Let \(U \subseteq \Rd\) be *connected and open* in \(\Rd\), and let \(f \colon U \to \reals\) be real analytic (Krantz and Parks 2002).

Most of us learn in school that if \(f\) vanishes on an open set, then \(f\) is zero on its whole domain: this is called the Identity Theorem (see also MSE).

When \(f\) is univariate, that statement is often accompanied by stronger versions where “open set” is replaced by “a set which has an accumulation point”.

For the case where \(f\) is multivariate, it was asked on Math Stack Exchange whether the identity theorem still works if, instead of knowing that \(f\) is zero on an open set, we have the weaker assumption that \(f\) is zero on more than a set of measure zero.

In one of the comments there, a user pointed to a neat two-page note posted to arXiv and then published by Mityagin (2020).

This post just aims to present Mityagin’s neat proof, written a bit differently.

Notation:

- Natural numbers: \(\mathbb{N} = \{0, 1, 2, \ldots\}\).
- Multi-index partial derivative: given \(t \in \mathbb{N}^d\), let \(|t| = t_1 + \cdots + t_d\), and \(\partial_t f = \frac{\partial^{|t|} f}{\partial x_1^{t_1} \cdots \partial x_d^{t_d}}\).

**Lemma 1** The set \(W = \{ x \in U : \partial_t f(x) = 0 \textrm{ for all } t \textrm{ with } |t| \geq 1 \}\) is either empty or all of \(U\).

*Proof*. Since \(U\) is connected, the only subsets of \(U\) that are both open and closed are the empty set and \(U\) itself. The set \(W\) is closed in \(U\) since it is defined by a countable number of continuous equality conditions. It remains to show that \(W\) is open. To this end, note that for each \(x \in U\) the function \(f\) is equal to its Taylor series in a neighborhood of \(x\). But if \(x\) is in \(W\) then that Taylor series is equal to \(f(x)\), and hence \(f\) is constant on a neighborhood of \(x\). As a result, all derivatives of \(f\) vanish on that neighborhood, so that \(W\) contains this neighborhood of \(x\). Thus, the set \(W\) is also open. (That argument tracks the standard proof of the identity theorem.)

From Lemma 1, we can get the usual corollary, namely, assume \(f\) vanishes on an open set, deduce that all of its derivatives vanish on that set too, and therefore they vanish everywhere by the lemma; conclude that \(f\) is constant (equal to zero). The argument of Mityagin (2020) takes us to a stronger corollary, via another lemma.

**Lemma 2** Given a \(C^1\) function \(g \colon U \to \reals\), let \(\nabla g(x)\) denote its gradient at \(x\). The set \[M = \{ x \in U : g(x) = 0 \textrm{ and } \nabla g(x) \neq 0 \}\] is an embedded submanifold of co-dimension 1 in \(\Rd\). In particular, it has measure zero.

*Proof*. Under the stated conditions, the differential of \(g\) at each \(x \in M\) has maximal rank, namely, one. Thus, the Constant Rank Theorem implies the claim about \(M\): see (Lee 2012, Thm. 5.12) or any other differential geometry book. (We call \(g\) a local defining function for \(M\).)

**Theorem 1 (Mityagin (2020))** If \(f\) is not identically zero, then \[Z = \{ x \in U : f(x) = 0 \}\] is contained in a countable union of manifolds of co-dimension 1. In particular, \(Z\) has measure zero.

*Proof*. Fix an arbitrary \(x \in Z\). Since \(f\) if not identically zero, we know from Lemma 1 that we can choose a multi-index \(t\) such that \(\partial_t f(x) \neq 0\) and \(|t| \geq 1\). Among all such \(t\), choose one such that \(|t|\) is minimal. Pick a coordinate \(i \in \{1, \ldots, d\}\) for which \(t_i \geq 1\). Let \(s\) be the same as \(t\), except \(s_i = t_i - 1\). There are two possibilities:

- Either \(|s| = 0\). In this case, we know \(\partial_s f(x) = f(x) = 0\).
- Or \(|s| \geq 1\). In this case, \(\partial_s f(x) = 0\) because \(|s| < |t|\) (this is why we chose \(|t|\) to be minimal).

In both cases, \(\nabla(\partial_s f)(x) \neq 0\) because the \(i\)th coordinate of that gradient is \(\partial_t f(x)\). Thus, the point \(x\) belongs to the set \[ W_s = \{ x \in U : (\partial_s f)(x) = 0 \textrm{ and } \nabla(\partial_s f)(x) \neq 0 \}. \] Each such set is an embedded submanifold of co-dimension 1 owing to Lemma 2, and each \(x \in Z\) belongs to at least one of them. It follows that \[ Z \subseteq \bigcup_{s \in \mathbb{N}^d} W_s, \] where the right-hand side is structured as announced.

## References

*A Primer of Real Analytic Functions*. 2nd ed. Birkhäuser Advanced Texts: Basler Lehrbücher/Birkhäuser Boston, Inc.

*Introduction to Smooth Manifolds*. 2nd ed. Vol. 218. Graduate Texts in Mathematics. Springer-Verlag New York. https://doi.org/10.1007/978-1-4419-9982-5.

*Mathematical Notes*107 (3–4): 529–30. https://doi.org/10.1134/s0001434620030189.