singular integral is a conditional one, guaranteeing the boundedness on {L^p} for a range {1<p\leq q}, on the assupution that the boundedness on {L^q} is already known; the most important instance of this occurs when {q=2}. In keeping with this, we consider bounded linear transformation {T} from {L^2({\mathbb R}^n)} to itself that commute with translation. As is well known, such operator are characterized by the existence of a bounded function {m} on {{\mathbb R}^n} (the “multiper”), so that {T} can be realized as,

\displaystyle \widehat{Tf(\xi)}=m(\xi)\widehat f(\xi) \ \ \ \ \ (9)

Where {\widehat{}} denotes the Fourier transform. Alternatively, at least on test function {f\in S}, {T} can be realized in terms of convolution with a kernel {K},

\displaystyle Tf=f*K \ \ \ \ \ (10)

 

Where {K} is the distribution given by {\hat K=m}. We shall now examine how the theorem with condition on singular integral weill lead to some result of this type of operator. Roughly speaking, it is due to now we know the boundedness on {L^2}, for technique condition, we need to assume the distribution {K} agree away from the origin with a function that is locally integrable away from the origin with a function that is locally integrable away from the origin; in this case we define the function by {K(x)}. Then 10 implies that,

\displaystyle Tf(x)=\int K(x-y)f(y)dy,\ for \ a.e. x\notin supp f. \ \ \ \ \ (11)

Whenever {f} is in {L^2} and {f} has campact support. Tis is the representation of singular integral in the present context. Next, the crucial hormander condition is then equivalent with,

\displaystyle \int_{|x|\geq c|y|}|K(x-y)-K(x)|dx\leq A \ \ \ \ \ (12)

 

for all {y\neq 0}, where {c>1}. In this case, the condition 12 have a further understanding, in fact,

Lemma 5

\displaystyle |(\frac{\partial}{\partial x}^{\alpha}K(x))|\leq A_{\alpha}|x|^{-n-|\alpha|},\ for\ all \ \ \alpha \ \ \ \ \ (13)

 

or its weaker form, (here {\gamma>0} is fixed )

\displaystyle |K(x-y)-K()|\leq A\frac{|y|^{\gamma}}{|x|^{n+\gamma}}, whenever \ |x|\geq c|y|. \ \ \ \ \ (14)

imply the hormander condition 12

Proof: Integral by part. \Box

So, now the key point is how do {K}, satisfied such conditions, come about? It turns out that, toughly speaking, such condition on {K} have equivalent versions when sated in terms of the Fourier transform of {K}, namely the multiper {m}. This is transform the difficulties from physics space to fractional space In the future note, we will find a proof of the following Theorem:

Theorem 6 For {m=\hat K}.

If we assume that,

\displaystyle |(\frac{\partial}{\partial \xi}^{\alpha}m(\xi))|\leq A'_{\alpha}|\xi|^{-n-|\alpha|},\ for\ all \ \ \alpha \ \ \ \ \ (15)

holds for all {\alpha}, then {K} satisfied 5 for all {\alpha}.

If we assume that {m} satisfied the above inequality for all {0\leq |\alpha| \leq l}, where {l} is the smallest integer {>\frac{n}{2}}, then {K} satisfied 12

Remark 2 The multiplier {m} satisfied the second part condition of 6, are called Marcinkiewicz mulltiplier.

 

4. Maximal function, singular integral, and square functions.

 

Periodic orbits and Sturm–Liouville theory

I thinks there is some problem related to the solution of a 2 order differential equation given by Sturm-Liouville system which is nontrivial.

It is well-know that the power of Sturm-Liouville theory see  wiki, is due to it is some kind of “spectral decomposition” in the solution space.

Two kind of problem is interesting, one is the eigenvalue estimate, both upper bound and lower bound, this already investigated in ESTIMATING THE EIGENVALUES OF STURM-LIOUVILLE. PROBLEMS BY APPROXIMATING THE DIFFERENTIAL EQUATION.

I post two problem here, this is a product due to a random walk along the boundary of topology and the analysis,

Problem 1.

Fix a set A=\{k_1<k_2<...<k_l\}, is there a 2 order ordinary differential equation given by Sturm–Liouville theory  such that the eigenfunction f_{k} is periodic if and only if k\in A?

There is also some weak version of this and a infinity version of this.

Of course we have the following map, from the high order ordinary differential equation to the 1 order differential equation in high dimension. But the key point is that it is not a bijection! The Frobenius condition play a crucial role.

Problem 2.

There is a homotopy in the moduli space of differential equation, and we could define a direct product operator in this space, and we consider the topology defamation of the eigenfunction, could there be some equality, one side of it explain the topology information, the other side explain the spectral (or analysis) information?

There is another interesting problem.

Calderon-Zygmund theory of singular integrals.

1. Calderon-Zygmund decomposition

The Calderon-Zygmund decomposition is a key step in the real variable analysis of singular integrals. The idea behind this decomposition is that it is often useful to split an arbitrary integrable function into its “small” and “large” parts, and then use different technique to analyze each part.

The scheme is roughly as follows. Given a unction { f} and an altitude { \alpha}, we write { f=g+b}, where { |g|} is point wise bounded by a constant multiple {\alpha}. While { b} is large, it does enjoy two redeeming features: it is supported in a set of reasonable small measure, and its mean value is zero on each of the ball that constitute its support. To obtain the decomposition { f=g+b}, one might be tempted to “cut” { f} at the height { \alpha}; however, this is not what works. Instead, one bases the composition on the set where the maximal function of { f}has height { \alpha}.

Theorem 1 (Calderon-Zygmund decomposition)

Suppose we are given a function { f\in L^1} and a positive number { \alpha}, with {\alpha>\frac{1}{\mu(R^n)}\int_{R^n}|f|d\mu}. Then there exists a decomposition of { f}, {f=g+b}, with { b=\sum_{k}b_k}, and a sequences of balls {\{B_k^*\}}, so that,

  1. { |g(x)|\leq c\alpha}, for a.e. { x}.
  2. Each {latex b_k} is supported in {B_k^*},{ \int|b_k(x)|d\mu(x)\leq c\alpha\mu(B_k^*)}, and { \int b_k(x)d\mu(x)=0}.
  3. { \sum_k\mu(B_k^*)\leq \frac{c}{\alpha}\int|f(x)|d\mu(x)}.

 

Before proof this theorem, I explain the geometric intuition why this theorem could be true first. Merely speaking, this is just base on cut off the function into two part, the part with high altitude and the part with low altitude and extension the part with high altitude to make the extension one satisfied the condition 2 and 3.

Proof: In fact this decomposition have a good geometric explain, we just divide the part {\{x: |f(x)|>\alpha\}} and extension it carefully to make they behaviour like several balls, to satisfied the special condition on this part. \Box

Remark 1 Remark 1: A Calderon-Zygmund decomposition for {L^p} function was done in Charlie Fefferman’s thesis; see Section II of ams.org/mathscinet-getitem?mr=257819  One can also find this in Loukas Grafakos’s Classical Fourier Analysis Classical Fourier Analysis page 303 exercise 4.3.8. The question is broken up into parts that should be easy to handle.

Several people have considered with this question. An excellent paper that comes to mind is Anthony Carbery’s Variants of the Calderon–Zygmund theory for { L^p}-spaces which appeared in Revista Matematica Iberoamericana, Volume 2, Number 4 in 1986. There are also several useful references that appear in Carbery’s paper.

Remark 2 We could also consider a variant of Calderon-Zygmund decomposition, such as equipped with a nontrivial weight function { w} or find some different way to decomposition for some special purpose.

Remark 3 Consider suitable decomposition of the physics space or even both the physics space and fractional space try to gain some reasonable estimate is a fundamental philosophy in harmonic analysis, beside the Calderon-Zegmund decomposition,

Whitney decomposition. Which is important trick in the proof of fefferman-stein restriction theorem and differential topology.

Wave packet decomposition. The wave packet decomposition. This decomposition underlies the proof of Carleson’s theorem (this is more explicit in Fefferman’s proof than Carelson’s original proof), Lacey and Thiele’s proof of the boundedness of the bilinear Hilbert transform, as well as a host of follow-up work in multilinear harmonic analysis. The idea of the wave packet decomposition is to decompose a function/operator in terms of an overdetermined basis. This allows one to preserve symmetries (such as modulation symmetries) that aren’t preserved by a classical Calderon-Zygmund decomposition (which endows the frequency with a distinguished role). One might consider using a wave packet decomposition if is working with an operator that has a modulation symmetry. This is discussed in more detailed in Tao’s blog post on the trilinear Hilbert transform.

Polynomial decomposition. The application of polynomial decomposition to harmonic analysis is more recent, and its full potential still seems unclear. Applications include Dvir’s proof of the finite field Kakeya conjecture, Guth’s proof of the endpoint multilinear Kakeya conjecture (and, indirectly, the Bourgain-Guth restriction theorems), Katz and Guth’s proof of the joints problem and Erdos distance problem, among many other results. Generally, the idea behind the polynomial decomposition is to partition a subset of a vector space over a field into a finite number of cells each of which contains roughly the same fraction of the original set. One further wishes that no low degree algebraic variety can intersect too many of the cells. In Euclidean space, the polynomial ham sandwich decomposition does exactly this. This allows one to, for instance, control linear (or, more generally, `low algebraic degree’) interactions between points in distinct cells. This has so far proven the most useful in incidence-type problems, but many problems in harmonic analysis, thanks to the translation symmetry of the Fourier transform, are inextricably linked with such incidence-type problems. See (again) Tao’s survey of this topic for a more detailed account.

 

2. Singular integrals

Have the Calderon-Zegmund decomposition in hand, now we proof a conditional one bounded result for singular integrals.

The singular integral one is interested in are operator { T}, expressible in the form

\displaystyle (Tf)(x)=\int_{R^n}K(x,y)f(y)d\mu(y) \ \ \ \ \ (1)

 

Where the kernel { K} is singular near { x=y}, and so the expression is meaningful only if { K} is treated as a distribution or in some limiting sense. Now the particular regularization of { (Tf)(x)} may be appropriate depends much on the context, and a complete treatment of the issues thereby raised take us quite far afield.

Let us limit ourselves to two closely related ways of dealing with the questions concerning the definability of the operator. One is to prove estimates for the (dense) subspace where the operator is initially defined. The other is to regularize the given operators by replacing it with a suitable family, and to prove the uniformly estimates for this family. This idea is similar occurring in spectral geometry when we wish to investigate the spectrum of some operator we try to consider some deformation, so deduce to control the spectrum of a seres of paramatrix, for example, consider the wave kernel or heat kernel rather than the passion kernel itself. Common to both methods is a priori approach: We assume some additional properties of the kernel, but then prove estimates that are independent of these “regularity” properties.

We now carry out the first approach in detail. There will be two kinds of assumptions made about the operator. The first is quantitative: we assume that we are given a bound { A}, so that the operator { T} is defined and bounded on { L^q} with norm { A}; that is,

\displaystyle \|T(f)\|_q\leq A\|f\|_q, \forall f, f\in L^q \ \ \ \ \ (2)

 

Moreover, we assume that there is associated to { T} a measurable function { K} (that plays the role of its kernel), so that for the same constant { A} and some constant { c>1},

\displaystyle \int_{R^n-B(y,c\delta)}|K(x,y)-K(x,\bar y)|d\mu(x)\leq A, \forall \bar y\in B(y,\delta)  \ \ \ \ \ (3)

 

for all { y\in R^n, \delta>0}.

The further regularity assumption on the kernel { K} is that for each { f} in {L^q} that has compact surppot, the integral coverages absolutely for almost all { x } in the complement of the support of { f}, and that equality holds for these { x}.

Theorem 2 (Bounded of singular integral with condition)

Under the condition 1 and 3 made above on { K}, the operator { T} is bounded in { L^p} norm on { L^p\cap L^q}, when { 1<p<q}. More precisely,

\displaystyle \|T(f)\|_p\leq A_p\|f\|_p

For { f\in L^p\cap L^q} with { 1<p<q}, where the bound { A_p} depends only on the constant { A} appearing in 1 and 3 and on { p}, but not on the assumed regularity of { K}, or on { f}.

 

Proof:

Now let us begin to prove the conditional theorem. The key point is to use the potential of {T} has been a bounded operator from {L^q\rightarrow L^q}. Said, it already assumed {\exists A>0} such that {\forall f\in L^q} we have {\|T(f)\|_q\leq A\|f\|_q}. Now let us look at the singular integral expression:

\displaystyle (Tf)(x)=\int_{R^n}K(x,y)f(y)d\mu(y). \ \ \ \ \ (4)

 

 

The key point is to proof the mapping {f\rightarrow T(f)} is a weak-type {1-1}; that is,

\displaystyle \mu\{x:|Tf(x)|>\alpha\}\leq \frac{A'}{\alpha}\int |f|d\mu. \ \ \ \ \ (5)

 

At once we establish 5, then the theorem followed by interpolation. Now we use theorem 1 on {f} get {f=g+b}, thanks to the triangle inequality and something similar we have {g,b \in L^q}, in fact {R^n= A\amalg B, B\cup_{k}B_k}, {g=\chi_A g+\chi_{B}g, b=\chi_A b+\chi_{B} b}, by triangle inequality and {f=g+b}, to proof {g,b \in L^q}, we only need to proof {\chi_A g, \chi_B g, \chi_A b, \chi_B b\in L^q}, but this is easy to proof.

Now we know the {L^q} bounded of {g,b}, we divide the difficult of establish the weak 1-1 bound of {f} into the difficult of establish the weak 1-1 bound for {g} and {b}. i.e.

\displaystyle \mu\{x:|Tf(x)|>\alpha\}\leq \mu \{x:|Tg(x)|>\alpha\} +\mu\{x:|Tb(x)|>\alpha\} \ \ \ \ \ (6)

 

For {g}, if this weak 1-1 bound is not true, we have,

\displaystyle \mu \{x:|Tg(x)|>\alpha\}\geq \frac{A'}{\alpha}\int |g|d\mu \ \ \ \ \ (7)

 

thanks to the trivial estimate {\|g\|_q \leq c\alpha^{q-1}\|g\|_1 }. combine this two estimate we have:

\displaystyle c\alpha^{q-1}\|g\|_1\geq \|g\|^q_q\geq c\|Tg\|^q_q \geq A'\alpha^{q-1} \|g\|_1 \ \ \ \ \ (8)

 

The first estimate is true on {A} due to {|g|\leq \alpha, a.e. x\in R^n}. But compare the left and the right of 8 lead a contradiction, so 7 follows. For {b}, the thing is more complicated and in fact really involve the structure of the convolution type of the singular integral. The key point is controlling near the diagonal of {K(x,y)}. we warm up with a more refine decomposition {b=\sum b_k}, {\forall k, b_k=b\cdot \chi_{B_k}}. For a large constant {c>>1} choose later define {B^*_k=c B_k}. We know {b\in L^q}, but the really difficult thing occur in the how to combine the following 5 condition to lead a contradiction:

  1. {\|Tb\|_q\leq \|b\|_q}.
  2. property come from the Calderon-Zegmund decomposition, {\int_{B_k}\|b\|\leq c\alpha \mu(B_k),\forall k} and {\int_{B_k}b=0}.
  3. Hormander condition 3 , {\int_{R^n-B(y,c\delta)}|K(x,y)-K(x,\bar y)|d\mu(x)\leq A, \forall \bar y\in B(y,\delta)}
  4. the reverse of weak 1-1 of {b}, {\mu\{x:b(x)>\alpha\}> \frac{A'}{\alpha}\|b\|_1}.
  5. the structure {Tb(x)=\int_{R^n} K(x,y)b(y)dy}

The first step is to break {b} into {b_k}, and reduce the case of several balls to the case of only one ball, this could be done by triangle inequality or more may be we could do it derectly, but any way it is not difficult.

Then the thing become intersting, we focus on {b_1}, divide {Tb_1=T\chi_{B_1} b_1+ T\chi_{{\mathbb R}^n-B_1}b}. thanks to the hormander condition 3 we have good control on {T\chi_{{\mathbb R}^n-B_1^*}}, in fact we can proof a weak 1-1 bound on it,

\displaystyle \mu\{x:|T_{{\mathbb R}^n-B_1^*}b_1|>\alpha\}< \frac{A'}{\alpha}\|b_1\|_1 \ \ \ \ \ (9)

\displaystyle \begin{array}{rcl} T_{{\mathbb R}^n-B_1^*}b_1(x) & = & \int_{{\mathbb R}^n-B_1^*}K(x,y)b_1(y)dy\\ & = & \int_{{\mathbb R}^n-B_1^*}[K(x,y)-K(x,\bar y)]b_1(y)dy+\int_{{\mathbb R}^n-B_1^*}K(x,\bar y)b_1(y)dy\\ & \leq & \int_{{\mathbb R}^n-B_1^*}Ab_1(y)dy+\int_{{\mathbb R}^n-B_1^*}K(x,\bar y)b_1(y)dy \end{array}

So we conclude,

\displaystyle \begin{array}{rcl} \|T_{{\mathbb R}^n-B_1^*}b_1\|_1 & = & \int_{{\mathbb R}^n}|\int_{{\mathbb R}^n-B_1^*}K(x,y)b_1(y)dy|dx\\ & = & \int_{{\mathbb R}^n}\int_{{\mathbb R}^n-B_1^*}|[K(x,y)-K(x,\bar y)]b_1(y)dy|dx+\int_{{\mathbb R}^n}|\int_{{\mathbb R}^n-B_1^*}K(x,\bar y)b_1(y)dy|dx\\ & \leq & A\int_{{\mathbb R}^n}b_1(y)dy+\int_{{\mathbb R}^n-B_1^*}K(x,\bar y)b_1(y)dy=A\int_{{\mathbb R}^n}b_1(y)dy \end{array}

The last equality used the condition {\int b_1=0}.

\Box

 

The large sieve and the Bombieri-Vinogradov theorem

-1.Motivation-

Large sieve a philosophy reflect as a large group of inequalities which is very effective on controlling some linear sum or square sum of some correlation of arithmetic function, some idea of which could have originated in harmonic analysis, merely rely on almost orthogonality.

One fundamental example is the estimate of the quality,

\sum_{n\leq x}|\Lambda(n)\overline{\chi(n)}|

One naive idea of control this quality is using Cauchy-schwarz inequality. But stupid use this we gain something even worse than trivial estimate. In fact by triangle inequality and trivial estimate we gain trivial bound: \sum_{n\leq x}|\Lambda(n)\overline{\chi(n)}|\leq x. But by stupid use Cauchy we get following,

\sum_{n\leq x}|\Lambda(n)\overline{\chi(n)}|\leq ((\sum_{n\leq x}|\Lambda(n)|^2)(\sum_{n\leq x}|\chi(n)|^2))^{\frac{1}{2}}\leq xlog^{\frac{1}{2}}x

But this does not mean Cauchy-Schwarz is useless on charge this quality, we careful look at the inequality and try to understand why the bound will be even worse. Every time we successful use Cauchy-Schwarz there are two main phenomenon, first, we lower down the complexity of the quantity we wish to bound, second we almost do not loss any thing at all. So we just reformulate the quantity and find it lower down the complexity and the change is compatible with the equivalent condition of Cauchy-Schwarz. For example we have following identity,

\sum_{n\leq x}|\Lambda(n)\overline{\chi(n)}|=\sqrt{ \sum_{n\leq x}|\Lambda(n)\overline{\chi(n)}| \sum_{m\leq x}|\Lambda(m)\overline{\chi(m)}|}=\sqrt{ \sum_{k_1,k_2\in \mathbb F_p^{\times}}\sum_{n',m'\leq \frac{x}{p}}|\Lambda(n')\Lambda(m')\overline{\chi(k_1)\chi(k_2)}| }

So we could understand this quality as the Variation of primes in arithmetic profession constructed by \{pn+b| b\in\{1,2,...,p-1\}\}. But this is still difficult to estimate, merely because of we need to control the variation of convolution of \Lambda with itself on \mathbb F_p^{\times}\simeq \{pn+b| b\in\{1,2,...,p-1\}\}.

Now we change our perspective, recall a variant of Cauchy-Schwarz inequality, which called Bessel inequality, as following,

Bessel inequality

Let {g_1,\dots,g_J: {\bf N} \rightarrow {\bf C}} be finitely supported functions obeying the orthonormality relationship,

\displaystyle \sum_n g_j(n) \overline{g_{j'}(n)} = 1_{j=j'}

for all {1 \leq j,j' \leq J}. Then for any function {f: {\bf N} \rightarrow {\bf C}}, we have,

\displaystyle (\sum_{j=1}^J |\sum_{n} f(n) \overline{g_j(n)}|^2)^{1/2} \leq (\sum_n |f(n)|^2)^{1/2}.

Pf: The proof is not very difficult, we just need to keep an orthogonal picture in our mind, consider \{g_{j}(n)\}, 1\leq j\leq J to be a orthogonal basis on l^2(\mathbb N), then this inequality is a natural corollary.

Have this inequality in mind, by the standard argument given by transform from version of orthogonal to almost orthogonal which was merely explained in the previous note.  We could image the following corresponding almost orthogonal variate of “Bessel inequality” is true:

Generalised Bessel inequality

Let {g_1,\dots,g_J: {\bf N} \rightarrow {\bf C}} be finitely supported functions, and let {\nu: {\bf N} \rightarrow {\bf R}^+} be a non-negative function. Let {f: {\bf N} \rightarrow {\bf C}} be such that {f} vanishes whenever {\nu} vanishes, we have

\displaystyle (\sum_{j=1}^J |\sum_{n} f(n) \overline{g_j(n)}|^2)^{1/2} \leq (\sum_n |f(n)|^2 / \nu(n))^{1/2} \times ( \sum_{j=1}^J \sum_{j'=1}^J c_j \overline{c_{j'}} \sum_n \nu(n) g_j(n) \overline{g_{j'}(n)} )^{1/2}

for some sequence {c_1,\dots,c_J} of complex numbers with {\sum_{j=1}^J |c_j|^2 = 1}, with the convention that {|f(n)|^2/\nu(n)} vanishes whenever {f(n), \nu(n)} both vanish.

 

Linear metric on F2, free group with two generator.

img_0515.jpg

I may have made a stupid mistake, but if not, we could construct a metric by pullback a metric on a suitable linear normalized space H which we carefully constructed. Let we define the generators of free group F_2 by a,b.

Step 1.

Constructed the linear normalized space H. the space H was spanned by basis \Lambda=\Lambda_a \coprod \Lambda_b, \Lambda_a, \Lambda_b are defined by look at the Cayley graph of F_2, there is a lot of vertical vector and horizontal vector in the Cayley graph, for every level set of vertical vector we put a basis in \Lambda_a, because there is only countable many vertical vectors (for example, a,a^2,a^{-5} are in the same vertical level, bab^{-1}, ba^{10}b^{-1} are in the same vertical level, bab^{-1},a are not in the same vertical level), we put a basis in \Lambda_a for every vertical level and claim we accomplished the construct of \Lambda_a, we do the same operation for \Lambda_b but only change the vertical level with horizontal level. Now we accomplished the construction of \Lambda, We spanned this with coefficient \mathbb Z and we get a linear space V. by Zorn’s lemma there exists a norm on the space, take one norm \|\cdot\| we accomplished the construction of H=(V,\|\cdot\|).

Step 2:

Pullback the norm \|\cdot\| on H to the free group F_2. In fact there is a natural bijection T: F_2\to H, which is given by following: On the Cayley graph (imaged it is embedding in \mathbb R^2), identity 1 in the group F_2 corresponding to the original, and more general every element in F_2 exactly identify with a point in the Cayley graph, thanks to there is no relation between a,b. And then there is of course infinity many of path from original to the point, but there is only one shortest path , thanks to there is no loop in the Cayley graph. We identify the elements in F_2 with the point in Cayley graph with the shortest path. Now we could explain why the path lies H. This path only across to finite vertical level and horizontal level and on every level it only pass finite step, this already given a representation \sum_{e_i\in \Lambda}c_i\cdot e_i, c_i\in \mathbb Z, the key point is there is only finite c_i\neq 0. So we have defined the bijection T:F_2\to H, and we could use the bijection to pullback the norm on H to a norm on F_2.

Step 3:

Now we begin to proof the norm we get by pullback satisfied the condition we need. We need only to proof the condition of linear growth and triangle inequality. The conjugation invariance is automatically by linear growth by the comments of Tobias Fritz. The triangle inequality is automatically, due to the bijection T stay the structure in fact, the multiplier of elements x_1,x_2 \in_2 could be view as put the two path together but this  is not true… merely because of the addition operation is not commutative.

The space we should consider is the path space equipped with the composition operation. I image there exists a “big space” such that the natural metric on the “big space” restrict on the embedding image of \mathbb F_2 is a linear growth metric.

Almost orthogonality

 

Motivation and Cotlar’s lemma

We always need to consider a transform T on Hilbert space l^2(\mathbb Z) (this is a discrete model), or a finite dimensional space V. If under a basis T is given by a diagonal matrix this story is easy,

\displaystyle A = \begin{pmatrix} \Lambda_1 & 0 & \ldots & 0 \\ 0 & \Lambda_2 & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & \Lambda_n \end{pmatrix} \ \ \ \ \ (5)

Then ||T||=\max_{i}\lambda_i.

In fact, for T is a transform of a finite dimensional space, T is given by (a_{ij})_{n\times n} by duality we have ||T||=||TT^*||, so we have,

||T||=||TT^*||=|(\sum a_{ij}x_j)y_i|\leq |\sum_{i,j}\frac{1}{2}(|a_{ij}(|x_i|^2+|y_j|^2)|\leq M

If we have given \sum_{i}|a_{ij}|\leq M and \sum_{j}|a_{ij}|\leq M \forall i,j\in \{1,2,...,n\}.

But in application of this idea, the orthogonal condition always seems to be too restricted and due too this we have the following lemma which is follow the idea but change the orthogonal condition by almost orthogonal.

Lemma(Catlar-Stein)

Let \{T_j\}_{j=1}^N be finitely many operators on some Hilbert space H. Such that for some function \gamma : \mathbb Z\to R^+ one has,

||T_j^*T_k||\leq \gamma^2(j-k),||T_jT_k^*||\leq \gamma^2(j-k)

for any 1\leq j,k\leq N. Let \sum_{l=-\infty}^{\infty}\gamma(l)=A<\infty. then ,

||\sum_{j=1}^NT_j||\leq A

Pf:

tensor power trick + duality ||T||=||TT^*||^{\frac{1}{2}}.

Singular integrals on L^2

 

Lemma(Schur)

Define T is a operator on measure space X\times Y equipped positive product measure \mu\wedge \nu, via,

(Tf)(x)=\int_YK(x,y)f(y)\nu(dy)

K is a measurable kernel, then,

1). ||T||_{1\to 1}\leq \sup_{y\in Y}\int_{X}|K(x,y)|\mu(dx)=:A.

2). ||T||_{\infty\to \infty}\leq \sup_{x\in X}\int_{Y}|K(x,y)|\nu(dy)=:B.

3). ||T||_{p\to p}\leq A^{\frac{1}{p}}B^{\frac{1}{p'}},  \forall 1\leq p\leq \infty.

4). ||T||_{1\to \infty}\leq ||K||_{L^{\infty}(X\times Y)}.

Pf:

1),2),4) merely due to Fubini theorem and Bath lemma.

 

3) proof by the interpolation and combine 1) and 2).

Theorem

Let K be a Calderon-Zegmund operator, with the additional assumption

that |\nabla K(x)|\leq B|x|^{-d-1}. Then

||T||_{2\to 2} \leq CB

with C = C(d).

Caldero ́n–Vaillancourt theorem

 

Hardy’s inequality

Theorem(Hardy inequality)

For any 0 \leq s < \frac{d}{2} there is a constant C(s, d) with the prop-

arty that,

|||x|^{-s} f||_2 \leq C(s,d)||f||_{H^s(R^d)}

for all f \in H^s(R^d).

 

The correlation of Mobius function and nil-sequences in short interval

I wish to establish the following estimate:

Conjecture :(correlation of Mobius function and nil-sequences in short interval)

\lambda(n) is the liouville function we wish the following estimate is true.

\int_{0\leq x\leq X}|\sup_{f\in \Omega^m}\sum_{x\leq n\leq x+H}\lambda(n)e^{2\pi if(x)}|dx =o(XH).

Where we have H\to \infty as x\to \infty, \Omega^m=\{a_mx^m+a_{m-1}x^{m-1}+...+a_1x+a_0 | a_m,...,a_1,a_0\in [0,1]\} is a compact space.

I do not know how to prove this but this is result is valuable to consider, because by a Fourier identity we could transform the difficulty of (log average) Chowla conjecture to this type of result.

There is some clue to show this type of result could be true, the first one is the result established by Matomaki and Raziwill in 2015:

Theorem (multiplication function in short interval)

f(n): \mathbb N\to \mathbb C is a multiplicative function, i.e. f(mn)=f(n)f(m), \forall m,n\in \mathbb N. H\to \infty as x\to infty, then we have the following result,

\int_{1\leq x\leq X}|\sum_{x\leq n\leq x+H}f(n)|=o(XH).

And there also exists the result which could be established by Vinagrodov estimate and B-S-Z critation :

Theorem(correlation of multiplication function and nil-sequences in long interval)

f(n): \mathbb N\to \mathbb C is a multiplicative function, i.e. f(mn)=f(n)f(m), \forall m,n\in \mathbb N. g(n)=a_n^m+...+a_1n+a_0 is a polynomial function then we have the following result,

\int_{1\leq n \leq X}|f(n)e^{2\pi i g(n)}|=o(X).