Checking real analyticity on surfaces

TL;DR

The paper proves that a real-valued function on a real analytic surface is analytic if its restrictions to all 2-sphere submanifolds are analytic, extending a classical complex analysis result to real manifolds.

Contribution

It establishes a new criterion for analyticity of functions on real surfaces based on restrictions to 2-spheres, analogous to Hartogs' theorem in complex analysis.

Findings

Functions are analytic if restrictions to all 2-sphere submanifolds are analytic.

No continuity assumption is needed for the function.

Extends classical Hartogs theorem to real analytic manifolds.

Abstract

We prove that a real-valued function (that is not assumed to be continuous) on a real analytic manifold is analytic whenever all its restrictions to analytic submanifolds homeomorphic to the 2-sphere are analytic. This is a real analog for the classical theorem of Hartogs that a function on a complex manifold is complex analytic iff it is complex analytic when restricted to any complex curve.

Equations

$$\tau_{r}(p):=\tfrac{d^{r}f}{dt^{r}}(tp)\bigr{|}_{t=0}\quad\mbox{and set}\quad\tau_{r}(0)=0.$$

$$\mu:(x_{1},\dots,x_{n})\mapsto\bigl{(}\tfrac{x_{1}}{\sum x_{i}^{2}},\dots,\tfrac{x_{n}}{\sum x_{i}^{2}}\bigr{)}$$

$$S:=\bigl{(}h-\ell_{1}=\ell_{2}=\cdots=\ell_{n-2}=0\bigr{)},$$

$$f:=\frac{x_1\cdots x_n}{x_1^{2n}+\cdots +x_n^{2n}}\text{mbox}for(x_1,\dots ,x_n)\ne (0,\dots ,0)\text{mbox}andf(0,\dots ,0)=0$$

$$g(x,y,z)=\frac{x^8+y(x^2-y^3{)}^2+z^4}{x^{10}+(x^2-y^3{)}^2+z^2}\text{mbox}for(x,y,z)\ne (0,0,0)\text{mbox}andg(0,0,0)=0.$$

Full text

By a theorem of Hartogs, a function $f:{\mathbb{C}}^{n}\to{\mathbb{C}}$ (that is not assumed to be continuous) is analytic iff it is analytic when restricted to any translate of the coordinate axes; see [Har1906] or [BM48, p.140]. An analogous claim does not hold for real analytic functions; see Example 10 for a simple non-continuous function $f\colon{\mathbb{R}}^{3}\to{\mathbb{R}}$ that is real analytic on every translate of the coordinate hyperplanes and a more complicated non-continuous function $g\colon{\mathbb{R}}^{3}\to{\mathbb{R}}$ that is real analytic on every smooth analytic curve in ${\mathbb{R}}^{3}$.

Bochnak and Siciak proved in 1971 [BS71, BS18] that a function $f:{\mathbb{R}}^{n}\to{\mathbb{R}}$ is real analytic iff its restriction to any 2-plane is analytic. They also conjectured that a similar result holds on any real analytic manifold using restrictions to $2$-dimensional compact analytic submanifolds.

The aim of this note is to prove a stronger variant of this, using an improvement of [BS18, Thm.1] that uses fewer 2-planes.

Working on local coordinate charts, this is an immediate consequence of the following more precise version.

Here we use 2-sphere in the most restrictive sense, that is, the intersection of an $(n-1)$-sphere \bigl{(}\textstyle{\sum}_{i}x_{i}^{2}=\textstyle{\sum}_{i}c_{i}x_{i}\bigr{)} with a vector subspace of dimension 3.

Terminology. From now on by a function we mean an arbitrary real-valued function and real analytic is shortened to analytic. To simplify notation, we let $g|_{Y}$ denote the restriction of $g$ to Y\cap(\mbox{domain of g}).

Before we start the proof we need some preparation. A subset $C\subset{\mathbb{R}}^{n}$ is a cone if ${\mathbb{R}}C=C$. (This is slightly more convenient for us than the more usual variant ${\mathbb{R}}^{+}C=C$.) A cone $C$ is called open if $C\setminus\{0\}$ is open in ${\mathbb{R}}^{n}\setminus\{0\}$. A cone-neighborhood of a set $W$ is an open cone that contains $W$.

We will work both with vector subspaces $V\subset{\mathbb{R}}^{n}$ and affine-linear subspaces $Q\subset{\mathbb{R}}^{n}$. A property ${\mathcal{P}}$ holds for all vector subspaces near $V$ if there is a cone-neighborhood $C\supset V$ such that ${\mathcal{P}}$ holds for all vector subspaces (of the same dimension) contained in $C$. A property ${\mathcal{P}}$ holds for all affine-linear subspaces near $Q$ if there is a choice of the origin $0\in Q$, a cone-neighborhood $C\supset Q$ and $\epsilon>0$ such that ${\mathcal{P}}$ holds for all affine-linear subspaces (of the same dimension) contained in $C\cup B(\epsilon)$, where $B(\epsilon)$ denotes the ball of radius $\epsilon$ centered at [math]. For a cone $W$ and $\epsilon>0$ we write $W(\epsilon):=W\cap B(\epsilon)$. We need the following obvious fact.

By Lagrange interpolation, a 1-variable polynomial of degree $d$ is uniquely determined by $d+1$ of its values. Equivalently, a 2-variable homogeneous polynomial of degree $d$ on ${\mathbb{R}}^{2}$ is uniquely determined by its restriction to $d+1$ vector lines.

More generally, the following is easy to prove by induction on $d$.

Proof. By induction on $n$, starting with $n=2$ when there is nothing to prove. For $n\geq 3$ pick any $V^{\prime}\subset C_{V}$ such that $\dim(V^{\prime}\cap V)=1$ and let $H$ be a vector hyperplane that contains $V$ but does not contain $V^{\prime}$. For every $i=0,\dots,d$ let $V_{i}\subset H_{i}$ be a general perturbation of $V\subset H$ such that $V_{i}\subset C_{V}$, the intersections $V^{\prime}\cap V_{i}$ are distinct lines and $V^{\prime}\cap H_{i}=V^{\prime}\cap V_{i}$.

By induction, there are degree $d$ homogeneous polynomials $g_{i}$ defined on $H_{i}$ such that $g_{i}$ agrees with $\sigma$ on $H_{i}\cap C_{V}$. By Claim 4 there is thus a unique degree $d$ homogeneous polynomial $g$ on ${\mathbb{R}}^{n}$ such that $g|_{H_{i}}=g_{i}$ for every $i$.

Thus $g|_{V^{\prime}}$ and $\sigma|_{V^{\prime}}$ are 2 homogeneous polynomials of degree $d$ that agree on the $d+1$ lines $V^{\prime}\cap H_{i}$. So $g|_{V^{\prime}}=\sigma|_{V^{\prime}}$. ∎

Proof. For any vector line $L\subset C_{V}$ the function $f|_{L}$ is analytic at the origin. Thus, for $0\neq p\in L$ we can define

[TABLE]

By assumption $f|_{V^{\prime}}$ is analytic at the origin for every vector 2-plane $V^{\prime}\subset C_{V}$, hence $\tau_{r}|_{V^{\prime}}$ is a degree $r$ homogeneous polynomial. Furthermore, $f|_{L}=\textstyle{\sum}_{r}\tfrac{1}{r!}\tau_{r}$ near the origin. By Lemma 5 there is a degree $r$ homogeneous polynomial $T_{r}$ on ${\mathbb{R}}^{n}$ that agrees with $\tau_{r}$ on $C_{V}\subset C$.

Consider the series of homogeneous polynomials

[TABLE]

As we noted, this sum converges in some neighborhood of the origin $0\in L(\epsilon_{L})\subset L$ for every vector line $L\subset C_{V}$. Therefore the series (6.1) defines an analytic function $T$ in a neighborhood $B(\eta)\ni 0$ by [BS18, Lem.3]. Clearly $T$ has the required properties. ∎

Remark. It should be noted that the convergence of a series of homogeneous polynomials is quite different from the usual convergence of power series. In particular, if the series (6.1) absolutely converges at a point $(p_{1},\dots,p_{n})$, it does not imply that it also converges at all points $(x_{1},\dots,x_{n})$ with $|x_{i}|<|p_{i}|$. Thus [BS18, Lem.3] is quite a subtle tool.

Now we are ready to prove the following stronger form of [BS18, Thm.1].

Proof. The question is local, so we may assume that $0\in Q\cap U$ and work near it. Lemma 6 gives a function $T$ that is analytic in a ball $B(\eta)$, and a cone-neighborhood $C$ of $Q$, such that the restrictions $f|_{L}$ and $T|_{L}$ agree in a neigborhood of the origin for every vector line $L\subset C$. We may assume that $B(\eta)\subset U$, thus $f$ and $T$ are both defined and analytic on $L(\eta)$. Hence $f$ and $T$ agree on $C(\eta)=C\cap B(\eta)$.

It remains to show that $T=f$ in a neighborhood of [math]. Fix an affine line $\ell\subset Q$ not passing through the origin such that $\ell\cap C(\eta)$ is not empty. If $p\in{\mathbb{R}}^{n}$ is close to [math] then the affine 2-plane $Q_{p}$ obtained as the span of $p$ and $\ell$ is close to $Q$. Thus $T$ and $f$ restrict to analytic functions on the convex open set $Q_{p}\cap B(\eta)$ that agree on the nonempty open subset $Q_{p}\cap C(\eta)$. Hence they agree everywhere in a neigborhood of $Q\cap U$. ∎

[TABLE]

maps the punctured unit ball \bigl{(}0<\textstyle{\sum}x_{i}^{2}<1\bigr{)} to the outside of the unit ball \bigl{(}\textstyle{\sum}x_{i}^{2}>1\bigr{)} and gives a one-to-one correspondence between the 2-spheres $S^{2}\subset B(1)$ that pass through the origin and 2-planes contained in the outside of the unit ball.

For every p\in\bigl{(}\textstyle{\sum}x_{i}^{2}>1\bigr{)} there is an affine 2-plane p\in Q_{p}\subset\bigl{(}\textstyle{\sum}x_{i}^{2}>1\bigr{)} and every affine 2-plane near $Q_{p}$ is also contained in \bigl{(}\textstyle{\sum}x_{i}^{2}>1\bigr{)}. Thus $f\circ\mu^{-1}$ is analytic by Theorem 7, so $f$ is analytic, except possibly at the origin.

To see what happens at the origin, we use inversion $\mu_{\mathbf{p}}({\mathbf{x}}):=\mu({\mathbf{x}}-{\mathbf{p}})+{\mathbf{p}}$ centered at another point $0\neq{\mathbf{p}}\in B(1)$. This maps $B(1)\setminus\{{\mathbf{p}}\}$ to an open set $U_{\mathbf{p}}\subset{\mathbb{R}}^{n}$. The 2-planes that pass through $\mu_{\mathbf{p}}(0)$ and are contained in $U_{\mathbf{p}}$ are in one-to-one correspondence with the 2-spheres $S^{2}\subset B(1)$ passing through both the origin and ${\mathbf{p}}$. Thus $f\circ\mu^{-1}_{\mathbf{p}}$ is analytic on these planes. We already know that $f\circ\mu^{-1}_{\mathbf{p}}$ is analytic away from $\mu_{\mathbf{p}}(0)$, thus it is also analytic at $\mu_{\mathbf{p}}(0)$ by Theorem 7. So $f$ is also analytic at the origin. ∎

The following is another variant of Theorem 2.

[TABLE]

where the $\ell_{i}:=\sum_{j}a_{ij}x_{j}$ are linear.

Then $f$ is analytic in a punctured neighborhood of $p\in M$.

Proof. We argue as in Paragraph 8, but use the map \bigl{(}\tfrac{x_{1}}{h},\dots,\tfrac{x_{n}}{h}\bigr{)}, which sends the submanifolds $S\subset M$ to affine 2-planes in ${\mathbb{R}}^{n}$.∎

[TABLE]

is analytic on every translate of the coordinate hyperplanes, but not even bounded at the origin. Thus we definitely need more planes than in the complex case.

Let $g\colon{\mathbb{R}}^{3}\to{\mathbb{R}}$ be the function defined by

[TABLE]

Then $f$ is analytic on every nonsingular analytic curve in ${\mathbb{R}}^{3}$, but $f$ is not even continuous at $(0,0,0)$. See also [BMP91] for an even stronger example.