Hadamard's lemma

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 $0$.

Going deeper

For a student

In terms of germs, 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.