Special Topic: Elliptic Operators

Introduction¶

We discuss how to use Fourier series to prove the regularity of Elliptic operators, following this excellent note https://math.mit.edu/~vwg/classnotes-spring05.pdf

We use a different normalization convention than other notes for Fourier transform. Recall that by derivative theorem we have

We adopt two normalization conventions to simplify the notations:

• We define the Fourier transform to be $\mathcal Ff(\xi) = \int_{\mathbb{R}^n} f(x)e^{-ix\xi} dm(x)$, here we write $x\xi$ instead of $\langle x,\xi\rangle$ for simplicity. Consequently, let $T$ be a $2\pi$-

High-dimensional Fourier series¶

In our study of Fourier series we can summerize the theory development as follows:

• For every smooth periodic function $f$ on $\mathbb{R}$ with period $2\pi$, the Fourier transform $\mathcal Ff = \sum_{k\in \mathbb{Z}} \hat{f}(k)\delta_k$, where $\hat{f}(k) = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-ikx}$.
• The sequence $\hat f(k)$ satisfies the rapidly decreasing condition: For every $r$, there exists constant $C_r$ such that $|\hat{f}(k)|\leq C_r |k|^{-r}$. We also call such a sequence smooth.
• The inversion theorem: $f(x) = \mathcal F^{-1}\mathcal Ff(x) = \mathcal F^{-1}\sum_{k\in \mathbb{Z}}\hat f(k)\delta_k = \sum_{k\in \mathbb{Z}}\hat f(k)\mathcal F^{-1}\delta_k = \sum_{k\in \mathbb{Z}}\hat f(k)e^{ikx}$.

The theory generalizes to $n$-dimensional case as well. We restrict our discussion to smooth case.

Differential operators¶

A differential operator of order $r$ is an opeartor $P:\mathscr{C}^{\infty}(T) \to \mathscr{C}^{\infty}(T)$ of the form $Pu = \sum_{|\alpha|\leq r} a_\alpha (x) D^{\alpha}u$.

Main Theorem¶

Let $P$ be an elliptic operator on $T^n$. Then $P$ is right-invertible module smoothing operator, $PQ=I+\text{smoothing}$. Precisely, there exists an operator $Q:\mathscr{C}^{\infty}(X)\to \mathscr{C}^{\infty}(X)$ and a smoothing operator $T_K$ such that $PQ = I-T_K$.

We know that after Fourier transform, a differential operator becomes an algebraic object. To construct such an operator, we take Fourier transform, do some algebra on the “frequency domain” and construct a corresponding algebraic object of $Q$. Then we are done by inversion. The algebraic object corresponding to differential operators in the frequency domain are symbols.

Some Calculus of Symbols¶

Motivation¶

Lemma¶

$\square$

Let $P = \sum_{|\alpha|\leq m} a_\alpha(x)D^{\alpha}$ be a differential operator. Let $f$ be a smooth function on $T^n$, then we identify $f$ with $\mathcal F^{-1}\mathcal F f = \sum_{k\in \mathbb{Z}^n}\hat f(k) e^{-ikx}$. Then

Let $p(x,\xi) = \sum_{|\alpha|\leq m}a_\alpha(x)\xi^{\alpha}$ be the polynomial in RHS, we can write $Pf = \sum_{k\in \mathbb{Z}^n}a(x,k)\hat{f}(k)e^{ikx}$. Therefore, via Fourier transform, we can identify $P$ with its polynomial $p(x,\xi)$.

Conversely, with a polynomial $\sum_{|\alpha|\leq m}a_\alpha(x)\xi^{\alpha}$ we can define a differential operator by replacing $\xi^\alpha$ by $D^\alpha$.

Observe that for a general function $a(x,\xi)$, not necessarily a polynomial, we can still define an operation $T_a f = \sum_{k\in \mathbb{Z}^n}a(x,k)\hat{f}(k)e^{ikx}$ for $f\in \mathscr{C}^{\infty}(T^n)$, and $T_a f$ will be $\mathscr{C}^{\infty}(T)$ when the ξ part satisfy sufficient decreasing condition. Such a function $a(x,\xi)$ is called a symbol, and the corresponding $T_a$ is called a pseudo differential operator.

Symbols and pseudodifferential operators¶

Definition¶

A function $a:T^n \times \mathbb R^n\to \mathbb{C}$ is a symbol of degree $m$, which we denote by $S^m$, provided that for all multiindices $\alpha,\beta$, $|D_x^\alpha D_\xi^\beta a(x,\xi)| \leq C_{\alpha,\beta}\langle\xi \rangle^{m - \beta|}$.

• We use $\langle \xi \rangle:=\sqrt{1+|\xi|^2}$ instead of $|\xi|$ to measure the growth condition to avoid the blowing up issue of $|\xi|^{\text{negative}}$.

Example¶

• $p(x,\xi) = \sum_{|\alpha|\leq m} a_\alpha(x)\xi^\alpha$ is in $S^m$.
• $(\sqrt(1+|\xi|^2))^m$ is in $S^m$, which is not a polynomial.

Note that $m$ can take nevative value. Acutally this is the main interesting case. If $a\in S^{m}$ for positive $m$, then $D^\alpha_\xi a \in S^{0}$ for $|\alpha|$ big enough. Also, $(S^m)_{m\in \mathbb{Z}}$ is a decreasing family. $S^m\supset S^{m-1}\supset ....$. Let $S^{-\infty}$ be the decreasing limit $\cap_{m\in \mathbb{Z}}S^{m}$.

$\square$

Lemma¶

• $a\in S^l, b\in S^m \implies ab\in S^{m+l}$.
• $a\in S^m \implies D_x^{\alpha}D_{\xi}^\beta a \in S^{m - |\beta|}$. Taking derivatives decrease growth condition. Note that as stated, taking derivative on $x$ is irrevalent.

$\square$

Definition¶

A pseudodifferential operator is an operator $T:\mathscr{C}^{\infty}(T^n)\to \mathscr{C}^{\infty}(T^n)$ given by some $a\in S^m$ for some $m$, so that $Tf = \sum_{k\in \mathbb{Z}}\hat{f}(k)a(x,k)e^{ikx}$. We call $a$ the symbol of $T$. We usually let $T_a$ to denote a pseudodifferential operator with symbol $a$ specified.

$\square$

It follows that a differential operator is a pseudodifferential operator with positive degree. But how about negative degree? Consider the example $a(x,k) = k^{-2}$, so $\widehat{T_af}(k) = k^{-2}\hat{f}(k)$, so $\hat f(k) = \widehat{D^2T_af}(k)$ by derivative theorem. This indicates that if $a$ is negative degree, then $T_a$ will make the function more regular. In particular, if $a\in S^{-\infty}$, then $T_a$ will be a “smoothing operator”. We’ll make these intuitions precise later. Now let’s study a first algebraic property of pseudodifferential operators.

Composition of pseudodifferential operators¶

The idea to construct the right inverse $Q$ is to find out the symbol of $Q$. We already know the symbol of $P$ is $p(x,\xi)$, so the first question is to understand what is $PQ$ in the symbol world.

Lemma¶

Let $P$ be a differential operator, $T_a$ be a pseudodifferential operator for $a\in S^r$. Then $PT_a$ is a pseudodifferential operator with symbol

To find the symbol of $PT_a$ we need to find its action on the Fourier coefficients of a function. For $f\in \mathscr{C}^{\infty}(T^n)$, we have

It follows that by definition, the symbol of $PT_a$ is $e^{-i\xi x}\cdot P[a(x,\xi)e^{-i\xi x}]$. It sufficies to show that the latter is given by equation $(cmp)$ above.

We make a more general computation which will be used later. Consider $P(e^{itf}u)$ for smooth $f$ and $u$. First we look at $D_j$, we define $\delta_j$ such that $D_j(e^{itf}u) = e^{itf}\delta_j u$. This allows us to simply switch the $e^{itf}$ out to avoid extra term. It’s straightforward to find that $\delta_j u = D_j u + t \frac{\partial f}{\partial x_j} \cdot u$, and we just write $\delta_j = D_j + t \frac{\partial f}{\partial x_j}$. Similarly, $D_jD_r(e^{itf}u) = D_j(e^{itf}\delta_r u) = e^{itf}\delta_j\delta_r u$. It follows that

Now we apply the formula to the case $f(x) = \xi x = \xi_1x_1+\cdots+\xi_n x_n$ and $t = 1$ we get

Recall that $p$ is a polynomial function of ξ, so (totally algebraic) global Taylor expansion givecs $p(x,\eta+\xi) = \sum_{|\beta|\leq m} \frac{1}{\beta_1!\cdots \beta_n!} \partial_\xi^{\beta}p(x,\xi) \eta^{\beta}$ (as polynomial), so note that the $D$ is taking on $x$ variable,

This proves the symbol of $P\circ T_a$ is $\sum_{|\beta|\leq m} \frac{1}{\beta!}\partial_\xi^{\beta} p(x,\xi)D^{\beta}_x a(x,\xi).$

Regularity of pseudodifferential operator¶

Let’s think about pseudodifferential operators again. The action of $T_a$ is given by “modifying Fourier coefficients” by multiplication, we know that such operation is essentially convolution, by convolution theorem. In particular, if $a$ doesn’t depend on $x$, then $T_a$ is precisely convolution with the inverse Fourier transform of $(a(k))_{k\in \mathbb{Z}}$, in other words, $T_a f = \int_{T^n}[\sum_{k\in \mathbb{Z}}a(k)e^{-ik(x-y)}]f(y)d\mu(y)$.

Lemma¶

The series $\sum_{k\in \mathbb{Z}^n}\langle k \rangle^{-m}$ converges if $m>n$ (note that $>$ is required not ≥).

Proof. By comparing with integrals it sufficies to show that $\int_{\mathbb R^n} \langle \xi \rangle^{-m}dm(\xi)$ converges. Use polar coordinate and change of variable formula we get $dm(\xi) = r^{n-1}\sin(\varphi_1)\cdots \sin(\varphi_{n-2})$ (see Lecture 20 for the notations). Then

And the RHS is integrable on $\mathbb{R}$ when $n-1-m<-1$ so $m>n$.

$\square$

Lemma¶

For $a\in S^l$, let $K_a(x,y) = \sum_{k\in \mathbb{Z}^n}a(x,k)e^{ik(x-y)}$.

• If $l<-n$ then $T_a f = \int_{T^n}K_a(x,y)f(y)d\mu(y)$.
• If $l<-(m+n)$ then $K_a(x,y)\in \mathscr{C}^{m}(T^n\times T^n)$.

Proof.

It follows that when $|\alpha|+|\beta|\leq m \implies l+|\alpha|+|\beta|<-n$, so $D_x^{\alpha}D_y^{\beta}K_a(x,y)$ converges absolutely. This implies $D_x^{\alpha}D_y^{\beta}K_a(x,y)$ is continuous on $T^n\times T^n$ for every $|\alpha|+|\beta|\leq m$, i.e. $K_a(x,y)$ is in $\mathscr{C}^{m}(T^n\times T^n)$.

Smoothing operator¶

If $a\in S^{-\infty}$ then $T_a$ is a smoothing operator.

Construction of the inverse: bootstrap¶

We want to prove that for elleptic differential operator $P$ of order $m$, there exists a pseudodifferential operator $T_a$ which is almost right inverse of $P$ (i.e. $PT_a = I-\text{smoothing }$). Since $P$ is positive order, then to be inverse, $T_a$ should be $S^{-m}$.

Let $p_m(x,\xi)$ be the top homogeneous term of $p$, $p_m(x,\xi) = \sum_{|\alpha| = m}a_\alpha(x)\xi^{\alpha}$. $P$ is elliptic implies $p_m(x,\xi)\neq 0$ when $\xi\neq 0$. Remember we want to find some symbol with order $-m$, the raw ingadient we have is $\frac{1}{p_m(x,\xi)}$, the growth condition is $\langle \xi\rangle^{-m}$. To make it a symbol we need to deal with the issue at zero, we can modify it with a smooth “high-pass” filter. Let ρ be a smooth cut off function on $\mathbb{R}$ with $\rho(t)\equiv 0$ for $t<1$ and $\rho(t)\equiv 1$ for $t>2$. Then $a_0:=\rho(|\xi|)\frac{1}{p_m(x,\xi)}$ is a symbol in $S^{-m}$.

Let $T_{r_1} = I - PT_{a_0}$, or $r_1=1-p\circ a_0$. Then $r_1$ is order -1, because the degree $m$ part of $T_{a_0}$ and $P$ are reverse to each other. One can also use the composition formula to see it. By (cmp), the symbol of $PT_{a_0}$ is

Observe that, if we let $a_1 = r_1a_0$, $r_2 = r_1 - p\circ(r_1a_0) = 1-p\circ a_0-p\circ a_1$, then $r_2$ is order -2, as stated in the lemma below.

Lemma¶

For $b\in S^i$, $b - p\circ(a_0b)$ is in $S^{i-1}$.

The prove of this lemma is the same as the case $p\circ a_0$. Here we need $P$ to be elliptic to make sure the $m$-order term of $P$ cancel with $a_0$.

Continue this process we get a sequence $r_i\in S^{-(i+1)}$ and $a_i\in S^{m-(i+1)}$ so that $r_i =1-[ p\circ a_0 + \cdots + p\circ a_{i-1}]$. Eventually we expect to arrive at some $a = \sum_{i=0}^{\infty}a_i$ and $r\in S^{-\infty}$ such that $r = 1-p\circ a$. Then we are done if we can show such $a$ is in $S^{m}$. The lemma below is slightly different from the ideal case but satisfies our need.

Lemma (The Asymptotic summation theorem)¶

Let $(b_i)_{i=0}^{+\infty}$ be a sequence of symbols such that $b_i\in S^{m-i}$. Then there exists $b\in S^{m}$ such that for every $N$, $b-S_N\in S^{m-(N+1)}$, where $S_N = \sum_{i=0}^{N}b_i$ is the $N$-th partial sum.

Proof. Fix $0<\epsilon < 1$. Since $b_i\in S^{m-i}$, then by definition $b_i$ has the growth condition

Then we can choose $\lambda_i$ big enough such that when $|\xi|>\lambda_i$, $\langle\xi \rangle^{\epsilon} > 2^{i}$, so $|b_i(x,\xi)|<\cdot\frac{1}{2^{i}}\cdot\langle \xi\rangle^{m+\epsilon-i}$.

The idea is to take the small part for $b_i$ for every $i$, and use smooth high-pass filter to cut off the remaining part, then sum them up. Let $\rho\in \mathscr{C}^{\infty}(\mathbb R)$ be a smooth function such that

Then $\rho(\frac{|\xi|}{\lambda_i})$ is 0 on $|\xi|<\lambda_i$ and 1 on $|\xi|>2\lambda_i$, so $\rho(\frac{|\xi|}{\lambda_i})b_i(x,\xi)$ is a filtered symbol which takes only small parts of $b_i$.

Let $b = \sum_{i=0}^{\infty}\rho(\frac{|\xi|}{\lambda_i})b_i(x,\xi)$. Then we have by direct computation

where $r_{N-1} = \sum_{i=0}^{N-1}(\rho(\frac{|\xi|}{\lambda_i})-1)b_i+\rho(\frac{|\xi|}{\lambda_N})b_N+\sum_{i=N+1}^{+\infty}\rho(\frac{|\xi|}{\lambda_i})b_i$. Next we show that the error term $r_{N-1}$ is in $S^{m-N}$.

The first term is compact supported so doesn’t matter, the second term is in $S^{m-N}$; The third term is dominated by $\sum_{i=N+1}^{+\infty}\frac{1}{2^{i}}\langle \xi \rangle^{m+\epsilon -i}\leq \langle \xi \rangle^{m+\epsilon}\frac{1}{2^{N+1}}\langle \xi\rangle^{-(N+1)}$ so it is $O(\langle \xi\rangle^{m-N} )$ because $\epsilon < 1$.

To show that the third term is in $S^{m-N}$ we need to look at derivatives as well.

Fix $M>0$, we can choose $\lambda_{i,M} > 0$ such that $|D_x^{\alpha}D_\xi^{\beta}b_i(x,\xi)|\leq \frac{1}{2^{i}}\langle \xi\rangle^{m+\epsilon-|\beta|-i}$ for every $|\alpha|+|\beta|\leq M$. The previous case is the $M=0$ case. So by replacing $\lambda_i$ with $\lambda_{i,M}$ we can get new $b$ and $r_N$ with

for every $|\alpha|+|\beta|\leq M$.

Next we need to eleminate dependence of $b$ on $M$. Run the argument for every $M$ we get a infinite matrix of positive numbers $\lambda_{i,M}$. Now we choose its diagnal, let $\lambda_i = \lambda_{i,i}$. Then create $b$ with such $\lambda_i$ will give us desired decreasing property for every $\alpha, \beta$, hence the $r_N$ of this $b$ satisfiess $r_N\in S^{m-N}$.

$\square$

There exists $a\in S^{-m}$ such that $a= \sum_{i=0}^{+\infty}a_i + S^{-\infty}$. This means $a = \sum_{i=0}^{+\infty} a_i + r$ for some $r\in S^{-\infty}$. It follows that $PT_a = 1-T_r$ for some $r\in S^{-\infty}$.

Since $T_r$ is smoothing operator we have finished the proof.

The $*$Kernel-Range relation¶

Let $W$ be an inner product space. $T:W\to W$ be a linear operator, and $T^*$ be its adjoint operator given by requiring $(Tf,g) = (f,T^*g)$ on $W$. Let $\mathrm{Ker}(T):=\{w\in W:Tw=0\}$ be the kernel of $T$ and $\mathrm{Ran}(T):=\{w\in W:w=Tw' \text{ for some }w'\in W\}$ be the range of $T$. Then we automatically have

this is pure linear algebra and no Hilbert space theory involved.

Taking $\perp$ both side we get $\mathrm{Ker}(T^*)^{\perp} = (\mathrm{Ran}(T^*)^{\perp})^{\perp}$, where the RHS is equal to $\overline{\mathrm{Ran}(T)}$ when $W$ is a Hilbert space. But we are looking for a condition to imply $\mathrm{Ker}(T^*)^{\perp} = \mathrm{Ran}(T)$, with no closure allowed, since $W$ will be space of smooth functions and taking closure in $L^2$ will violate regularity. Here is a handy lemma which satisfies our needs.

Lemma (Key Linear algebra)¶

If there exists a finite dimensional subspace $V\subset W$ such that $V^{\perp}\supset \mathrm{Ran}(T)$, then

(1) $\mathrm{Ker}(T^*) = (\mathrm{Ran}(T)\cap V)^{\perp _V}$, here $\perp_V$ means taking orthogonal complement in $V$. It is equal to $(\mathrm{Ran}(T)\cap V)^{\perp}\cap V$. By direct sum decomposition of the finite dimensional $V$ we get $V = \mathrm{Ker}(T^*)\oplus (\mathrm{Ran(T)\cap V}).$

(2) Consequently, $\mathrm{Ker}(T^*)^{\perp} = \mathrm{Ran}(T)$.

Proof.

(1)

• First we show that $\mathrm{Ker}(T^*) = \mathrm{Ran}(T)^{\perp}\cap V$.

Let $h_1,\dots,h_r$ be an orthonormal basis of $V$. For any $g\in \mathrm{Ker}(T^*)=\mathrm{Ran}(T)^{\perp}$, let $g = P_V g+g_2$, where $P_V g = \sum_{i=1}^r (g,h_i)h_i$ be the projection of $g$ to $V$. Then $g_2\perp V \implies g_2\in \mathrm{Ran}(T)$. So $g_2 = 0$. It follows that $g = P_V g \implies g\in V$. So $g\in \mathrm{Ran}(T)^{\perp}\cap V$, this proves $\subset$ direction.

By the $^*$kernel-range relation we have $\mathrm{Ker(T^*)} = \mathrm{Ran}(T)^{\perp}\supset \mathrm{Ran}(T)^{\perp}\cap V$. So $\mathrm{Ker}(T^*) = \mathrm{Ran}(T)^{\perp}\cap V$ proved.

• Next we show that $\mathrm{Ran}(T)^{\perp}\cap V = (\mathrm{Ran}(T)\cap V)^{\perp V}$.

For $f\in \mathrm{Ran}(T)$, write $f=P_V f+f_2$. Then $f_2\perp V\implies f_2\in \mathrm{Ran}(T)$, so $P_V f \in \mathrm{Ran}(T)\cap V$. If $g\in V$ and $g\perp \mathrm{Ran}(T)\cap V$, then $g\perp P_V f$. But $f_2\perp V$, so $g\perp f_2$. So $g\perp f$. We have showed that $g\in (\mathrm{Ran}(T)\cap V)^{\perp V}\implies g\in \mathrm{Ran}(T)^{\perp}\cap V$, the $\supset$ direction.

The converse inclusion is automatic because $(\mathrm{Ran}(T)\cap V)^{\perp}$ is bigger than $\mathrm{Ran}(T)^{\perp}$.

(2) The $*$kernel-range relation implies $\mathrm{Ran}(T)\subset \mathrm{Ker}(T^*)^{\perp}$.

Just need to prove $\mathrm{Ker}(T^*)^{\perp}\subset \mathrm{Ran}(T)$. For $f\in W$ such that $f\perp \mathrm{Ker}(T^*)$, write $f = P_V f + f_2$ so $f_2\in \mathrm{Ran}(T)$. Then $f_2\perp \mathrm{Ker}(T^*)$, so $P_V f \perp \mathrm{Ker}(T^*)$. By (1) we have $P_V f\in \mathrm{Ran}(T)$. This implies $f\in \mathrm{Ran}(T).$

The Fredholm theorem for Elliptic operators¶

Theorem¶

Let $T_K$ be a smoothing operator on $X$. Then $I-T_K$ satisfies

(1) $\mathrm{Ker}(I-T_K)$ is finite dimensional.

(2) For $f\in \mathscr{C}^{\infty}(X)$, $f\in \mathrm{Ran}(I-T_K)$ if and only if $f\perp \mathrm{Ker}(I-T_{K^*})$.

Finite rank case¶

A smoothing operator $T_F$ is called finite rank provided $F(x,y) = \sum_{i=1}^N f_i(x)g_i(y)$.

Lemma¶

Let $F$ be a finite rank smoothing operator. Then

(1) $\mathrm{Ker}(I-T_F)$ is finite dimensional.

(2) If $(f,g_i) = 0$ for every $i=1,\dots,N$, then $f\in \mathrm{Ran}(I-T_F)$.

Approximation by finite rank¶

Lemma¶

There exists smoothing $L$ such that $I-T_{L}$ is invertible and

(1) $(I-T_L)(I-T_K) = I-T_{F_1}$

(2) $(I-T_K)(I-T_L) = I-T_{F_2}$

for some finite rank smothing $F_1,F_2$.

Proof of theorem¶

Let $L,F_1,F_2$ be the one in the approximation by finite rank lemma above.

Proof of (1):

If $f\in \mathrm{Ker}(I-T_K)$, then $(I-T_L)(I-T_K)f = (I-T_L) 0 = 0$, so $f\in \mathrm{Ker}(I-T_{F_1})$. By the finite rank case $\mathrm{Ker}(1-T_F)$ is fintie dimensional. Since $I-T_L$ is invertible this implies $\mathrm{Ker}(I-T_K)$ is also finite dimensional.

Proof of (2):

It’s sufficient to prove $f\perp\mathrm{Ker}(I-T_{K^*})\implies f\in \mathrm{Ran}(I-T_K)$. The other direction is automatic by the $*$kernel-range relation.

Let $F_2 = \sum_{i=1}^N f_i(x)g_i(y)$. It’s sufficient to prove that $f\perp \mathrm{Ker}(I-T_{K^*})$ implies $f\in \mathrm{Ran}(I-T_F)$. Because $f\in \mathrm{Ran}(1-T_{F_2}) \implies f = (I-T_K)(I-T_{L})g$ for some $g$, so $f\in \mathrm{Ran}(I-T_K)$.

Take $V = \mathrm{span}\{g_1,\dots,g_N\}$. By the lemma in finite rank case, $f\perp V$ implies $f\in \mathrm{Ran}(I-T_{F_2})\implies f\in \mathrm{Ran}(I-T_K)$. Then apply the Key linear algebra lemma above to $T = I-T_{K}$ and $V$ we conclude that $f\in \mathrm{Ran}(I-T_{F_2})$.