# 数学分析笔记——An Introduction to Manifolds（1）

## Chapter 1 Euclidean Spaces

### §1 Smooth Functions on a Euclidean Space

#### 1.1 $C^{\infty}$ Versus Analytic Functions

Definition. Let $k$ be a nonnegative integer. A real-valued function $f:U\to\mathbb{R}$ is said to be $C^k$ at $p\in U$ if its partial derivatives
$$\frac{\partial^j f}{\partial x^{i_1}\cdots\partial x^{i_j}}$$
of all orders $j\leq k$ exist and are continuous at $p$. The function $f:U\to\mathbb{R}$ is $C^\infty$ at $p$ if it is $C^k$ for all $k\geq 0$; in other words, its partial derivatives $\partial^j f/\partial x^{i_1}\cdots\partial x^{i_j}$ of all orders exist and are continuous at $p$. A vector-valued function $f:U\to\mathbb{R}^m$ is said to be $C^k$ at $p$ if all of its component functions $f^1,\cdots,f^m$ are $C^k$ at $p$. We say that $f:U\to\mathbb{R}^m$ is $C^k$ on $U$ if it is $C^k$ at every point in $U$. A similar definition holds for a $C^\infty$ function on an open set $U$. We treat the terms “$C^\infty$” and “smooth” as synonymous.

#### 1.2 Taylor’s Theorem with Remainder

We say that a subset $S$ of $\mathbb{R}^n$ is star-shaped with respect to a point $p$ in $S$ if for every $x$ in $S$, the line segment from $p$ to $x$ lies in $S$.

Lemma (Taylor’s theorem with remainder). Let $f$ be a $C^{\infty}$ function on an open subset U of $\mathbb{R}^n$ star-shaped with respect to a point $p=(p^1,p^2,\cdots,p^n)$ in $U$. Then there are functions $g_1(x),\cdots,g_n(x)\in C^{\infty}(U)$ such that
$$f(x)=f(p)+\sum\limits_{i=1}^n(x^i-p^i)g_i(x),\quad g_i(p)=\frac{\partial f}{\partial x^i}(p).$$

Proof. Since $U$ is star-shaped with respect to $p$, for any $x$ in $U$ the line segment $p+t(x-p),0\leq t\leq 1$, lies in $U$. So $f(p+t(x-p))$ is defined for $0\leq t\leq 1$.

By the chain rule,
$$\frac{\mathrm{d}}{\mathrm{d}t}f(p+t(x-p))=\sum(x^i-p^i)\frac{\partial f}{\partial x^i}(p+t(x-p)).$$
If we integrate both sides with respect to $t$ from $0$ to $1$, we get
$$\left.f(p+t(x-p))\right|^1_0=\sum(x^i-p^i)\int_0^1\frac{\partial f}{\partial x^i}(p+t(x-p))\mathrm{d}t.$$
Let
$$g_i(x)=\int_0^1\frac{\partial f}{\partial x^i}(p+t(x-p))\mathrm{d}t.$$
Then $g_i(x)$ is $C^\infty$ and we get
$$f(x)-f(p)=\sum(x^i-p^i)g_i(x).$$
Moreover,
$$g_i(p)=\int_0^1\frac{\partial f}{\partial x^i}(p)\mathrm{d}t=\frac{\partial f}{\partial x^i}(p).$$