---
tag: XT034AHK9E
kind: statement
role: lemma
name: Hadamard's lemma
msc: ["26E10", 58A05]
uses: [0V33BNKZ9S, 09EGC6F8PT, 7SWSRGDT4A]
about: [0V33BNKZ9S]
sources:
  - url: https://en.wikipedia.org/wiki/Hadamard%27s_lemma
same_as:
  wikidata: Q2074239
  wikipedia: Hadamard's lemma
---

## Statement

Let $U \subseteq \R^n$ be an open set that is star-shaped with respect to a point $a \in U$.
This means that the segment from $a$ to $x$ lies in $U$ for every $x \in U$.
Let $f \colon U \to \R$ be a [[smooth function|0V33BNKZ9S]].
Then there are smooth functions $g_1, \dots, g_n \colon U \to \R$ such that
$$f(x) = f(a) + \sum_{i=1}^{n} (x_i - a_i)\, g_i(x) \quad \text{for all } x \in U,$$
and $g_i(a) = \dfrac{\partial f}{\partial x_i}(a)$ for $i = 1, \dots, n$, where $\partial f / \partial x_i$ is the [[partial derivative|09EGC6F8PT]].

## Remarks

The proof is one line of calculus.
For $x \in U$, the fundamental theorem of calculus and the chain rule give
$$f(x) - f(a) = \int_0^1 \frac{d}{dt} f(a + t(x - a))\, dt = \sum_{i=1}^n (x_i - a_i) \int_0^1 \frac{\partial f}{\partial x_i}(a + t(x - a))\, dt.$$
Set $g_i(x) = \int_0^1 \frac{\partial f}{\partial x_i}(a + t(x - a))\, dt$.
Differentiation under the integral sign shows that $g_i$ is smooth, and $g_i(a) = \frac{\partial f}{\partial x_i}(a)$.

Example: for $f(x) = x^3$ on $\R$ and $a = 1$, one has $x^3 = 1 + (x - 1)(x^2 + x + 1)$, so $g_1(x) = x^2 + x + 1$ and $g_1(1) = 3 = f'(1)$.

If $f$ is only of class $C^k$ with $k \ge 1$, the same formula gives functions $g_i$ of class $C^{k-1}$ only.
For example, $f(x) = x \abs{x}$ is $C^1$ on $\R$, and with $a = 0$ the formula gives $g_1(x) = \abs{x}$, which is not differentiable at $0$.

## Explanations

### L1 human

A smooth function that vanishes at a point can be divided by the coordinates of that point, and the quotients are still smooth.
In one variable it says: if $f(a) = 0$, then $f(x) = (x - a) g(x)$ with $g$ smooth.
This is the smooth analogue of the factor theorem for polynomials.

### L2 human

In terms of [[germs|7SWSRGDT4A]], the lemma says that the maximal ideal $\mathfrak{m}_a$ of the ring of smooth germs at $a$ is generated by the germs of $x_1 - a_1, \dots, x_n - a_n$.
Applying the lemma twice gives $f(x) = f(a) + \sum_i \frac{\partial f}{\partial x_i}(a)(x_i - a_i) + \sum_{i,j} (x_i - a_i)(x_j - a_j) h_{ij}(x)$ with $h_{ij}$ smooth.
This is the key step in the proof that derivations of the ring of germs are exactly the tangent vectors, and in the proof of the Morse lemma.