3.3: Approximation of CR Functions

The following theorem (proved circa 1980) holds in much more generality, but we state its simplest version. One of the simplifications we make is that we consider only smooth CR functions here, although the theorem holds even for continuous CR functions where the CR conditions are interpreted in the sense of distributions.

Theorem $\PageIndex{1}$

Baouendi–Trèves

Suppose $M \subset \mathbb{C}^n$ is a smooth real hypersurface, $p \in M$ is a point, and $z=(z_1,\dots,z_n)$ are the holomorphic coordinates of $\mathbb{C}^n$. Then there exists a compact neighborhood $K \subset M$ of $p$, such that for every smooth CR function $f \colon M \to \mathbb{C}$, there exists a sequence $\{ p_j \}$ of polynomials in $z$ such that $$p_j(z) \to f(z) \qquad \text{uniformly in $K$.}$$

A key point is that $K$ cannot be chosen arbitrarily, it depends on $p$ and $M$. On the other hand it does not depend on $f$. Given $M$ and $p \in M$ there is a $K$ such that every CR function on $M$ is approximated uniformly on $K$ by holomorphic polynomials. The theorem applies in one dimension, although in that case the theorem of Mergelyan (see Theorem B.31) is much more general.

Example $\PageIndex{1}$

Let us show that $K$ cannot possibly be arbitrary. For simplicity $n=1$. Let $S^1 \subset \mathbb{C}$ be the unit circle (boundary of the disc), then every smooth function on $S^1$ is a smooth CR function. Let $f$ be a nonconstant real function such as $\Re z$. Suppose for contradiction that we could take $K = S^1$ in the theorem. Then $f(z) = \Re z$ could be uniformly approximated on $S^1$ by holomorphic polynomials. By the maximum principle, the polynomials would converge on $\mathbb{D}$ to a holomorphic function on $\mathbb{D}$ continuous on $\overline{\mathbb{D}}$. This function would have nonconstant real boundary values, which is impossible. Clearly $K$ cannot be the entire circle.

The example is easily extended to $\mathbb{C}^n$ by considering $M = S^1 \times \mathbb{C}^{n-1}$, then $\Re z_1$ is a smooth CR function on $M$ that cannot be approximated uniformly on $S^1 \times \{ 0 \}$ by holomorphic polynomials.

The technique of the example above will be used later in a more general situation, to extend CR functions using Baouendi–Trèves.

Remark $\PageIndex{1}$

It is important to note the difference between Baouendi–Trèves (and similar theorems in complex analysis) and the Weierstrass approximation theorem. In Baouendi–Trèves we obtain an approximation by holomorphic polynomials, while Weierstrass gives us polynomials in the real variables, or in $z$ and $\bar{z}$. For example, via Weierstrass, every continuous function is uniformly approximable on $S^1$ via polynomials in $\Re z$ and $\Im z$, and therefore by polynomials in $z$ and $\bar{z}$. These polynomials do not in general converge anywhere but on $S^1$.

Exercise $\PageIndex{1}$

Let $z=x+iy$ as usual in $\mathbb{C}$. Find a sequence of polynomials in $x$ and $y$ that converge uniformly to $e^{x-y}$ on $S^1$, but diverge everywhere else.

The proof is an ingenious use of the standard technique used to prove the Weierstrass approximation theorem. Also, as we have seen mollifiers before, the technique will not be completely foreign even to the reader who does not know the Weierstrass approximation theorem. Basically what we do is use the standard convolution argument, this time against a holomorphic function. Letting $z=x+iy$ we only do the convolution in the $x$ variables keeping $y=0$. Then we use the fact that the function is CR to show that we get an approximation even for other $y$.

In the formulas below, given a vector $v = (v_1,\ldots,v_n)$, it will be useful to write $$[v]^2 \overset{\text{def}}{=} v_1^2 + \cdots + v_n^2 .$$

The following lemma is a neat application of ideas from several complex variables to solve a problem that does not at first seems to involve holomorphic functions.

Lemma $\PageIndex{1}$

Let $W$ be the set of $n \times n$ complex matrices $A$ such that $$||(\Im A)x|| < ||(\Re A)x||$$ for all nonzero $x \in \mathbb{R}^n$ and $\Re A$ is positive definite. Then for all $A \in W$, $$\int_{\mathbb{R}^n} e^{-{[Ax]}^2} \det A \, dx = \pi^{n/2} .$$

Proof

Suppose $A$ has real entries and $A$ is positive definite (so $A$ is also invertible). By a change of coordinates $$\int_{\mathbb{R}^n} e^{-{[Ax]}^2} \det A \, dx = \int_{\mathbb{R}^n} e^{-{[x]}^2} \, dx = \left(\int_{\mathbb{R}} e^{-x_1^2} \, dx_1 \right) \cdots \left(\int_\mathbb{R} e^{-x_n^2} \, dx_n \right) = {(\sqrt{\pi})}^n .$$ Next suppose $A$ is any matrix in $W$. There is some $\epsilon > 0$ such that $||(\Im A) x||^2 \leq (1-\epsilon^2) ||(\Re A) x||^2$ for all $x \in \mathbb{R}^n$. That is because we only need to check this for $x$ in the unit sphere, which is compact (exercise). By reality of $\Re A$, $\Im A$, and $x$ we get ${[(\Re A)x]}^2 = ||(\Re A)x||^2$ and ${[(\Im A)x]}^2 = ||(\Im A)x||^2$. So $$\left| e^{-{[Ax]}^2}\right| = e^{-\Re {[Ax]}^2} \leq e^{-{[(\Re A)x]}^2 + {[(\Im A)x]}^2} \leq e^{-\epsilon^2 {[(\Re A)x]}^2} .$$ Therefore, the integral exists for all $A$ in $W$ by a similar computation as above.

The expression $$\int_{\mathbb{R}^n} e^{-{[Ax]}^2} \det A \, dx$$ is a well-defined holomorphic function in the entries of $A$, thinking of $W$ as a domain (see exercises below) in $\mathbb{C}^{n^2}$. We have a holomorphic function that is constantly equal to $\pi^{n/2}$ on $W \cap \mathbb{R}^{n^2}$ and hence it is equal to $\pi^{n/2}$ everywhere on $W$.

Exercise $\PageIndex{2}$

Prove the existence of $\epsilon > 0$ in the proof above.

Exercise $\PageIndex{3}$

Show that $W \subset \mathbb{C}^{n^2}$ in the proof above is a domain (open and connected).

Exercise $\PageIndex{4}$

Prove that we can really differentiate under the integral to show that the integral is holomorphic in the entries of $A$.

Exercise $\PageIndex{5}$

Show that some hypotheses are needed for the lemma. In particular, take $n=1$ and find the exact set of $A$ (now a complex number) for which the conclusion of the lemma is true.

Given an $n \times n$ matrix $A$, let $||A||$ denote the operator norm, $$||A|| = \sup_{||v||=1} ||Av|| = \sup_{v \in \mathbb{C}^n, v\not= 0} \frac{||Av||}{||v||} .$$

Exercise $\PageIndex{6}$

Let $W$ be as in Lemma $\PageIndex{1}$. Let $B$ be an $n \times n$ real matrix such that $||B|| < 1$. Show that $I + iB \in W$.

We will be using differential forms, and the following lemma says that as far as the exterior derivative is concerned, all CR functions behave as restrictions of holomorphic functions.

Lemma $\PageIndex{2}$

Let $M \subset \mathbb{C}^n$ be a smooth real hypersurface, $f \colon M \to \mathbb{C}$ be a smooth CR function, and $(z_1,\ldots,z_n)$ be the holomorphic coordinates of $\mathbb{C}^n$. Then at each point $p \in M$, the exterior derivative $df$ is a linear combination of $dz_1,\ldots,dz_n$, thinking of $z_1,\ldots,z_n$ as functions on $M$.Namely, $$d(f \, dz) = df \wedge dz = 0.$$

Recall the notation $dz = dz_1 \wedge dz_2 \wedge \cdots \wedge dz_n$.

Proof

After a complex affine change of coordinates, we simply need to prove the result at the origin. Let $\xi_1,\ldots,\xi_n$ be the new holomorphic coordinates and suppose the $T^{(1,0)}_0 M$ tangent space is spanned by $\frac{\partial}{\partial \xi_1}\big|_0, \ldots, \frac{\partial}{\partial \xi_{n-1}}\big|_0$, and such that $\frac{\partial}{\partial \Re \xi_n}\big|_0$ is tangent and $\frac{\partial}{\partial \Im \xi_n}\big|_0$ is normal. At the origin, the CR conditions are $\frac{\partial f}{\partial \bar{\xi}_k}(0) = 0$ for all $k$, so $$df(0) = \frac{\partial f}{\partial \xi_1}(0) \, d\xi_1(0) + \cdots + \frac{\partial f}{\partial \xi_{n-1}}(0) \, d\xi_{n-1}(0) + \frac{\partial f}{\partial \Re \xi_{n}}(0) \, d(\Re \xi_{n})(0) .$$ Also, at the origin $d\xi_n(0) = d(\Re \xi_n)(0) + i d(\Im \xi_n)(0) = d(\Re \xi_n)(0)$. So $df(0)$ is a linear combination of $d\xi_1(0),\ldots,d\xi_n(0)$. As $\xi$ is a complex affine function of $z$, then each $d\xi_k$ is a linear combination of $dz_1$ through $dz_n$, and the claim follows. So if $f$ is a CR function, then $d(f\,dz) = df \wedge dz = 0$ since $dz_k \wedge dz_k = 0$.

Proof of the theorem of Baouendi–Trèves. Suppose $M \subset \mathbb{C}^n$ is a smooth real hypersurface, and without loss of generality suppose $p=0 \in M$. Let $z=(z_1,\ldots,z_n)$ be the holomorphic coordinates, write $z=x+iy$, $y=(y',y_n)$, and suppose $M$ is given by $$y_n = \psi(x,y') ,$$ where $\psi$ is $O(2)$. The variables $(x,y')$ parametrize $M$ near 0: $$z_j = x_j+iy_j , \quad \text{ for $j = 1,\ldots,n-1$,} \quad \text{and} \quad z_n = x_n + i \psi(x,y') .$$ Define $$\varphi(x,y') = \bigl(y_1,\ldots,y_{n-1},\psi(x,y')\bigr) .$$ Write $(x,y') \mapsto z = x + i\varphi(x,y')$ as the parametrization. That is, think of $z$ as a function of $(x,y')$.

Let $r > 0$ and $d > 0$ be small numbers to be determined later. Assume they are small enough so that $f$ and $\varphi$ are defined and smooth on some neighborhood of the set where $||x|| \leq r$ and $||y'|| \leq d$.

There exists a smooth $g \colon \mathbb{R}^n \to [0,1]$ such that $g \equiv 1$ on $B_{r/2}(0)$ and $g \equiv 0$ outside of $B_{r}(0)$. See Figure $\PageIndex{1}$. Explicit formula can be given. Alternatively we obtain such a $g$ by use of mollifiers on a function that is identically one on $B_{3r/4}(0)$ and zero elsewhere. Such a $g$ is commonly called a cutoff function.

Figure $\PageIndex{1}$

Exercise $\PageIndex{7}$

Find an explicit formula for $g$ without using mollifiers.

Let $$K' = \bigl\{ (x,y') : ||x|| \leq \frac{r}{4} , ||y'|| \leq d \bigr\} .$$ Let $K = z(K')$, that is the image of $K'$ under the mapping $z(x,y')$.

Consider the CR function $f$ a function of $(x,y')$ and write $f(x,y')$. For $\ell \in \mathbb{N}$, let $\alpha_{\ell}$ be a differential $n$-form defined (thinking of $w \in \mathbb{C}^n$ as a constant parameter) by $$\begin{align}\begin{aligned} \alpha_{\ell}(x,y') & = {\left(\frac{\ell}{\pi}\right)}^{n/2} e^{-\ell [w - z]^2} g(x) f(x,y') \, dz \\ & = {\left(\frac{\ell}{\pi}\right)}^{n/2} e^{-\ell [w - x-i\varphi(x,y')]^2} g(x) f(x,y') \\ & \qquad \qquad (dx_1 + idy_1) \wedge \cdots \wedge (dx_{n-1} + i dy_{n-1}) \wedge \bigl(dx_{n} + i d \psi (x,y') \bigr) . \end{aligned}\end{align}$$

The key is the exponential, which looks like the bump function mollifier, except that now we have $w$ and $z$ possibly complex. The exponential is also holomorphic in $w$, and that will give us entire holomorphic approximating functions.

Fix $y'$ with $0 < ||y'|| < d$ and let $D$ be defined by $$D = \bigl\{ (x,s) \in \mathbb{R}^n \times \mathbb{R}^{n-1} : ||x|| < r \text{ and } s = t y' \text{ for $t \in (0,1)$} \bigr\} .$$ $D$ is an $(n+1)$-dimensional “cylinder.” We take a ball in the $x$ directions, then take a single fixed point $y'$ in the $s$ variables and make a cylinder. See Figure $\PageIndex{2}$.

Figure $\PageIndex{2}$

Orient $D$ in the standard way as if it sat in the $(x,t)$ variables in $\mathbb{R}^n \times \mathbb{R}$. Stokes’ theorem says $$\int_D d \alpha_{\ell} (x,s) = \int_{\partial D} \alpha_{\ell} (x,s) .$$ Since $g(x) = 0$ if $||x|| \geq r$, $\alpha_\ell$ is zero on the sides of the cylinder $D$, so the integral over $\partial D$ only needs to consider the top and bottom of the cylinder. And because of $g$, the integral over the top and bottom can be taken over $\mathbb{R}^n$. As is usual in these sorts of arguments, we do the slight abuse of notation where we ignore that $f$ and $\varphi$ are undefined where $g$ is identically zero:

$$\begin{align} \label{eq:1} \int_{\partial D} \alpha_{\ell}(x,s) = & {\left(\frac{\ell}{\pi}\right)}^{n/2} \int_{x \in \mathbb{R}^n} \!\!\! e^{-\ell [w - x-i\varphi(x,y')]^2 } g(x) f(x,y') \, dx_1 \wedge \cdots \wedge dx_{n-1} \wedge \bigl(dx_{n} + i d_x \psi (x,y') \bigr) \\ & - {\left(\frac{\ell}{\pi}\right)}^{n/2} \int_{x \in \mathbb{R}^n} \!\!\! e^{ -\ell [w - x-i\varphi(x,0)]^2 } g(x) f(x,0) \, dx_1 \wedge \cdots \wedge dx_{n-1} \wedge \bigl(dx_{n} + i d_x \psi (x,0) \bigr) , \nonumber \end{align}$$ where $d_x$ means the derivative in the $x$ directions only. That is, $d_x \psi = \frac{\partial \psi}{\partial x_1} dx_1 + \cdots + \frac{\partial \psi}{\partial x_n} dx_n$.

We will show that as $\ell \to \infty$, the left-hand side of $\eqref{eq:1}$ goes to zero uniformly for $w \in K$ and the first term on the right-hand side goes to $f(\tilde{x},y')$ if $w = z(\tilde{x},y')$ is in $M$. Hence, we define entire functions that we will show approximate $f$: $$f_\ell(w) = {\left(\frac{\ell}{\pi}\right)}^{n/2} \int_{x \in \mathbb{R}^n} e^{ -\ell [w - x-i\varphi(x,0)]^2 } g(x) f(x,0) \, dx_1 \wedge \cdots \wedge dx_{n-1} \wedge \bigl(dx_{n} + i d_x \psi (x,0) \bigr) .$$ Clearly each $f_\ell$ is holomorphic and defined for all $w \in \mathbb{C}^n$.

In the next claim it is important that $f$ is a CR function.

Claim $\PageIndex{1}$

We have $$d \alpha_\ell(x,s) = {\left(\frac{\ell}{\pi}\right)}^{n/2} e^{-\ell [w - z(x,s)]^2} f(x,s) \, dg(x) \wedge dz(x,s) ,$$ and for sufficiently small $r>0$ and $d>0$, $$\lim_{\ell\to\infty} {\left(\frac{\ell}{\pi}\right)}^{n/2} \int_{(x,s)\in D} e^{-\ell [w - z(x,s)]^2} f(x,s) \, dg(x) \wedge dz(x,s) = 0$$ uniformly as a function of $w \in K$ and $y' \in B_d(0)$ (recall that $D$ depends on $y'$).

Proof

The function $(x,s) \mapsto e^{-\ell [w - z(x,s)]^2}$ is CR (as a function on $M$), and so is $f(x,s)$. Therefore, using Lemma $\PageIndex{2}$, $$d \alpha_{\ell}(x,s) = {\left(\frac{\ell}{\pi}\right)}^{n/2} e^{-\ell [w - z(x,s)]^2 } f(x,s) \, dg(x) \wedge dz(x,s) .$$ Since $dg$ is zero for $||x|| \leq \frac{r}{2}$, the integral $$\int_D d \alpha_\ell(x,s) = {\left(\frac{\ell}{\pi}\right)}^{n/2} \int_D e^{ -\ell [w - z(x,s)]^2} f(x,s) \, dg(x) \wedge dz(x,s)$$ is only evaluated for the subset of $D$ where $||x|| > \frac{r}{2}$.

Suppose $w \in K$ and $(x,s) \in D$ with $||x|| > \frac{r}{2}$. Let $w = z(\tilde{x},\tilde{s})$. We need to estimate $$\left|e^{ -\ell {[w - z(x,s)]}^2 }\right| = e^{ -\ell \Re {[w - z(x,s)]}^2 } .$$ Then $$-\Re {[w - z]}^2 = -||\tilde{x}-x||^2 + ||\varphi(\tilde{x},\tilde{s})-\varphi(x,s)||^2 .$$ By the mean value theorem $$||\varphi(\tilde{x},\tilde{s})-\varphi(x,s)|| \leq ||\varphi(\tilde{x},\tilde{s})-\varphi(x,\tilde{s})|| + ||\varphi(x,\tilde{s})-\varphi(x,s)|| \leq a ||\tilde{x}-x|| + A ||\tilde{s}-s|| ,$$ where $a$ and $A$ are $$a = \sup_{||\hat{x}|| \leq r, ||\hat{y}'|| \leq d} \left| \left| \,\left[\frac{\partial \varphi}{\partial x}(\hat{x},\hat{y}')\right]\, \right| \right| , \qquad A = \sup_{||\hat{x}|| \leq r, ||\hat{y}'|| \leq d} \left| \left| \,\left[\frac{\partial \varphi}{\partial y'}(\hat{x},\hat{y}'))\right]\,\right| \right| .$$ Here $\bigl[ \frac{\partial \varphi}{\partial x} \bigr]$ and $\bigl[ \frac{\partial \varphi}{\partial y'} \bigr]$ are the derivatives (matrices) of $\varphi$ with respect to $x$ and $y'$ respectively, and the norm we are taking is the operator norm. Because $\bigl[ \frac{\partial \varphi}{\partial x} \bigr]$ is zero at the origin, we pick $r$ and $d$ small enough (and hence $K$ small enough) so that $a \leq \frac{1}{4}$. We furthermore pick $d$ possibly even smaller to ensure that $d \leq \frac{r}{32A}$. We have that $\frac{r}{2} \leq ||x|| \leq r$, but $||\tilde{x}|| \leq \frac{r}{4}$ (recall $w \in K$), so $$\frac{r}{4} \leq ||\tilde{x}-x|| \leq \frac{5r}{4} .$$ Also, $||\tilde{s}-s|| \leq 2d$ by triangle inequality.

Therefore, $$\begin{align}\begin{aligned} -\Re {[w - z(x,s)]}^2 & \leq - ||\tilde{x}-x||^2 + a^2 ||\tilde{x}-x||^2 + A^2 ||\tilde{s}-s||^2 + 2aA ||\tilde{x}-x|| \: ||\tilde{s}-s|| \\ & \leq \frac{-15}{16} ||\tilde{x}-x||^2 + A^2 ||\tilde{s}-s||^2 + \frac{A}{2} ||\tilde{x}-x|| \: ||\tilde{s}-s|| \\ & \leq \frac{-r^2}{64} . \end{aligned}\end{align}$$

In other words, $$\left| e^{-\ell[w-z(x,s)]^2}\right| \leq e^{-\ell r^2 / 64} ,$$ or $$\left| {\left(\frac{\ell}{\pi}\right)}^{n/2} \int_{(x,s)\in D} e^{-\ell [w - z(x,s)]^2} f(x,s) \, dg(x) \wedge dz(x,s) \right| \leq C \ell^{n/2} e^{-\ell r^2 / 64} ,$$ for some constant $C$. Note that $D$ depends on $y'$. The set of all $y'$ with $||y'|| \leq d$, is a compact set, so we can make $C$ large enough to not depend on the $y'$ that was chosen. The claim follows.

Claim $\PageIndex{2}$

For the given $r>0$ and $d>0$, $$\begin{align}\begin{aligned} \lim_{\ell\to\infty} {\left(\frac{\ell}{\pi}\right)}^{n/2} \int_{x \in \mathbb{R}^n} e^{ -\ell [\tilde{x}+i\varphi(\tilde{x},y') - x-i\varphi(x,y')]^2 } g(x) f(x,y') dx_1 \wedge \cdots \wedge dx_{n-1} \wedge \bigl(dx_{n} + i d_x \psi (x,y') \bigr) \\ = f(\tilde{x},y')\end{aligned}\end{align}$$ uniformly in $(\tilde{x},y') \in K'$.

That is, we look at $\eqref{eq:1}$ and we plug in $w = z(\tilde{x},y') \in K$. The $g$ (as usual) makes sure we never evaluate $f$, $\psi$, or $\varphi$ at points where they are not defined.

Proof

The change of variables formula implies $$dx_1 \wedge \cdots \wedge dx_{n-1} \wedge \bigl(dx_{n} + i d_x \psi (x,y') \bigr) = d_x z(x,y') = \det \left[\frac{\partial z}{\partial x}(x,y')\right] dx ,$$ where $\bigl[\frac{\partial z}{\partial x}(x,y')\bigr]$ is the matrix corresponding to the derivative of the mapping $z$ with respect to the $x$ variables evaluated at $(x,y')$.

Let us change variables of integration via $\xi = \sqrt{\ell} ( x-\tilde{x})$: $$\begin{align}\begin{aligned} &{\left(\frac{\ell}{\pi}\right)}^{n/2} \int_{x \in \mathbb{R}^n} e^{ -\ell [\tilde{x}+i\varphi(\tilde{x},y') - x-i\varphi(x,y')]^2 } g(x) f(x,y') \det \left[\frac{\partial z}{\partial x}(x,y')\right] dx = \\ &{\left(\frac{1}{\pi}\right)}^{n/2} \int_{\xi \in \mathbb{R}^n} e^{-{\left[\xi + i\sqrt{\ell}\left(\varphi\left(\tilde{x}+\frac{\xi}{\sqrt{\ell}},y'\right) - \varphi(\tilde{x},y')\right)\right ]}^2} g\left(\tilde{x}+\frac{\xi}{\sqrt{\ell}}\right) f\left(\tilde{x}+\frac{\xi}{\sqrt{\ell}},y'\right) \det \left[\frac{\partial z}{\partial x}\left(\tilde{x}+\frac{\xi}{\sqrt{\ell}},y'\right)\right] d\xi .\end{aligned}\end{align}$$ We now wish to take a limit as $\ell \to \infty$ and for this we apply the dominated convergence theorem. So we need to dominate the integrand. The second half of the integrand is uniformly bounded independent of $\ell$ as $$x \mapsto g(x) f(x,y') \det \left[\frac{\partial z}{\partial x}(x,y')\right]$$ is a continuous function with compact support (because of $g$). Hence it is enough to worry about the exponential term. We also only consider those $\xi$ where the integrand is not zero. Recall that $r$ and $d$ are small enough that

$$\sup_{||\hat{x}|| \leq r, ||\hat{y}'|| \leq d} \left| \left| \,\left[ \frac{ \partial \varphi}{\partial x}(\hat{x},\hat{y}') \right]\,\right| \right| \leq \frac{1}{4} ,$$ and as $||\tilde{x}|| \leq \frac{r}{4}$ (as $(\tilde{x},y') \in K$) and $\left| \left| \tilde{x}+\frac{\xi}{\sqrt{\ell}}\right| \right| \leq r$ (because $g$ is zero otherwise), then

$$\left| \left| \varphi\left(\tilde{x}+\frac{\xi}{\sqrt{\ell}},y'\right) - \varphi(\tilde{x},y')\right| \right| \leq \frac{1}{4} \left| \left| \tilde{x}+\frac{\xi}{\sqrt{\ell}}-\tilde{x}\right| \right| = \frac{|| \xi|| }{4 \sqrt{\ell}} .$$

So under the same conditions we have $$\begin{align}\begin{aligned} \left| e^{-{\left[\xi + i\sqrt{\ell}\left(\varphi\left(\tilde{x}+\frac{\xi}{\sqrt{\ell}},y'\right) - \varphi(\tilde{x},y')\right)\right]}^2}\right| & = e^{-\Re {\left[\xi + i\sqrt{\ell}\left(\varphi\left(\tilde{x}+\frac{\xi}{\sqrt{\ell}},y'\right) - \varphi(\tilde{x},y')\right)\right]}^2} \\ & = e^{-||\xi||^2 + \ell \left| \left| \varphi\left(\tilde{x}+\frac{\xi}{\sqrt{\ell}},y'\right) - \varphi(\tilde{x},y')\right| \right| ^2} \\ & \leq e^{-(15/16)|| \xi|| ^2} . \end{aligned}\end{align}$$

And that is integrable. Thus, take the pointwise limit under the integral to obtain $${\left(\frac{1}{\pi}\right)}^{n/2} \int_{\xi \in \mathbb{R}^n} e^{-{\left[\xi + i\left[ \frac{\partial \varphi}{\partial x}(\tilde{x},y') \right] \xi \right]}^2} g(\tilde{x}) f(\tilde{x},y') \det \left[\frac{\partial z}{\partial x}(\tilde{x},y')\right] d\xi .$$ In the exponent, we have an expression for the derivative in the $\xi$ direction with $y'$ fixed. If $(\tilde{x},y') \in K'$, then $g(\tilde{x}) = 1$, and so we can ignore $g$.

Let $A = I + i \bigl[ \frac{\partial \varphi}{\partial x}(\tilde{x},y') \bigr]$. Lemma $\PageIndex{1}$ says $${\left(\frac{1}{\pi}\right)}^{n/2} \int_{\xi \in \mathbb{R}^n} e^{-{\left[\xi + i\left[ \frac{\partial \varphi}{\partial x}(\tilde{x},y') \right] \xi \right]}^2} f(\tilde{x},y') \det \left[\frac{\partial z}{\partial x}(\tilde{x},y')\right] d\xi = f(\tilde{x},y') .$$

That the convergence is uniform in $(\tilde{x},y') \in K'$ is left as an exercise.

Exercise $\PageIndex{8}$

In the claim above, finish the proof that the convergence is uniform in $(\tilde{x},y') \in K'$. Hint: It may be easier to use the form of the integral before the change of variables and prove that the sequence is uniformly Cauchy.

We are essentially done with the proof of the theorem. The two claims together with $\eqref{eq:1}$ show that $f_\ell$ are entire holomorphic functions that approximate $f$ uniformly on $K$. Entire holomorphic functions can be approximated by polynomials uniformly on compact subsets; simply take the partial sums of Taylor series at the origin.

Exercise $\PageIndex{9}$

Explain why being approximable on $K$ by (holomorphic) polynomials does not necessarily mean that $f$ is real-analytic.

Exercise $\PageIndex{10}$

Suppose $M \subset \mathbb{C}^n$ is given by $\Im z_n = 0$. Use the standard Weierstrass approximation theorem to show that given a $K \subset \subset M$, and a smooth CR function $f \colon M \to \mathbb{C}$, then $f$ can be uniformly approximated by holomorphic polynomials on $K$.